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# Understanding the non-trivial automorphism which exists on $G$

Assume that $G$ is a finite group such that $|G| \gt 1$ and there exists $x \in G$ such that $x^2 \neq e$. Prove that there exists a non-trivial automorphism on $G$.

Note: This question has been asked and answered here. But, I can't understand that answers. Both of them use the concept of inner-automorphism. We haven't studied this in the class, So i'm not allowed to use the term 'Inner-Automorphism'. Also, I don't understand that what is their automorphism doing. We are given a group. Maybe its abelian. Maybe it's not. We should provide a function which takes an element of $G$ and maps it to another element of $G$. Which element is that another element?

If $G$ is abelian, then $\phi:g \mapsto g^{-1}$ is a non-trivial automorphism because $\phi(x)\ne x$.
If $G$ is not abelian, there are elements $a,b \in G$ such that $ab\ne ba$. Then $\phi:g \mapsto a^{-1}ga$ is a non-trivial automorphism because $\phi(b)\ne b$.
Thus, the existence of $x \in G$ such that $x^2 \neq e$ in only used in the abelian case, but as the original answer notes, it is not really needed.
• @llrf If G is not abelian , in your example just take $x \notin Z(G)$. – Riju May 12 '17 at 14:06

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Question 1: Find the locus of a point equidistant from the point $(2,4)$ and the y-axis.

Let $P(h, k)$ be the point which is equidistant from point $( 2, 4)$ and y-axis.

Distant of point $( h, k)$ from y-axis is $h$.

$\therefore h = \sqrt{ (h-2)^2 + ( k-4)^2}$

$\Rightarrow h^2 = (h-2)^2 + ( k-4)^2$

$\Rightarrow h^2 = h^2 + 4 - 4 h + k^2 + 16 - 8k$

$\Rightarrow k^2 - 4h - 8k + 20 = 0$

Hence the locus of $(h, k)$ is $y^2 - 4x-8y + 20 = 0$

$\\$

Question 2: Find the equation of the locus of a point which moves such that the ratio of its distances from $(2,0)$ and $(1,3)$ is $5 :4$.

Given $A( 2, 0), B( 1, 3)$. Let $P(h,k)$ be the point which moves such that the ratio of its distances from $(2,0)$ and $(1,3)$ is $5 :4$

Given $\frac{PA}{PB}$ $=$ $\frac{5}{4}$

$\frac{\sqrt{ (h-2)^2 + ( k-0)^2}}{\sqrt{ (h-1)^2 + ( k-3)^2}}$ $=$ $\frac{5}{4}$

$\Rightarrow 16 ( h^2 + 4 - 4h + k^2) = 25 ( h^2 + 1 - 2h +k^2 + 9 - 6k)$

$\Rightarrow 16h^2 +64 - 64 h + 16k^2 = 25h^2 - 50h + 25k^2 + 250 - 150k$

$\Rightarrow 9h^2 + 9k^2 + 14h - 150 k + 186 = 0$

Hence the locus of $(h, k)$ is $9x2 + 9y^2 + 14x - 150 y + 186 = 0$

$\\$

Question 3: A point moves as so that the difference of its distances from $(ae,0)$ and $(-ae,0)$ is $2a$, prove that the equation to its locus $b$ $\frac{x^2}{a}$ $-$ $\frac{y^2}{b}$ $= 1$, where $b^2 = a^2 ( e^2 -1)$.

Given $A(ae,0)$ and $B(-ae,0)$ is $2a$. Let $P(h, k)$ be the point.

$\therefore PA- PB = 2a$

$\sqrt{(h-ae)^2+(k-0)^2} - \sqrt{(h+ae)^2+(k-0)^2} = 2a$

$\sqrt{(h-ae)^2+k^2} = 2a + \sqrt{(h+ae)^2+k^2}$

Squaring both sides we get

$(h-ae)^2+k^2 = 4a^2 + (h+ae)^2 + k^2 + 4a \sqrt{(h+ae)^2+k^2}$

$h^2 + a^2e^2- 2aeh + k^2 = 4a^2 + h^2 + a^2e^2 + 2aeh + k^2 + 4a \sqrt{(h+ae)^2+k^2}$

$-4aeh-4a^2 = 4a \sqrt{(h+ae)^2+k^2}$

$-eh-a = \sqrt{(h+ae)^2+k^2}$

Squaring once again we get

$e^2 h^2 + a^2 + 2aeh = h^2 + a^2 e^2 + 2aeh + k^2$

$h^ ( e^2 - 1) = a^2 ( e^2 - 1) + k^2$

$h^2 = a^2 +$ $\frac{k^2}{e^2 - 1}$

$\frac{h^2}{a^2}$ $-$ $\frac{k^2}{a^2(e^2-1)}$ $= 1$

$\frac{h^2}{a^2}$ $-$ $\frac{k^2}{b^2}$ $= 1$

Hence locus of $(h, k)$ is $\frac{h^2}{x^2}$ $-$ $\frac{y^2}{b^2}$ $= 1$ where $b^2 = a^2(e^2-1)$

$\\$

Question 4: Find the locus of a point such that the sum of its distances from $(0,2)$ and $(0, -2)$ is $6$.

Given $A(0,2)$ and $B(0, -2)$. Let $P(h,k)$ be the point such that the sum of its distances from $(0,2)$ and $(0, -2)$ is $6$.

$\therefore \sqrt{ (h-0)^2+(k-2)^2} + \sqrt{ (h-0)^2+(k+2)^2} = 6$

$\Rightarrow \sqrt{ h^2+(k-2)^2} = 6 - \sqrt{ h^2+(k+2)^2}$

Squaring both sides

$h^2 + k^2 + 4 - 4k = 36 + h^2 + k^2 + 4 + 4k - 12\sqrt{ h^2+(k+2)^2}$

$\Rightarrow -36 -8k = 12 \sqrt{ h^2+(k+2)^2}$

$\Rightarrow -9 -2k = 3 \sqrt{ h^2+(k+2)^2}$

Squaring once again we get

$\Rightarrow 81+4k^2 + 36k = 9 ( h^2 + k^2 + 4 + 4k)$

$\Rightarrow 81+4k^2 + 36k = 9h^2 + 9k^2 + 36 + 36k$

$\Rightarrow 9h^2 + 5k^2 - 45 = 0$

Hence the locus of $(h,k)$ is $9x^2 + 5y^2 - 45 = 0$

$\\$

Question 5: Find the locus of a point which is equidistant from $(1, 3)$ and x-axis.

Given $A(1,3)$ and x-axis. Let $P(h,k)$ be the point.

$AP = k \Rightarrow AP^2 = k^2$

$\Rightarrow (h-1)^2 + (k-3)^2 = k^2$

$\Rightarrow h^2 + 1 - 2h + k^2 + 9 -6k = k^2$

$\Rightarrow h^2 -2h - 6k +10 = 0$

Hence the locus of $(h,k)$ is $x^2 -2x - 6y +10 = 0$

$\\$

Question 6: Find the locus of a point which moves such that its distance from the origin is three times its distance from x-axis.

Let $P(h,k)$ be the point. The distance of $P(h, k)$ from x-axis is $k$

$\therefore OP = k \Rightarrow OP^2 = k^2$

$\Rightarrow (h-0)^2 + ( k-0)^2 = ( 3k)^2$

$\Rightarrow h^2 + k^2 = 9k^2$

$\Rightarrow h^2 = 8k^2$

Hence the locus of $(h, k)$ is $x^2 = 8y^2$

$\\$

Question 7: $A (5, 3), B (3, - 2)$ are two fixed points; find the equation to the locus of a point $P$ which moves so that the area of the $\triangle PAB$ is $9$ units.

Given $A (5, 3), B (3, - 2)$. Let $P(h,k)$ be the point.

Area of $\triangle ABP = \frac{1}{2} \{ x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3( y_1 - y_2) \}$

$\Rightarrow 9 = \frac{1}{2} | 5( -2-k) + 3(k-3) + h ( 3+2) |$

$\Rightarrow | 5h - 2k - 19 | = 18$

Therefore $5h - 2k - 19 = 18$      or      $5h - 2k - 19 = -18$

$\Rightarrow 5h -2k - 37 = 0$      or      $5h-2k-1=0$

Hence the locus of $(h, k) is 5x-2y - 37 = 0$      or      $5x - 2y - 1 = 0$

$\\$

Question 8: Find the locus of a point such that the line segments having end points $(2, 0)$ and $(-2, 0)$ subtend a right angle at that point.

Given $(2, 0)$ and $(-2, 0)$. Let $P(h,k)$ be the point such that $\angle APB = 90^{\circ}$

$\therefore AB^2 = AP^2 + BP^2$

$\Rightarrow ( 2+2)^2 = ( h - 2)^2 + k^2 + ( h+2)^2 + k^2$

$\Rightarrow 16 = h^2 + 4 - 4h + k^2 + h^2 + 4 + 4h + k^2$

$\Rightarrow 8 = 2h^2 + 2k^2$

$\Rightarrow h^2 + k^2 = 4$

Hence the locus of $(h, k)$ is $x^2 + y^2 = 4$

$\\$

Question 9: If $A (-1, 1)$ and $B (2, 3)$ are two fixed points, find the locus of a point $P$ so that the area of $\triangle PAB=8$ sq. units.

Given $A (-1, 1)$ and $B (2, 3)$. Let $P(h,k)$ be the point.

Therefore are of $\triangle PAB = \frac{1}{2} \{ x_1(y_2-y_3) + x_2 ( y_3-y_1) + x_3( y_1 - y_2) \}$

$\Rightarrow 8 \times 2 = | -1( 3-k) + 2 ( k-1) + h ( 1 - 3) |$

$\Rightarrow 16 = | 2h - 3k + 5|$

$\therefore 2h - 3k + 5 = 16$      or      $2h - 3k + 5 = -16$

$\Rightarrow 2h - 3k -11 = 0$      or      $2h - 3k + 17 = 0$

Hence the locus of $( h, k)$ is $2x - 3y - 11 = 0$      or    $2x-3y+17 = 0$

$\\$

Question 10: A rod of length $l$ slides between the two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio $7 :2$.

Let $x$ and $y$ axis be the two perpendicular lines.

Let $A(a,0)$ and $B ( 0, b)$ be the two intercepts on the axes.

Let $P(h, k)$ be the point that divides $AB$ in $1:2$ ratio.

$\therefore$ $\frac{BP}{AP}$ $=$ $\frac{1}{2}$

$\therefore ( h, k) = \Big($ $\frac{a+0}{3}$ $,$ $\frac{0+2b}{3}$ $\Big ) = \Big($ $\frac{a}{3}$ $,$ $\frac{2b}{3}$ $\Big)$

$\Rightarrow a = 3h$ and $h =$ $\frac{3k}{2}$

Now $AB = l$

$\therefore \sqrt{a^2 + b^2} = l$

$\Rightarrow a^2 + b^2 = l ^2$

$\Rightarrow (3h)^2 + \Big($ $\frac{3k}{2}$ $\Big )^2 = l^2$

$h^2 +$ $\frac{k^2}{4}$ $=$ $\frac{l^2}{9}$

Hence locus of $(h, k)$ is $x^2 +$ $\frac{y^2}{4}$ $=$ $\frac{l^2}{9}$

$\\$

Question 11: Find the locus bf the mid-point of the portion of the line $x \cos \alpha + y sin \alpha = p$ which is intercepted between the axes.

Given line is $x \cos \alpha + y \sin \alpha = p$

When $x = 0 , y =$ $\frac{p}{\sin \alpha}$ $= p \mathrm{cosec} \alpha$

When $y = 0, x =$ $\frac{p}{\cos \alpha}$ $= p \sec \alpha$

Let $P(h, k)$ be th emid point of $AB$

$\therefore h =$ $\frac{p \sec \alpha+0}{2}$           $h =$ $\frac{0 + p \mathrm{cosec} \alpha}{2}$

We know $\cos^2 \alpha + \sin^2 \alpha = 1$

$\Rightarrow$ $\frac{p^2}{4h^2}$ $+$ $\frac{p^2}{4k^2}$ $= 1$

$\Rightarrow$ $\frac{1}{h^2}$ $+$ $\frac{1}{k^2}$ $=\frac{4}{p^2}$

Hence the locus of $(h, k)$ is $\frac{1}{x^2}$ $+$ $\frac{1}{y^2}$ $=$ $\frac{4}{p^2}$

$\\$

Question 12: If $O$ is the origin and $Q$ is a variable point on $y^2 =x$. Find the locus of the mid-point of $OQ$.

Let $Q$ be $( a, b)$. It lies on $y^2 = x$

$\Rightarrow b^2 = a$     … … … … … i)

Let $P(h,k)$ be the mid point of $OQ$

$\therefore h =$ $\frac{0+a}{2}$     and     $k =$ $\frac{0+b}{2}$

$\Rightarrow a = 2h$    and      $b = 2k$

Substituting in i) we get $( 2k)^2 = 2h \Rightarrow 2k^2 = h$

Hence the locus of $( h, k)$ is $2y^2 = x$


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## Tuesday, August 2, 2011

### Last Non Zero Digit Of a factorial

This is an oft repeated concept in different examinations but if you can grasp the following algorithm solving problems is a one min task

Concept

Lets say D(N) denotes the last non zero digit of factorial, then the algo says
D(N)=4*D[N/5]*D(Unit digit of N)[If tens digit of N is odd]
D(N)=6*D[N/5]*D(Unit digit of N)[If tens digit of N is even]; Where [N/5] is greatest Integer Function

#Problem 18
Find the last non zero digit of 26!*33!.

Solution Scheme and Approach
D(26)=6*D[26/5]*D(6)=6*D(5)*D(6)=6*2*2=4[D(5) means last non zero digit of 5!=120 which is 2, same for D(6)]
D(33)=4*D[33/5]*D(3)=4*D(6)*D(3)=4*2*6=8

Hence last non aero digit of 26!*33!=4*8=2
Remember
D(1)=1 D(2)=2 D(3)=6 D(4)=4 D(5)=2 D(6)=2 D(7)=4 D(8)=2 D(9)=8

Now solve:
(1)Last Non Zero Digit of 99!*26!.
(2)How many zeros are at the end of 13!+14!+15!+16!+17!+18!

## Wednesday, July 27, 2011

### After One Year

#Problem 17

If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)
(a)11/5 (b)17/3 (c)2 (d)11/9

Solution Scheme And Approach:

a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)>a/(a+b+d+c)+ b/(b+c+a+d) + c/(b+c+d+a) + d/(a+c+d+b)=1
a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)<a/(a+b)+ b/(a+b)+ c/(c+d)+ d/(c+d) =2
Hence the range of the expression is (1,2)

Only option (d) suffice the condition.

## Thursday, April 15, 2010

### Digital root

#Problem 16
Prove any perfect number ,except 6, has digital root 1.
Solution scheme and approach:
We already know that what is digital root(look under concept of number theory) and we also learned that dividing any number by 9 we get digital root.
Let, the number is
dkdk-1dk-2…….d2d1
Now,
dkdk-1dk-2…….d2 d1                                                                                = dk *10^(k-1)+ dk-1*10^(k-2)+dk-2*10^(k-3)…….d2*10 + d1(mod 9)                                            =dk +dk-1+dk-2…….+d2+d1(mod 9) [As 10^k=1(mod9)]                                                                                All perfect numbers can be written as 2p-1(2p-1) [Where p is a prime number]                  for p=3, 2p-1(2p-1)=28 which is a perfect square and 28=1(mod9)..(i)                                                                                for p>3, we can write p=6k+1 or p=6k+5                                                                      Case 1: p=6k+1
As 26k=1(mod 9) and 26k+1-1=1(mod 9) we can write,
26k(26k+1-1) =1(mod 9)….(ii)
Case 2: p=6k+5
As 26k+4= 24(mod 9)=7(mod 9) and 26k+5-1=25-1(mod 9)= 4 we can write,
26k+4(26k+5-1) =7*4(mod 9)=1(mod 9)……….(iii)
From (i),(ii) and (iii) we can conclude any perfect number ,without 6, will be having digital root 1.(Q.E.D.)

## Wednesday, April 14, 2010

### Breaking a long overdue

#Problem 15
Find the total number of solutions of | [x] - 2x |=4.
(a)1,(b)2,(c)4,(d)6,(e)None of these [Where [x] is Greatest Int Func]
Solution scheme and approach:
Case 1 :( x is a whole number)
so [x]=x
=>|-x|=4 =>x=4 or x=-4
Case 2: ( x is a fraction)
so x=a+f (f is the fractional portion)
then the equation gets converted into
|a-(2a+2f)|=4
=>|-a-2f|=4
=>a=4-2f =>a=3 (when f=1/2) =>x=3.5
or a=-4-2f =>a=-5(when f=1/2)=>x=-4.5
Total four solutions.

Hence option (c)

## Saturday, February 27, 2010

### Polynomial

#Problem 14
Let f be a polynomial of degree 98, such that f(k)=1/k. [k=1,2,3,4...99]. Find the value of f(100).[USAMath Talent Search 2009]
Solution scheme and approach
Consider a function g(k) such that
g(k)=kf(k)-1
As f(k) is a polynomial of deg 98 g(k) should be a polynomial of deg 99.
=>g(k)=C*(k-1)*(k-2)*(k-3)*...*(k-99)
=>kf(k)-1=C*(k-1)*(k-2)*(k-3)*...*(k-99)............(1)
by putting k=0
=>C=1/99!
=>kf(k)-1=[(k-1)*(k-2)*(k-3)*...*(k-99)]/99!
If k=100
=>100f(100)-1=1
=>f(100)=1/50

## Thursday, February 11, 2010

### Number Of Solutions

This is a typical pattern. If you can remember different formats of this specific problem, cracking questions is just a matter of few seconds. Let's have a look.

#Problem 13.1
Let n ≥ 1 be an integer and let t denote the number of positive integer divisors of n^2. Show
that the number of positive integer solutions (a, b) of the equation 1/a−1/b = 1/n is precisely
equal to (t − 1)/2.
Solution scheme and approach:
1/a-1/b=1/n..............(1)
From the above equation it's clear that 1/a>1/n =>n>a
Now we can consider an integer x such that
a=n-x
From equation number (1) we can also write
1/b=1/(n-x)-1/n=x/(n-x)*n
=>b=(n^2-nx)/x=n^2/x-n
To count the total number of solutions we have to count total numbers of possible values of x.
As b is an integer x has to be an integer which must be a factor of n^2 such that x
Now think, n^2 is a perfect square which has a total of t divisors. out of these factors one is n another (t-1)/2 is less than n each of them is paired with other (t-1)/2 number of factors which are greater than n.
[Suppose for 16=4x4=2x8=1x16]
Hence, Total number of solutions are (t-1)/2. Q.E.D.
[N.B: Total number of ordered pair(a,b) would be (t-1)]

Food for thought:
Let n ≥ 1 be an integer and let t denote the number of positive integer divisors of n^2. Show that the number of positive integer solutions (a, b) of the equation 1/a+1/b = 1/n is precisely equal to t /2.

#Problem 13.2
Find the smallest positive integer n with the property that the equation 1/a − 1/b = 1/n has exactly 2010 different solutions in positive integers a and b.
(a)6^2010,(b)2^2010,(c)1,(d)10^2010,(e)Cannot be determined
Solution scheme and approach
From the above equation it's clear that
n^2 should have t number of divisors where (t-1)/2=2010=>t=4021
Suppose
n=p1^a*p2^b*p3^c*...
=>n^2=p1^2a*p2^2b*p3^2c*...pn^2n
It has (2a+1)*(2b+1)*(2c+1)*....*(2n+1)=4021 number of divisors
Now the catch is 4021 is a prime number.=>n^2=p1^2a
=>(2a+1)=4021
=>a=2010
As we need to find out smallest possible value of p1 that means p1=2
So the required number is 2^2010
Hence, option (b) is the right option.

#Problem 13.3
If Ramu and Krishna work on alternate days to complete a work, then the work gets completed in exactly 24 days. If R and K denote the number of days required by Ramu and Krishna respectively to complete the work independently, then how many ordered pairs of integral values of R and K are possible?
(a)14, (b)8, (c)15, (d)7,(e)16                                                [Career Launcher Mock CAT]
Solution scheme and approach
Ramu and Krishna work 1/R and 1/K th work per day respectively
According to the question as they are working on alternative day.We may say if they work together work must be completed on 12th day
=>12(1/R+1/K)=1
=>(1/R+1.K)=1/12
As, 12^2=2^4*3^2 has [(4+1)*(2+1)]/2=15/2 number of solutions.So total number of ordered pair(R,K) solutions are 15. (refer Food for thought)
Hence, Option (c).

Hope, this session has helped you a little bit to clear these things.
All the best!!

## Tuesday, February 9, 2010

### Powered Modulo

When we need to calculate the modulus (or remainder) of a nth powered number, we have to find out out the pattern.

#Problem 12.1
if a^2+b^2 is a multiple of 7^2009 then prove that ab is a multiple of 7^2010.[USA Math Talent Search]
Solution scheme and approach
a^2+b^2=7^2009*k
As right hand side is divisible by 7 left hand should be divisible by 7.
Now we can deduce the the following table
Number-----Square-------remainder (modulo 7)
0-------------0-------------0
1-------------1-------------1
2-------------4-------------4
3-------------9-------------2
4------------16-------------2
5------------25-------------4
6------------36-------------1
From the table it's clear that as sum of two numbers (>=1) gives 7 so both of them should be divisible by 7.
Let a=7a1,b=7b1.
=>a^2+b^2=7^2009*k
a1^2+b1^2=7^2007*k
Likewise, let a1=7a2 and b1=7b2
a2^2+b2^2=7^2005*k
=>a=7^2*a1 and b=7^2*b1
and so on ...................
at last we get
a1004^2 + b1004^2=7k
=>a=7^1005*a1 and b=7^1005*b1
=>ab=7^2010*(a1b1)
=>ab is a multiple of 7^2010. Q.E.D.

#Problem 12.2
A set consists of 143 natural numbers each of them is a perfect cube. If they are divided by 13, how many maximum numbers within the set are possible, such that one will get the same reminder in each case.
(a)11, (b)31, (c)33, (d)36, (e)29 [AIMCAT]

Solution scheme and approach
When any cube(0^3,1^3,2^3,...,12^3) divides by 13 it generates 5 different numbers 0,1,5,8,12. Now at least [143/5]+1=29 numbers are possible which will generate same remainder.
Hence, option (e) is the answer

## Monday, February 1, 2010

### CAT CLASSIC : Number Theory

Today I talked to one of my Pagalguy friend. He was right in saying that my blog is not good for all. I agree. What I try to discuss, here, is obviously not for all and sometimes the concept becomes too tough to understand. And there are a lot of people who are unable to attend coaching classes due to different reasons. So being a pagalguy member I should help them. And if this section can help you a little, I will consider myself blessed. There will be two sections CAT CLASSIC and CAT SHOT, which will only concentrate on cracking CAT. Though this is, once I have written in my another blog The Unfinished Evening
Here it goes.

Introduction
In ancient Babylon, a sexagesimal(base 60) was in use.In India a decimal system was in place during Harappan period.
It is generally acknowledged that concept of zero,crucial to development of science,is India’s contribution to the world, which was given to Europe through the Arabs.The ancient India astronomer Brahmagupta is credited with having put forth the concept of zero first time.
Another famous Indian mathematician developed the concept of pi as 3.1416(circa 500 AD) claiming,for the first time it was an approximation.Pi represents the number reflecting how many times a circle’s diameter may be divided into its circumference

In the following problem we may get confused:
1+a+a^2+….n terms = (a^n-1)/(a-1) ; if a=1 then 0/0=n, so where lies the pitfall?
No man! It’s correct. 0/0 takes any given value for different n. So it is undefined.We can say something either exists, or it doesn’t. If it exists, then it has a quality that we call number associated with it, and if it doesn’t exist we call this absence, zero.

145 = 1! + 4! + 5! only number which can be expressed as the factorial summation of it’s individual digit.
————————————————————————————————————————————————————Number theory is playing a pivotal role so far as management entrance examinations like CAT/XAT/JMET/IIFT/MAT are concerned. So it’s imperative that good grasp on this topic is essential to ace quantitative section. The whole chapter is divided into three topics.
Topic 1: Numbers and Divisibility

Divisibility Fundamentals
(i)Any composite number N=a^m * b^n * c^p [a,b,c all are prime numbers] has number of factors equals to (m+1)*(n+1)*(p+1)
(ii)sum of it’s divisors is given by (a^m+1 – 1)/(a-1)*(b^n+1 – 1)/(b-1)*(c^p+1 – 1)/(c-1)
(iii)Product of it’s divisors equals to N^[(m+1)*(n+1)*(p+1)/2]
(iv)The number of ways in which the number can be resolved into two factors are (m+1)*(n+1)*(p+1)/2 and if the number is a square number then number of ways would be [(m+1)*(n+1)*(p+1)+1]/2

Condition for two divisors of any number n to be co-prime to each other
(i)Let N=a^m*b^n has (m+1)(n+1)-1+mn numbers of factors which are co-prime to each other. If N=12, then as 12=2^2*3,so it has (2+1)*(1+1)-1+2=7 sets which are co-prime to each other.They are: (1,2),(1,3),(1,4),(1,6),(1,12),(2,3),(3,4).
(ii)Similarly we can deduce the formula for higher orders.If N=a^m*b^n*c^p then it will have (m+1)(n+1)(p+1)-1+mn+np+mp+3mnp co prime factors.Actually it is a deduction of previous formula.As N=a^m*b^n has (m+1)(n+1) -1+mn number of factors So N=d^[(m+1)(n+1)-1+mn]*c^p has[(m+1)(n+1)+mn-1+1](p+1) -1+[(m+1)(n+1)-1+mn]p number of factors which is equal to (m+1)(n+1)(p+1)-1+mn+np+mp+3mnp.

The number of ways to express a number by means of product of two co-prime numbers
2^(n-1) ways; where ‘n’ represents number of prime factors of that number. Let N=12=2^2 x 3.Therefore it can be expressed as 2^(2-1)=2 ways as the product of two co-prime numbers.They are (1,12) and (3,4).

Euler’s Function and it’s application
Euler function is related to prime numbers counting theorem.By using this we can count number of co-primes less than of a number.It is denoted by φ(N). φ(N) is known as Euler’s function of number N. N=a^m*b^n*c^p then φ(N) = N(1 – 1/a)(1 – 1/b)(1 – 1/c) and so on for any higher terms.[N.B.:a,b,c all are prime numbers] Let, we have to count all numbers which are co-prime to 12 and less than 12.Therefore we have to count φ(12).As 12=2^2*3, So φ(12)=12(1 – 1/2)(1 – 1/3)=12*(1/2)*(2/3)=4 and obviously pairs are (1,12),(5,12),(7,12),(11,12) or in simple terms 1,5,7,11 are those numbers which are less than 12 and co-prime to 12.

Dirichlet’s theorem
If a is positive and hcf(a,b)=1,then there are infinitely prime numbers in the form of an+b.

Elucid’s Theorem
If p is a prime number and it divides ab(product of two integers a&b) then either p divides a or p divides b.This proposition is known by Elucidian’s lemma.

Prime Number
Till now, all attempts made to arrive at a single formula to represent all prime numbers proved to be fruitless.The main reason behind of it is all because there is no symmetricity between the differences among them.One standard notation for prime number is N^2+N+41 where -39=<39.Some properties of prime number are elucidated here.
(i)All prime numbers end in 1,3,7 or 9.(except 2,5).
(ii)Fermat’s little theorem
if p is prime then a^p leaves a remainder a when divided by p or in other words a^p-a=0(mod p)
(iii)Wilson’s theorem
an integer p will be prime only when (p-1)!+1 is divisible by p.
(iv)Bertrand’s Postulate says, if n is a positive integer greater than 1, then there is always a prime number p n
<2n.
(v)All prime numbers greater than 3 can be expressed as 6K+1 or 6K-1. But converse is not true always.eg.:13(6*2+1) is prime but 25(6*4+1) is not prime.
(vi)1/p (p is prime other than 2 &5 ) always a recurring decimal of period (p-1).It is true for other bases q, provided p is not a prime factor of q.

Perfect Number
Perfect number is defined as a positive integer which is the sum of its proper positive divisors, that is, the sum of the positive divisors excluding the number itself.First four perfect numbers can be generated by 2n−1(2n − 1).[Where 2n − 1 is a prime number].This are all even perfect numbers.All even perfect numbers (except 6) digital root 1.

Armstrong Number
Armstrong numbers are the sum of their own digits to the power of the number of digits. example:153 = 1³ + 5³ + 3³[as sum of the digits are 3], 14 + 64 + 34 + 44 = 1634.

Rare Number
The numbers, which gives a perfect square on adding as well as subtracting its reverse are rare and hence termed as Rare Numbers.If R is a positive integer and R1 is the integer obtained from r by writing its decimal digits in reverse order, then R + R1 and R – R1 both are perfect square then R is termed as Rare Number. For example: R=65, R1=56 we can write R+R1 = 65+56 = 121 = 112 & R-R1 = 65 – 56 = 9 = 32.So 65 is a Rare Number.

Properties of co-prime numbers
(i)The numbers a and b are coprime if and only if there exist integers x and y such that ax + by = 1.
(ii)If a and b are coprime and bx = by (mod a), then x = y (mod a).
(iii)Two natural numbers a and b are coprime if and only if the numbers 2a-1 and 2b-1 are coprime.

Fermat’s last theorem
If x, y, z and n are integers, there are no solutions of x^n+y^n=z^n with n>2 and x,y,z>0.

Topic 2:Cyclicity, Factorial and Modulo Arithmetic

Cyclicity :
The last digit of the exponents of all nubers have a cyclicity.That is,every Nth power of the base shall have the same digit.Now we can consider the following table to reveal the pattern.

Consider exponents of 2. 2^1=2|2^2=4|2^3=8|2^4=16|2^5=32|2^6=64.So the same unit digit 2 comes again when the exponent is increased by 4.Same pattern is followed for other exponents also.So cyclicity or period of 2 is 4.
Similarly we can conclude: 0,1 5,6 have cyclicity 1, 2,3,7,8 have cyclicity 4,4,9 have cyclicity 2.
We can take an easy example to illustrate it.
Problem:
Find the unit digit of 7^23 x 4^21
7^23=7^(4*5+3)=7^4*5 x 7^3=(..1) x (343)=…3[As 7 has cyclicity 4, so 7^4n always yields unit digit 1.] Similarly 4^21=4^(2*10+1)=4^(2*10)*4=..1*4=..4[As 4 has cyclicity 2, so 4^2n always produces unit digit 1.] 7^23*4^21=(..3)x(..4)=…2. Required unit digit is 2.

How to find last two digits of a number
(x+a)^n= x^n+ nc1*x^n-1*y+nc2*x^n-2*y^2+….+y^n. By application of binomial expansion we can do any problem under this category. We can do this kind of problem by using modular arithmetic or by pattern method too.Barring all of these there is also another process to solve.I hope that procedure will help us to solve any problem in lesser time.So I think it would be better to discuss them in another page.

Factorial
Factorial is another important part of Number System.Generally question comes from either pattern recognition or from determining highest power diving a factorial number.
Thus approach to finding highest power of p(p is prime) in y! is [y/x]+[y/x^2]+[y/x^3]+…….where [] represents greater integer function or gint function.Which determines just integral part of [].It can be used to determine number of zeros in a factorial number.
P.S.:1!+2!+3!+4!+5!=1+2+6+24+120=153 so 1!+2!+3!+4!+…..+n! always yields unit digit 3 and it does not depend on n.Where n can be infinitely any.As 5! is the first number which gives unit digit 0. 10! is the first number which has last two digits 0 and so on & so forth.

Modulo Arithmetic And Remainder Theorems
Suppose n divides a and gives a remainder b then it can be expressed in the format of modulo arithmatic. a=b(mod n). Modular arithmetic follows addition, subtraction and multiplication rule.
If a1=b1(mod n) and a2=b2( mod n) then; (i)(a1+a2)=(b1+b2)(mod n) (ii)(a1-a2)=(b1-b2)(mod n),(iii)(a1.a2)=(b1.b2)(mod n).

Euler’s Remainder Theorem:
We alrealdy know what is Euler’s function.Now according to this theorem a^ φ(N)=1(mod N) where a and N are co prime to each other.φ(N) is also known as ORDER of a modulo N.
Corollary:
(a) 1/a=a^ [φ(N)-2](mod N) when a & N are co prime to each other.

(b) a and b are relatively prime positive integers,
then there exist integers m and n such that a^m + b^n = 1 (mod ab)

(ii)Fermat’s Little Theorem:
If p is a prime number,then a^(p-1)=1(mod p)

(iii)Chinese Remainder Theorem:
If N=a*b where a & b are co prime,M is such that M=r1(mod a) and M=r2(mod b) then M=a*r2*x+b*r1*y(mod N) where ax+by=1.

More concepts:
(1)gcd(a^m – 1,a^n – 1)=a^gcd(m,n) – 1 for all positive integers m,n & a>1.
(2)ax+by =n has an non negative integer solution if gcd(a,b)=n.
(3) 1/x+1/y=1/n is a equation where x,y,n are positive integers & n>1.Now we have to determine the number of ordered pairs of (x,y). For this at first we have to do prime factorization of n=a^p1*b^p2*c^p3*….Therefore the number of positive solution se would be (2p1+1)(2p2+1)(2p3+1)….. (4)Divisibility by 7, 11, and 13
Let a number be ….kjihgfedcba where a, b, c, d, are respectively units digits, tens digits, hundreds digits, thousands digits and so on. Starting from right to left, we make groups of three digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.
Grouping of the above number in groups of three, from right to left, is done in the following manner 0kj,ihg,fed,cba. Groups are denoted as (4,3,2,1).
We add the alternate groups (1 st , 3 rd , 5 th etc.. and 2 nd , 4 th , 6 th , etc..) to obtain two sets of numbers, N 1 and N 2 .
In the above example, N 1 = cba + ihg and N 2 = fed + kj
Let D be difference of two numbers, N 1 and N 2 i.e. D = N 1 – N 2 .
If D is divisible by 7, then the original number is divisible by 7.
If D is divisible by 11, then the original number is divisible by 11
If D is divisible by 13 then the original number is divisible by 13.

Corollary:
Any six-digit, or twelve-digit, or eighteen-digit, or any such number with number of digits equal to multiple of 6, is divisible by EACH of 7, 11 and 13 if all of its digits are same .
For example 666666, 888888888888 etc. are all divisible by 7, 11, and 13.

(5)Divisibility by 9
Sum up all digits of a number, divide it by 9.The remainder is equal to the remainder when the original number is divided by 9.

Properties of squares
(1)Any square number can be expressed as 3k or 3k+1.
(2)Square of an odd number can be expressed as 4k+1,square of an even number can be expressed as 4k.
(3)Last digit of a square number cannot be 2,3,7,8.
(4)Square of any prime number can be expressed as 6K+1(p>3)
(5)If last digit of a number is odd the previous number would be even.
(6)A square number cannot be a perfect number.
Seed Number or Digital Root
If we sum up all the digits of a number till then single digit is arrived, then that single digit would be the seed of that number
Properties:
(i)To determine a seed we have to divide the number by 9.The remainder obtained will be the seed. If the number becomes perfectly divisible 9 then seed number would be 9.
(ii)A perfect square cannot have any seed number among 2,3,5,6,8.
(iii)6! onwards all number has equal seed number 9.
(iv)The difference between one number and it’s sum of the digits is always divisible by 9.
e.g.:For 68 it’s sum of digit is 15. So difference is 68-14=54.Divisible by 9.
Hope these concepts would help us to open the knot of problems

## Sunday, January 31, 2010

### Concept 3: CATLAN numbers

Can you tell me what is the next number of this series?
1,2,5,14,42,?

You may face trouble in finding though. But here is the formula
Tn=1/(n+1)*2nCn=(2n)!/[(n+1)!*n!] [For n>=1]
So T5=132.
The number of this series is popularly known as CATLAN numbers. And in different counting problems. I will try to explore some of them.

(A)Mountain Ranges and Diagonal Avoiding Paths

(i)How many mountain ranges you can draw using n upstrokes(/) and n downstrokes{\)
Drawing the above phenomenon would reduce our burden.
n=1: /\ =>1 way
n=2: /\/\,  /\ => 2 ways
/   \                                     /\
n=3: /\ /\ /\,  /\          /\     /\/\        /   \  => 5 ways
/   \/\, /\/  \, /      \,    /     \

Going like this it will generate nothing but the series of CATLAN numbers. For n=4 obviously we will get 14 distinct mountain ranges and so on.

(ii)Diagonal Avoiding Paths
In a grid of n×n squares, how many paths are there of length 2n that lead from the upper right corner
to the lower left corner that do not cross the diagonal dotted line from upper right to lower left? In
other words, how many paths stay on or above the main diagonal.

Again answer is noting but 4'th CATLAN numbers of the series. i.e. 14. And look it's not different than the arrangement of mountain ranges. [Follow the arrowhead paths]

(B)Triangulation of a convex polygon

For a convex ploygon having n sides (n>=4) total number of triangulation will be T(n-2) th term of the CATLAN series.
For n=6 obviously we will get 14 triangulations.

(C)Handshaking Lemma
If, 2n people are seated around a circular table, then in Tn ways they can shake their hands simultaneously such that at no instance the arms will cross each other.
Application
If 16 people are sitting in a circular table,in how many ways can all of them be simultaneously
shaking hands with another person at the table in such a way that none of the arms cross each other?Solution:
2n=16 =>n=8
Total ways T8 =16!/9!*8!=1430.

Hope you have enjoyed this concept session.

## Friday, January 15, 2010

### Again a perfect square

#Problem11
49=7^2
4489=67^2
444889=667^2
............................
Now prove that if we insert 48 in the middle of the previous term of this series, like this way, in each case it will produce a perfect square.

Solution scheme and approach
Number is in this format
[4444...4](n+1 numbers)[8888...8](n numbers)9
So we can write the number in the following way
[4*10^(2n+1)+4*10^2n+4*10^(2n-1)+....4*10^(n+1)] + [8*10^n+8*10^(n-1)+8*10^(n-2)+...8*10]+9
=4*10^(n+1)*[10^(n+1)-1]/9+8*10*(10^n-1)/9+9
=1/9[4*10^(n+1)*{10^(n+1)-1}+80*(10^n-1)+81]
=1/9[{2*10^(n+1)}^2-4*10^(n+1)+8*10^(n+1)-80+81]
=1/9[{2*10^(n+1)}^2-4*10^(n+1)+1]
=1/9[2*10^(n+1)-1]^2
=>a perfect square for n>=0 Q.E.D.

### CAT SHOT-1

This section is dedicated to CAT. Here we deal with the problems which often appears in mocks and actual cat as well.Fasten your belt. Here we go.

(1)2^y+2*(3^y)>3*(4^y) and y=3x^2+2x-2
Which of the following is a possible value of x?
(a)-1.5 (b)-2.5 (c)-0.5 (d)0.7 (e)1.2

(2)For N values of t, where t<=250, the highest power of t in t! is t^5. Then which of the following is true? (a)N<=4 (b)N=6 (c)N=8 (d)N=10 (e)N>=12

## Thursday, January 14, 2010

### Concept 2:Diophantine Equations

Diophantine equation is an equation of polynomials where variables are integers only.

Linear Diophantine Equation

$ax+by=1\,$
from the name it's clear that it's an equation of degree 1.
ax+by=g.c.d.(a,b)=d ...(I)
is also an linear Diophantine equation.According to Bézout's identity if equation number (I) holds true.i.e. gcd(a,b)=d satisfies then, there should be an integer solution of this equation.It's true for more than two variables as well.
Lets take an example
14x+35y=7
As, gcd(14,35)=7 hence Bézout's identity holds true.
Now we can verify it.(x,y)::(-2,1) is an integer solution set of this equation.

Fermat's Last Theorem

$x^n+y^n=z^n \,$
If n=2 this equation becomes x^2+y^2=z^2 nothing but Pythagorian triplets of (x,y,z). Infinite numbers of solution exist.
But if n>=3 no solution exists. This is known as Fermat's Last Theorem

Pell's Equation

$x^2-ny^2=\pm 1\,$

this equation is known as Pell's equation, Where n is a non square integer.Trivially x=1, y=0 is one of the solutions of this equation.But there could be other values ,till infinity, as well.

Now by squaring we can write
(x^2-ny^2)^2=1
=>(x^2+ny^2)^2-n(2xy)^2=1
=>(2x^2+1)^2-n(2xy)^2=1 [If x^2-ny^2=-1]...(I)
or, (2x^2-1)^2-n(2xy)^2=1[If x^2-ny^2=1]....(II)
(I) and (II) are also another form of Pell's equation.

Erdős–Straus conjecture

$\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$
For n>=2, 4/n can be expressed as the summation of three fractional numbers.
Wacław Franciszek Sierpiński generalized this equation and stated that,
for any positive k there exists a number N such that, for all nN, there exists a solution in positive integers to k/n =1/x+1/y+1/z

### Infinite primes

#Problem 10
Prove that infinitely many prime numbers are possible.

Solution Scheme and approach

Let if possible there are a1,a2...an number of finite primes.
Consider another number
N=(a1*a2*a3*...*an)+1
Now we can say,
N is such a number which can be divide by all primes but always gives a remainder 1. Hence, contradiction.
Because any number>1 should be divisible by at least one prime number.

=>Infinite primes are possible. Q.E.D.

## Wednesday, January 13, 2010

### Maximum value of a variable

#Problem 9
a+b+c+d+e=8
a^2+b^2+c^2+d^2+e^2=16. [a,b,c,d,e are real numbers]

Find the maximum value of e.

Solution Scheme and approach:
This problem can easily be done with Cauchy-Schwartz  inequality. But will explain this in my next topic. Let's do it without Cauchy.

(a-r)^2+(b-r)^2+(c-r)^2+(d-r)^2+(e-r)^2
=(a^2+b^2+c^2+d^2+e^2)-2r(a+b+c+d+e)+5r^2
=16-16r+5r^2[ By putting the values]

we can write
(e-r)^2<=16-16r+5r^2
=>(e-r)=root[16-16r+5r^2] [equality holds when a=b=c=d=r]
=>e=root[16-16r+5r^2]+r=f(r)
as this function is increasing we have only local minima
f'(r)=0
=>(10r-16)/root(16-16r+5r^2)=0...(i)
=>r=2 or 6/5
but r=2 is extraneous solution.Does not satisfy eqn number (i)
So r=6/5
=>e is maximum when a=b=c=d=6/5
=>max(e)=f(6/5)=16/5.

## Monday, January 11, 2010

### PROBLEM OF THE WEEK-1

Here I start a new section called problem of the week. Solution of this question will be posted in the coming week. Till then have fun.

Alphonse and Beryl are back! They are playing a two person game with the following
rules:
• Initially there is a pile of N stones, with N >= 2.
• The players alternate turns, with Alphonse going first. On his first
turn, Alphonse must remove at least 1 and at most N −1 stones from
the pile.
• If a player removes k stones on their turn, then the other player must
remove at least 1 and at most 2k − 1 stones on their next turn.
• The player who removes the last stone wins the game.
(a) Determine who should win the game when N = 7, and explain the winning
strategy.[Easy]
(b) Determine who should win the game when N = 8, and explain the winning
strategy.[Medium]
(c) Determine all values of N for which Beryl has a winning strategy. Explain this
strategy.Where N<=100[Difficult]

### e is irrational

#Problem 8
Prove that e is irrational
Solution scheme and approach
e=1+1/1!+2/2!+3/3!+.....(i)
Let e is rational so e=p/q
multiplying equation number (i) by q! we get
q!*e=q!+q!/1!+q!/2!+q!/3!+....+q/q!+ other terms
Now as q!+q!/1!+...+q!/q! is an integer and q!*e is also an integer
So we can say other terms should be an integer. Let I signifies the rest terms whose summation is an integer.
I=q![1/(q+1)!+1/(q+2)!+1/(q+3)!+...]
=1/(q+1)+1/(q+1)*(q+2)+1/(q+1)*(q+2)*(q+3)+.....
<1/(q+1)+1/(q+1)^2+1/(q+1)^3+....=1/(q+1)/[1-1/(q+1)]=1/q<1
=>I< 1
So, e should be an irrational number. Q.E.D.
[This solution has been proposed by Fourier]

## Thursday, January 7, 2010

### Number's application

#Problem 7
Show that there is no integer a such that a^2-3a-19 is divisible by 289.[R.M.O. 2009]
Solution Scheme and Approach
a^2-3a-19
=a^2-3a-70+51
=(a+7)(a-10)+51 ...(i)

Now as (i) divides by 289 it should be divisible by 17 as well. So, it is clear at least one of (a+7) and (a-10) should be divisible by 17.
Let, a-10=17k =>a=10+17k therefore a+7=17(k+1)
Hence we can write a^2-3a-19=289k(k+1)+51
As 289k(k+1) is divisible by 289 so 51 has to be divisible by 289 as well..
There is no integer a such that a^2-3a-19 is divisible by 289.Q.E.D.

## Saturday, January 2, 2010

### Cubic Equation

#Problem 6
The cubic equation x^3+2x-1 = 0 has exactly one real root r. Note that 0.4 < r < 0.5.
(a)Prove that an increasing sequence of positive integers a1,a2,a3,.....exists, such that 1/2=r^a1+r^a2+r^a3+.... is possible
(b) Prove that the sequence that you found in part (a) is the unique increasing sequence
with the above property.
Solution Scheme and approach
(a)
As r is a root then we may write
r^3+2r-1=0 =>1-r^3=2r=>r/(1-r^3)=1/2=>r*(1-r^3)^-1=1/2
By expanding
r*(1+r^3+r^6+r^9+....)=1/2
As |r|<1 the series converges
=>r+r^4+r^7+r^10+....=1/2 equating with r^a1+r^a2+r^a3+..=1/2
we can write {a1,a2,a3....}::{1,4,7....}

If possible let another for another series b1,b2,b3....               r^b1+r^b2+r^b3+...=r^a1+r^a2+r^a3+...=1/2                                                                                canceling equal terms
r^x1+r^x2+r^x3+...=r^y1+r^y2+r^y3+...=N [N>0]............(1)
without losing the generality we can consider y1>x1 =>r^y1>r^x1
r^x1<=r^x1+r^x2+r^x3+..=r^y1+r^y2+r^y3+....
Dividing both sides by r^x1
1<=r^p1+r^p2+r^p3+......[yi-xi=pi]...........(2)
As 0<1/2 =>1<1/(1-r)<2 and p1>=1 then,                                                                                From (2)
1<=r^p1+r^p2+r^p3+...<=r+r^2+r^3+....=r/1-r=1/(1-r)-1<1
Hence only one sequence exists.

### Perfect square

This is one of the important topics which often comes in various competitive exams.

#Problem 5(A)
Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.
Solution scheme and approach
Let n^2+24n+21 is a perfect square.Hence we can write
n^2+24n+21=k^2
=>n^2+2*n*12+12^2-12^2+21=k^2
=>(n+12)^12-k^2=123
Case I
(n+12+k)(n+12-k)=3*41=(-3)*(-41)
So, (n+12+k)+(n+12-k)=3+41
=>2(n+12)=44 =>n=10
or,
2(n+12)=-44=>n=-34
Case II
(n+12+k)(n+12-k)=1*123=-1*-123
So, 2(n+12)=124=>n=49
or,
2(n+12)=-124=>n=-73
Therefore total 4 values of n ::{-73,-34,10,49} are possible.
Food for thought
Find the total number of values of n for which n^2+24n+21 is square of a prime number.Where n is natural number.
What would be the answer then.Understand the difference

#Problem 5(B)
Find all positive integers n such that n^2 + n + 2009 is a square.[Pomona-Wisconsin Math Talent Search 2009]
Solution scheme and approach
Let, n^2+n+2009=k^2
=>(n+1/2)^2-k^2=1/4-2009[Follow the same method as stated in 5(A)]
=>(n+1/2+k)(n+1/2-k)=-8035/4
=>[2*(n+1/2+k)][2*(n+1/2-k)]=-8035
(PS:As n is an integer we need to maintain the symmetry. So don't consider [4(n+1/2+k)]*(n+1/2-k) or (n+1/2+k)*[4(n+1/2-k)]. Each of the cases LHS becomes fraction hence contradicts)
=>(2n+1+2k)(2n+1-2k)=-8035
Now as 2n+1+2k>2n+1-2k
therefore, (2n+1+2k)(2n+1-2k)=1607*-5=8035*-1
=>2(2n+1)=1607-5=1602
=>n=400
or,
2(2n+1)=8034=>n=2008
Two values of n are possible.

## Friday, January 1, 2010

### Concept 1 :KiSsInG cIrClEs

Descartes' theorem (1964)
If four mutually tangent circles have curvature ki (for i = 1,...,4), (ki=1/ri, Where ri is radius)
 (1)
$(k_1+k_2+k_3+k_4)^2=2\,(k_1^2+k_2^2+k_3^2+k_4^2).$
When trying to find the radius of a fourth circle tangent to three given kissing circles, the equation is best rewritten as:
 (2)
$k_4 = k_1 + k_2 + k_3 \pm2 \sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}.$
The ± sign reflects the fact that there are in general two solutions. Ignoring the degenerate case of a straight line, one solution is positive and the other is either positive or negative; if negative, it represents a circle that circumscribes the first three (as shown in the diagram above).
Other criteria may favor one solution over the other in any given problem.

Special case
If one of the three circles is replaced by a straight line, then one ki, say k3, is zero and drops out of equation (1). Equation (2) then becomes much simpler:
 (3)
$k_4=k_1+k_2\pm2\sqrt{k_1k_2}.$
Example:
If three circles of radius 1 touches one another then find the radius of the circle which can be inserted within these circles, touching all the three.[CAT]

Solution:
k4=3+2root(3)[positive for inscribed circle] therefore radius=1/[3+2root(3)]=[2-root(3)]/root(3)

PS:One can use conventional method too.

## Thursday, December 31, 2009

### How many primes

#Problem 4
How many primes among the positive integers, written as usual in base 10, are such that their digits are altenating 1’s and 0’s, beginning and ending with 1? [Putnam]

Solution scheme and approach
Let,
X=1010101…1=1+10^2+10^4+…+10^2n=(10^2n-1)/99=(10^n+1)*(10^n-1)/99

For n=2 X=101 which is a prime number
For n>=3
If n is even
{
As (10^n+1)>99
(10^n-1)=(9+1)^n -1=0(mod 9) So,(10^n-1)=9a
(10^n-1)=(11-1)^n -1=0(mod 11)So (10^n-1)=11b
therefore we can write X=99*a*b*(10^n+1)=>Composite
}
If n is odd
{
As (10^n-1)>99
(10^n-1)=(9+1)^n -1=0(mod 9) So, (10^n-1)=9a[a>1]
(10^n+1)=(11-1)^n +1=0(mod 11) So,(10^n+1)=11b[b>1]
therefore X=99*a*b=>composite
}

So, there is only one prime number, 101, exists.Q.E.D.

### Ultimate Divisibility

This is one of the famous puzzles. Try it.

#Problem 3
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:
a. The number should be divisible by 9.
b. If the most right digit is removed, the remaining number should be divisible by 8.
c. If then again the most right digit is removed, the remaining number should be divisible by 7.
d. etc. until the last remaining number of one digit which should be divisible by 1.

Have fun.

## Wednesday, December 30, 2009

### Infinite descent

If x, y, z and n are integers, there are no solutions of x^n+y^n=z^n with n>2 and x,y,z>0.This is known as Fermat's last theorem.

This problem can be an extension of Fermat's Infinite Descent

#Problem2

Find all solutions to c2 + 1 = (a2 − 1)(b2 − 1), in integers a, b, and c.

Solution Scheme and approach
Clearly a = b = c = 0 is one integer solution.
Also, if c = 0, a^2 − 1 = ±1, and so a = 0; similarly b = 0.
Without loss of generality, we now seek a solution with c > 0.
Assume such a solution exists.
Consider the equation, modulo 4.
Since the square of any integer is either 4k or 4k+1

we have, in mod 4, (a^2 − 1) and (b^2 − 1) = −1 or 0(mod 4),So (a^2-1)(b^2-1)=0 or 1 and (c^2 + 1) =1 or 2(mod 4).
As, L.H.S cant be 2, Hence (a^2 − 1) = (b^2 − 1) = −1 (mod 4), and (c^2 + 1) = 1 (mod 4); that is, a, b, c are even.
Set a = 2x, b = 2y, c = 2z.
Then 4z^2 + 1 = (4x^2 − 1)(4y^2 − 1).
Simplifying, we have z^2 = 4x^2y^2 − x^2 − y^2.

Again considering mod 4, z^2=1 or 0(mod 4) and x^2=y^2=0,1(mod 4) So, R.H.S. can be -2,-1 or 0. Considering R.H.S we can decide z,x,y all are multiple of 4 i.e. even.

Now set z=2s, x=2r, y=2t. Then 16s^2+1=(16r^2-1)(16t^2-1) =>s^2=-16r^2t^2-r^2-t^2. Following the previous modulo 4 method again we can conclude that s,r,t all are even.
And this series continues till infinity. But as c>z>s>...>0 that means c is bounded. Which contradicts. Hence our assumption was wrong. There is no solution exists if c>0.

Therefore the only integer solution to c2 + 1 = (a2 − 1)(b2 − 1) is a = b = c = 0.Q.E.D.

### Number of solutions

Here I start my blog with a very interesting problem.

#Problem 1

Find the value of x and y. x^3+y^3=7163.[Where x and y are positive integers]

Solution scheme and approach 1

First of all one number should be even another is odd. Let, x=(2n+1) and y=2m
So,
x^3+y^3=7163
=>(2n+1)^3+(2m)^3=7163.....(i)
=>8n^3+12n^2+6n+1+(2m)^3=7163
=>2n(4n^2+6n+3)+8m^3=7162
=>n(4n^2+6n+3)+4m^3=3581

As RHS is odd ans 4m^3 is even therefore n(4n^2+6n+3) must be odd =>n is odd=>n=2a+1

From eqn (i) we can write
(4a+3)^3+(2m)^3=7163
=>64a^3+144a^2+108a+27+8m^3=7163
=>64a^3+144a^2+108a+8m^3=7136
=>a(16a^2+36a+27)+2m^3=1784

As 2m^3 is even therefore a(16a^2+36a+27) should be even=>a is even

Now by trial and error we have to move forward.
If a=2
then a(16a^2+36a+27)=326 =>m^3=729 =>m=9

So (a,m)::(2,9) =>(x,y)::(4*2+3,2*9)=>(x,y)::(11,18)

Solution scheme and approach 2

x^3+y^3=(x+y)(x^2-xy+y^2)=7163=13*19*29

As x and y are positive integers therefore x+y=1,x+y=7129 should be rejected. and we can also write that x^2+y^2-xy>x+y

3 cases are possible
(i)If x+y=13
then, x^2-xy+y^2=(x+y)^2-3xy=169-3xy=19*29=>3xy=-382 (impossible)

(ii)If x+y=19
then (x+y)^2-3xy=13*29=>361-3xy=13*29 =>3xy=-16(impossible)

(iii)If x+y=29
the (x+y)^2-3xy=29^2-3xy=13*19=>3xy=841-247=>3xy=594=>xy=198=11*18

(x,y)::(11,18)

Hope you have enjoyed the problem :)

Problem for practice:

If ${a^4} + {b^4}$ =4721. Then find the sum of all values of (a+b).
(a)9, (b)11, (c)13, (d)15, (e)None of the foregoing [Source of this question is totalgadha.com/rmo]


text:
# SM Energy’s Volatility Forecast: The Implication for SM Stock

By

Updated

## SM Energy’s implied volatility

As of August 2, 2016, SM Energy (SM) had an implied volatility of ~74.9%, which is ~26.8% below its 260-trading-day historical price volatility of ~102.2%. In the five days leading up to the earnings, SM Energy’s implied volatility declined from ~76.4% to ~74.9%.

## SM Energy’s 30-day stock price forecast by implied volatility

Assuming normal distribution of prices (bell curve model) and standard deviation of one, and based on its implied volatility of ~74.9%, SM Energy stock is expected to close between \$31.84 and \$20.58 after 30 calendar days. Based on the standard statistical formula, the stock will stay in this range ~68% of the time.

## Other upstream stocks

As of August 2, 2016, other upstream stocks such as Parsley Energy (PE), Gulfport Energy (GPOR), and Energen (EGN) have implied volatilities of ~38.3%, ~44.9%, and ~49.4%, respectively. The SPDR S&P 500 ETF (SPY) has an implied volatility of ~11.1%.

Implied volatility shows the Market’s opinion of the stock’s potential moves, but it doesn’t forecast direction. Implied volatility is derived from an option pricing model. This means the data are theoretical in nature and there’s no guarantee these forecasts will be correct.


text:
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# For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75

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For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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28 Mar 2016, 08:20
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75% (hard)

Question Stats:

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For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ $$\frac{0.75r}{2}$$ ]
B. 2.00 + 0.75[ $$\frac{r}{2}$$ ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]
Math Expert
Joined: 02 Aug 2009
Posts: 8309
Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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28 Mar 2016, 08:37
25
1
16
lpetroski wrote:
For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

A. 2.00 + [ $$\frac{0.75r}{2}$$ ]
B. 2.00 + 0.75[ $$\frac{r}{2}$$ ]
C. 2.00 + 0.75[ r ]
D. 2.00 + [1.5r ]
E. 2.00 + 0.75 [ 2r]

Hi,
Fixed fee= 2.00
fee dependent on miles = 0.75 per 1/2 mile or fraction there off
no of miles = r
no of 1/2 miles= 2r..
and to convert 2r into integer mile [2r]..
total cost on miles= 0.75[2r]..
total = 2.00 + 0.75 [ 2r]

E
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For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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28 Oct 2016, 23:33
4
4
I got it wrong after spending a lot of time for a mere calculation mistake. That too this was the first question in my practice test.
I hope my solution helps others. Its easy to solve if we pick the numbers.

Since it is nowhere mentioned that r is an integer, we can take r=4.5 miles for our convenience and apply back-solving approach.

Quote:
For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

as per the question charges= $$2.00$$ + $$0.75*(4.5*2)$$ = $$2.00$$+ $$6.75$$

Now let go to options

A. 2.00 + [ $$\frac{0.75r}{2}$$ ]
= 2.00 + [ $$\frac{(0.75*4.5)}{2}$$ ]
= 2.00 + [1.6875]
= 2.00 + 1.00

Since this value does not match with that of above one, this is incorrect.

B. 2.00 + 0.75[ $$\frac{r}{2}$$ ]
= 2.00 + 0.75[ $$\frac{4.5}{2}$$ ]
= 2.00 + 0.75[2.25]
= 2.00 + $$0.75*2$$
= 2.00 + 1.5

Since this value does not match with that of above one, this is incorrect.

C. 2.00 + 0.75[ r ]
= 2.00 + 0.75[4.5]
= 2.00 + 0.75*4
= 2.00 + 3.00

Since this value does not match with that of above one, this is incorrect.

D. 2.00 + [1.5r ]
= 2.00 + [1.5 * 4.5 ]
= 2.00 + [6.75]
= 2.00 + 6

Since this value does not match with that of above one, this is incorrect.

E. 2.00 + 0.75 [ 2r]
= 2.00 + 0.75[2*4.5]
= 2.00 + 0.75[9]
= 2.00 + 0.75*9
= 2.00 + 6.75

This is the value we got above in pre-analysis.

Refer below link for info on greatest integer function
http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1755424
##### General Discussion
Intern
Joined: 27 Jul 2011
Posts: 49
Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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05 Apr 2016, 14:38
2
Use substitution,
If
r=3.7; [r]=4
r=3.2; [r]=4

Therefore
Charges $0.75 = 0.5 mile$ y = 4 miles

y = (4 * $0.75 ) / 0.5 = r * 0.75 *2 Therefore Total = 2 + 0.75(2r) Intern Joined: 15 Oct 2015 Posts: 2 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 30 Apr 2016, 11:28 Hello, Can someone please try to explain to me and an easy way? I didn't understand the explanations. I appreciate any help !! The part that says "for every number x, [x] is defined to be the least integer greater than or equal to x" confuses me. Thank you ! Paula Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 59720 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 02 May 2016, 03:17 4 17 Intern Joined: 15 Oct 2015 Posts: 2 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 02 May 2016, 07:15 Bunuel thank you so much! That was very helpful!! Intern Joined: 05 Jan 2015 Posts: 49 Location: United States WE: Consulting (Consulting) Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 11 Jul 2016, 00:00 1 1 chetan2u wrote: lpetroski wrote: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long? A. 2.00 + [ $$\frac{0.75r}{2}$$ ] B. 2.00 + 0.75[ $$\frac{r}{2}$$ ] C. 2.00 + 0.75[ r ] D. 2.00 + [1.5r ] E. 2.00 + 0.75 [ 2r] Hi, Fixed fee= 2.00 fee dependent on miles = 0.75 per 1/2 mile or fraction there off no of miles = r no of 1/2 miles= 2r.. and to convert 2r into integer mile [2r].. total cost on miles= 0.75[2r].. total = 2.00 + 0.75 [ 2r] E Hi, Thanks for such a brilliant explanation. One doubt how no of 1/2 miles= 2r.. Intern Joined: 25 May 2016 Posts: 2 GMAT 1: 570 Q44 V25 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 19 Oct 2016, 06:45 1 riteshpatnaik wrote: chetan2u wrote: lpetroski wrote: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long? A. 2.00 + [ $$\frac{0.75r}{2}$$ ] B. 2.00 + 0.75[ $$\frac{r}{2}$$ ] C. 2.00 + 0.75[ r ] D. 2.00 + [1.5r ] E. 2.00 + 0.75 [ 2r] Hi, Fixed fee= 2.00 fee dependent on miles = 0.75 per 1/2 mile or fraction there off no of miles = r no of 1/2 miles= 2r.. and to convert 2r into integer mile [2r].. total cost on miles= 0.75[2r].. total = 2.00 + 0.75 [ 2r] E Hi, Thanks for such a brilliant explanation. One doubt how no of 1/2 miles= 2r.. If 1/2 a mile is r then a full mile will be 2r. Hope this helps! Intern Joined: 13 Sep 2016 Posts: 1 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 15 Oct 2017, 03:54 lpetroski wrote: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long? A. 2.00 + [ $$\frac{0.75r}{2}$$ ] B. 2.00 + 0.75[ $$\frac{r}{2}$$ ] C. 2.00 + 0.75[ r ] D. 2.00 + [1.5r ] E. 2.00 + 0.75 [ 2r] Hi EMPOWERgmatRichC Can you please help me here with Test It ? Thank you! Manager Joined: 11 Jun 2017 Posts: 62 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 24 Nov 2017, 05:05 2 1 Prashant420 wrote: lpetroski wrote: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long? A. 2.00 + [ $$\frac{0.75r}{2}$$ ] B. 2.00 + 0.75[ $$\frac{r}{2}$$ ] C. 2.00 + 0.75[ r ] D. 2.00 + [1.5r ] E. 2.00 + 0.75 [ 2r] Hi EMPOWERgmatRichC Can you please help me here with Test It ? Thank you! Hi Prashant420, Suppose if the number of miles covered are 5.2. Each 1/2 mile or fraction of half costs 0.75$. 5 miles consist of 10 half miles and 0.2 would account for another half. Total 11 halves*0.75$=$8.25 (for miles covered). Total cost = 10.25$. This can be obtained through option E: 2+0.75[2*5.2] = 2+0.75*[10.4] = 2+0.75*11 =$10.25.

Hope this helps.
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Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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07 Dec 2017, 15:17
3
1
Hi All,

This question can be solved by TESTing Values (although since the prompt discusses 'fractions of a 1/2 mile' iit's 'hinting' that we should be thinking about non-integers...

Let's TEST R = 2.1

According to the prompt, for a 2.1-mile trip, the taxi company will charge $2 +$0.75(5) = $5.75 When we plug R = 2.1 into the answers, and we follow the definition of [x], we get.... Answer A: 2 + 1 = 3 NOT a match Answer B: 2 + 1.5 = 3.5 NOT a match Answer C: 2 + 2.25 = 4.25 NOT a match Answer D: 2 + 4 = 6 NOT a match Answer E: 2 + 3.75 = 5.75 This IS a match Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ Intern Joined: 08 Dec 2017 Posts: 15 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 08 Dec 2017, 23:43 1 Ignoring the least integer part the equation is clearly 2+.75(2r). This can be 2+1.5r as well so it's between d and e. Going 1 mile should cost 1.50. With choice d, it costs 2 so e has to be the answer. One problem with choice d is that whenever one goes an odd number of miles the round up results in an overcharge. Posted from my mobile device ESADE School Moderator Joined: 23 Jan 2018 Posts: 30 Location: India Concentration: General Management, Marketing GMAT 1: 710 Q49 V39 GPA: 3.45 WE: Analyst (Computer Software) Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 12 Sep 2018, 14:01 1 Nevernevergiveup wrote: I got it wrong after spending a lot of time for a mere calculation mistake. That too this was the first question in my practice test. I hope my solution helps others. Its easy to solve if we pick the numbers. Since it is nowhere mentioned that r is an integer, we can take r=4.5 miles for our convenience and apply back-solving approach. Quote: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long? as per the question charges= $$2.00$$ + $$0.75*(4.5*2)$$ = $$2.00$$+ $$6.75$$ Now let go to options A. 2.00 + [ $$\frac{0.75r}{2}$$ ] = 2.00 + [ $$\frac{(0.75*4.5)}{2}$$ ] = 2.00 + [1.6875] = 2.00 + 1.00 Since this value does not match with that of above one, this is incorrect. B. 2.00 + 0.75[ $$\frac{r}{2}$$ ] = 2.00 + 0.75[ $$\frac{4.5}{2}$$ ] = 2.00 + 0.75[2.25] = 2.00 + $$0.75*2$$ = 2.00 + 1.5 Since this value does not match with that of above one, this is incorrect. C. 2.00 + 0.75[ r ] = 2.00 + 0.75[4.5] = 2.00 + 0.75*4 = 2.00 + 3.00 Since this value does not match with that of above one, this is incorrect. D. 2.00 + [1.5r ] = 2.00 + [1.5 * 4.5 ] = 2.00 + [6.75] = 2.00 + 6 Since this value does not match with that of above one, this is incorrect. E. 2.00 + 0.75 [ 2r] = 2.00 + 0.75[2*4.5] = 2.00 + 0.75[9] = 2.00 + 0.75*9 = 2.00 + 6.75 This is the value we got above in pre-analysis. Refer below link for info on greatest integer function http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1755424 Your approach really helped but just one correction: while eliminating option D you have missed out that it is the least integer greater than or equal to x which in this case is [6.75]=7. But in any case, the final answer by D is still wrong; E still holds true. Thank you for the approach! Found it the most helpful. Intern Joined: 28 Jul 2016 Posts: 1 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 17 Mar 2019, 07:33 Can someone explain why not A? Because 0.75 is also a fraction and it needs to be formatted according to the given rule, no? Manager Joined: 16 Jul 2018 Posts: 69 Re: For each trip, a taxi company charges a fixed fee of$2.00 plus $0.75 [#permalink] ### Show Tags 22 Nov 2019, 09:04 EMPOWERgmatRichC wrote: Hi All, This question can be solved by TESTing Values (although since the prompt discusses 'fractions of a 1/2 mile' iit's 'hinting' that we should be thinking about non-integers... Let's TEST R = 2.1 According to the prompt, for a 2.1-mile trip, the taxi company will charge$2 + $0.75(5) =$5.75

When we plug R = 2.1 into the answers, and we follow the definition of [x], we get....

Answer A: 2 + 1 = 3 NOT a match
Answer B: 2 + 1.5 = 3.5 NOT a match
Answer C: 2 + 2.25 = 4.25 NOT a match
Answer D: 2 + 4 = 6 NOT a match
Answer E: 2 + 3.75 = 5.75 This IS a match

GMAT assassins aren't born, they're made,
Rich

Hi Rich,
Instead of using a decimal I picked 2 and 4 as the TEST it numbers (I tested 4 except for 2, because 2 was valid for both D and E) was my approach incorrect? Was I lucky to get the correct answer? Is it necessary to use decimals in this specific problem?
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Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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22 Nov 2019, 11:45
1
Hi UNSTOPPABLE12,

If you chose R=4, then BOTH Answers D and E would have been a match for what you were looking for. Can you show me your work for those two calculations?

GMAT assassins aren't born, they're made,
Rich
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Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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22 Nov 2019, 11:55
EMPOWERgmatRichC wrote:
Hi UNSTOPPABLE12,

If you chose R=4, then BOTH Answers D and E would have been a match for what you were looking for. Can you show me your work for those two calculations?

GMAT assassins aren't born, they're made,
Rich

Yes you are right Rich , it turns up I got extremely lucky on this one, making a calculation mistake on D and picking E. Could you please explain why in your example you decided to choose 2.1 and why did you afterwards multiplied it by 5? (I have read your explanation but still haven't understood the reason )
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Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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22 Nov 2019, 15:46
1
Hi UNSTOPPABLE12,

If you start off by looking at the answer choices (before you read the prompt), you might recognize that, algebraically-speaking, Answers D and E appear to be the SAME ANSWER. That's 'weird' for a GMAT question - and without looking at the prompt, that pattern would either mean that neither of those answers is correct OR that that there's some 'twist' to the question that means the "format" of the answer will matter.

The prompt refers to 'fractions of a 1/2 mile' - and THAT means that the math involved is not the typical arithmetic/algebra that we'll see in most prompts. I want to reiterate that the prompt LITERALLY states "....FRACTIONS....", so that's a big clue to TEST a non-integer.

If we TEST R = 2.1

According to the prompt, for a 2.1-mile trip, we will be dealing with 4 "full 1/2 miles" and 1 "fraction of a 1/2 mile" (re: the extra 0.1 miles). Each of those 5 increments costs $0.75. Thus, the charge would be...$2 + $0.75(5) =$5.75

Based on how we are supposed to use the [ x ] symbol, there's only one answer that matches...

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Rich
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Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75  [#permalink]

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03 Dec 2019, 03:55
Nevernevergiveup wrote:
I got it wrong after spending a lot of time for a mere calculation mistake. That too this was the first question in my practice test.
I hope my solution helps others. Its easy to solve if we pick the numbers.

Since it is nowhere mentioned that r is an integer, we can take r=4.5 miles for our convenience and apply back-solving approach.

Quote:
For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75 for each 1/2 mile or fraction of 1/2 miles. If, for every number x, [ x ] is defined to be the least integer greater than or equal to x, then which of the following represents the company's charge, in dollars, for a trip that is r miles long?

as per the question charges= $$2.00$$ + $$0.75*(4.5*2)$$ = $$2.00$$+ $$6.75$$

Now let go to options

A. 2.00 + [ $$\frac{0.75r}{2}$$ ]
= 2.00 + [ $$\frac{(0.75*4.5)}{2}$$ ]
= 2.00 + [1.6875]
= 2.00 + 1.00

Since this value does not match with that of above one, this is incorrect.

B. 2.00 + 0.75[ $$\frac{r}{2}$$ ]
= 2.00 + 0.75[ $$\frac{4.5}{2}$$ ]
= 2.00 + 0.75[2.25]
= 2.00 + $$0.75*2$$
= 2.00 + 1.5

Since this value does not match with that of above one, this is incorrect.

C. 2.00 + 0.75[ r ]
= 2.00 + 0.75[4.5]
= 2.00 + 0.75*4
= 2.00 + 3.00

Since this value does not match with that of above one, this is incorrect.

D. 2.00 + [1.5r ]
= 2.00 + [1.5 * 4.5 ]
= 2.00 + [6.75]
= 2.00 + 6

Since this value does not match with that of above one, this is incorrect.

E. 2.00 + 0.75 [ 2r]
= 2.00 + 0.75[2*4.5]
= 2.00 + 0.75[9]
= 2.00 + 0.75*9
= 2.00 + 6.75

This is the value we got above in pre-analysis.

Refer below link for info on greatest integer function
http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1755424

The highlighted numbers in red are rounded down. According to question, it should be rounded up.

[6.75]=7 and not 6
Re: For each trip, a taxi company charges a fixed fee of $2.00 plus$0.75   [#permalink] 03 Dec 2019, 03:55
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text:
# How Does the CFA Institute® Grade CFA® Exams?

If you are a CFA exam candidate, you are probably concerned about how easy it is to pass your exam and how the CFA Institute grades the exams. The CFA Institute applies a unique methodology to come up with a minimum passing score for each exam.

Let us explore how the Institute grades the CFA exams.

## How is the CFA Exam Graded?

All three levels of the CFA exam have multiple-choice questions; while level III has essay-type questions. The machine grades the multiple-choice sections while CFA charterholders grade the constructed response sections. However, to ensure fairness, the graders work in teams. Each team grades only one question, and the names and testing centers of the candidates remain unknown to the graders. In addition, the CFA Institute provides the graders with guidelines and grading keys.

The grading process is thorough and involves checking to ensure high levels of quality and consistency. For instance, human graders check ten percent of the multiple-choice questions graded by the machine. The junior, senior, and grader of grades also counter-check each other’s work to ensure excellence. Next, someone else grades 50% of the distribution scores to give them another chance to pass. The graders do the third round of grading when the two rounds of grading give different results.

Finally, the grading is finalized and collected, and the minimum passing score for each exam is applied.

## What is the CFA Minimum Passing Score?

The minimum passing score is the mark used by the CFA Institute to establish whether a candidate has passed or failed the exam. The minimum passing score for each CFA level exam is determined through workshops and varies each year. The workshops involve various participants who analyze each exam question and make independent judgments on how averagely competent candidates could perform. Secondly, they aggregate and review these judgments and come up with a range for the scores of each exam.

The report from these workshops forms the foundation on which the charterholders on the CFA Institute board of governors set the minimum passing score. However, although the CFA Institute does not announce these scores, research estimates that they are between 60 and 70 percent.

## How does the CFA Institute give the Final Exam Grade?

The CFA Institute grades the exam by applying the minimum passing score. For instance, the candidate gets a pass if its score is above the minimum passing score. On the other hand, if the candidate scored below the minimum passing score, they receive a did not pass result.

In addition, candidates receive a summary of their performance on each topic and the maximum score possible on each test topic. However, the CFA Institute expresses the score in ranges, for example, more than 50% or more than 70%.

Here are some examples of what your score could look like using AnalystPrep‘s computer-based testing mock exams, which have been created to be as close as possible to the actual exam:

## I. Set up a Study Routine as Early as Possible

To pass your level I exam, you need to understand the material and have the skills to apply it. The exam will test how to apply what you learned to different scenarios. The CFA Institute recommends that you dedicate at least 300 hours of study for each exam level.

You should set apart a time to study and develop a study strategy to ensure that you get the most out of each session, and starting a study routine as early as possible will go a long way in helping you achieve this.

## II. Take Note of the Core Topics

During your study sessions, it is essential to understand the topics that carry more weight in the exams versus those that do not. For example, here are the topic weights for the level I exam.

$$\small{\begin{array}{l|c} \textbf{Level I Topic} & \textbf{Level I Weight} \\ \hline \text{Quantitative Methods} & 8-12\% \\ \hline \text{Economics} & 8-12\% \\ \hline \text{Financial Reporting and Analysis} & 13-17\% \\ \hline \text{Corporate Issuers} & 8-12\% \\ \hline \text{Equity} & 10-12\% \\ \hline \text{Fixed Income} & 10-12\% \\ \hline\text{Derivatives} & 5-8\% \\ \hline \text{Alternative Investments} & 5-8\% \\ \hline\text{Portfolio Management} & 5-8\% \\ \hline \text{Ethics} & 15-20\%\end{array}}$$

## III. You Should be Aware of the Specific Learning Outcomes

Specific learning outcomes (also known as learning outcome statements or LOS) refer to what you should be able to do by the end of your studies for each exam level. These are the actual knowledge, skills, and abilities that the CFA Institute intends for you to learn or develop. There are around 500 learning outcomes per level.

You should plan out your study sessions around the core topics and the specific learning outcomes. Planning study sessions around the core topics will help your study sessions be more focused and successful.

Here are some sample learning outcomes:

• Interpret interest rates as required rates of return, discount rates, or opportunity costs.
• Explain an interest rate as the sum of a real risk-free rate and premiums that compensate investors for bearing distinct types of risk.
• Calculate and interpret the effective annual rate, given the stated annual interest rate and the frequency of compounding.
• Calculate the solution for time value of money problems with different frequencies of compounding.

Note that the command words can either be descriptive in nature (interpret or explain) leading most often to non-mathematical questions or quantitative in nature (calculate) leading to the candidate having to perform a calculation.

## IV. Relate your Knowledge to Real Situations

The CFA exam tests not just your knowledge of the material but the skill to apply that knowledge to realistic situations. You need to apply what you have learned to solve problems rather than replicate memorized information.

It would be best to remember that you will need to use the material you learn to work in the real world, and relating it to real-life situations makes it easier to remember.

## V. Practice Taking the Exam

The last but equally important step in preparing for your CFA examinations is to practice taking the exam as much as possible. You can always buy additional resources from Analystprep.com.

Supplementing your studies with a CFA exam study package provided by Analystprep.com will provide you with extra learning materials and online resources to give you an edge, such as video lessons. Our experienced instructors will guide you through challenging concepts and help you gain the confidence to pass the exam.

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text:
# Projective resolution

I just started to learn homological algebra and I find it quite hard, so I am sorry if the question is unclear and confused. In fact I am confused.

Let $K^-(A)$, where $A$ is an abelian category with enough projectives, denote the category of complexes in $A$ with morphisms given by complex morphisms modulo homotopy. How to show that there is a quasi isomorphism from a complex of projectives to any complex $M$?

My teacher proved this in class and defined projective resolution in these terms, but I can not follow my notes.

However, looking into texts that do this the "classical way". I know that there exists a projective resolution of $M$ (in the classical sense), i.e., an exact sequence of projective elements in $K^-(A)$ with $P_0$ mapping surjectively to $M.$ However I can't seem to deduce my teacher's definition from this one. I know also that the projective objects of $K^-(A)$ are split exact complexes of projectives but how do I turn all of these exact sequences on the n:th place into one projective element of $A$ having isomorphic n:th homology with $M$?

• Dear harajm, just to make sure I got you right: you have an arbitrary bounded above complex $M$ in $A$ and you want to find a quasi-isomorphism $P\to M$ with $P$ bounded above with projective components? Could you also tell us your teacher's construction? – Hanno Jan 17 '15 at 8:06
• Yes, this is precisely what I mean. I can't say to much about my teachers construction since I don't really understand it. But he proves the statement by induction on the components of the chain, assuming we have a quasi isomorphism on the $n:th$ place. Actually I am very unsure what his argument really is, sorry. But he is not using that each component $M$ has a "projective resolution" in the classical sence, because he later deduces this statement from the earlier. – harajm Jan 17 '15 at 16:25
• But any way to prove this statement would help my enormously! – harajm Jan 17 '15 at 16:27


text:
# Quiz 5 - Bitcoin Mining

Bitcoin depends on miners to:
Click the card to flip 👆
1 / 35
Terms in this set (35)
Q. Which of the following are true about Bitcoin miners?

a. The target hash has become so small that the block header nonce alone isn't generally large enough to allow miners so search enough of the hash output space to find a valid block

b. Bitcoin miners can more efficiently mine for blocks by specifically targeting parts of the nonce search space that have more puzzle solutions

c. Over a 2 week period, the average time to mine a block is always 10 minutes

d. The mining difficulty is recomputed roughly every 2 weeks to keep the proof-of-work puzzle difficult
a. The target hash has become so small that the block header nonce alone isn't generally large enough to allow miners so search enough of the hash output space to find a valid block - correct;

d. The mining difficulty is recomputed roughly every 2 weeks to keep the proof-of-work puzzle difficult - correct;


text:
Question

# If the $$n^{th}$$ term of an arithmetic progression is $$15n + 10$$ then its fifth term is _______.

A
25
B
75
C
55
D
85

Solution

## The correct option is D $$85$$$$n^{th}$$ term of Arithmetic Progressive$$T_{n} = 15n + 10$$To find the fifth term of A.P.,take $$n = 5$$$$T_{5} = 15(5) + 10$$$$= 75 + 10$$$$= 85$$$$5^{th}$$ term of A.P. is $$85$$.Mathematics

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text:
Machine learning anomaly detection for continues data

I need to find anomalies in logs created by network monitor tool Bro. Right now I am referring to the example given by Bro Analyze Tool. But the example only uses recent logs.

So I need to make it work with full log meaning to train on past logs and perform anomaly detection in real time on each new log. In the example, as far as I understand, unsupervised machine learning with IsolationForest and KMeans in scikit-learn python are used.

So I have the following question.

For unsupervised machine learning is it possible to store the model with trained data which is already created on previous run?

If possible can I add new data with existing model and then perform anomaly detection in real time?

You can save sklearn models with pickle. This blog post shows how.

• Yes I have found it. So is it possible to append new data with trained model. – Haris Feb 23 '18 at 18:03

Sklearn comes with various ways to save and reuse models, see Model Persistence for more information.

However, it is worth mentioning that not all unsupervised outlier detection models can easily "make the prediction" for the unseen data. predict() method may not be always consistent and visible, such as LOF. As you can see, there is no "predict()" method but "fit()" and "fit_predict()" only. Alternatively, you should use an internal method _decision_function() for LOF in Sklearn instead.

def _decision_function(self, X):
"""Opposite of the Local Outlier Factor of X (as bigger is better,
i.e. large values correspond to inliers).

The argument X is supposed to contain *new data*: if X contains a
point from training, it consider the later in its own neighborhood.
Also, the samples in X are not considered in the neighborhood of any
point.
The decision function on training data is available by considering the
opposite of the negative_outlier_factor_ attribute.

This inconsistency is due to the unsupervised nature of most outlier detection methods. They are no difference for "training" and "test" since the ground truth is missing anyway. For some unsupervised outlier detection methods, saving the "trained" model does not help. It is case by case though. For efficient algorithms, it may be fine to combine the historical data with the new data and refit the model, given they come from the same distribution.

Thus, for different outlier detection methods in Sklearn, you should be careful with model persistence and APIs. If you are looking for more unified APIs to try various anomaly detection methods, I would recommend my python outlier detection toolbox PyOD; the documentation can be found here PyOD documentation


text:
# Mass of photon in motion

1. $$\dfrac{h\nu}{c^2}$$
2. $$\dfrac{h\nu}{c}$$
3. Zero

## Answer (Detailed Solution Below)

Option 1 : $$\dfrac{h\nu}{c^2}$$

## Detailed Solution

The correct answer is $$\dfrac{h\nu}{c^2}$$

Explanation:

A photon has energy E = hf where h is Planck’s constant and f is the frequency of the photon.

According to Einstein’s mass-energy equivalence E = mc2

but $$E = h\nu$$

⇒ $$h\nu =mc^2$$

$$m$$ $$=$$ $$\dfrac{h\nu}{c^2}$$ is the mass of a photon in motion.


text:
• Type Parameters:
E - the list element type
All Implemented Interfaces:
Iterable<E>, Collection<E>, Deque<E>, List<E>, Queue<E>

extends AbstractSequentialList<E>
implements Deque<E>
DoublyLinkedList implements a doubly linked List data structure, that exposes its ListNodes where the data is stored in.

An element holding ListNode can be removed or added to a DoublyLinkedList in constant time O(1). Other methods that operate on ListNodes directly also have constant runtime. This is also the case for methods that operate on the first(head) and last(tail) node or element. Random access methods have a runtime O(n) that is linearly dependent on the size of the DoublyLinkedList.

A DoublyLinkedList supports null elements but does not support null ListNodes. This class is not thread safe and needs to be synchronized externally if modified by concurrent threads.

The iterators over this list have a fail-fast behavior meaning that they throw a ConcurrentModificationException after they detect a structural modification of the list, that they're not responsible for.

This class is similar to LinkedList. The general difference is that the ListNodes of this List are accessible and can be removed or added directly. To ensure the integrity of the List nodes of this List have a reference to the List they belong to. This increases the memory occupied by this list implementation compared to LinkedList for the same elements. Instances of LinkedList.Node have three references each (the element, next and previous), instances of DoublyLinkedList.ListNode have four (the element, next, previous and the list).

Author:
Timofey Chudakov, Hannes Wellmann

• ### Method Detail

public void addNode(int index,
Inserts the specified node at the specified position in this list.

This method has a linear runtime complexity O(n) that depends linearly on the distance of the index to the nearest end. Adding node as first or last takes only constant time O(1).

Parameters:
index - index at which the specified node is to be inserted
node - the node to add
Throws:
IndexOutOfBoundsException - if the index is out of range (index < 0 || index > size())
IllegalArgumentException - if node is already part of this or another DoublyLinkedList
NullPointerException - if node is null

Inserts the specified node at the front of this list.

This method has constant runtime complexity O(1).

Parameters:
node - the node to add
Throws:
IllegalArgumentException - if node is already part of this or another DoublyLinkedList
NullPointerException - if node is null

Inserts the specified node at the end of this list.

This method has constant runtime complexity O(1).

Parameters:
node - the node to add
Throws:
IllegalArgumentException - if node is already part of this or another DoublyLinkedList
NullPointerException - if node is null

Inserts the specified node before the specified successor in this list.

This method has constant runtime complexity O(1).

Parameters:
node - the node to add
successor - ListNode before which the node is inserted
Throws:
IllegalArgumentException - if node is already contained in this or another DoublyLinkedList or successor is not contained in this list
NullPointerException - if successor or node is null
• #### getFirstNode

Returns the first node of this list.

This method has constant runtime complexity O(1).

Returns:
the first ListNode of this list
Throws:
NoSuchElementException - if this list is empty
• #### getLastNode

Returns the last node of this list.

This method has constant runtime complexity O(1).

Returns:
the last ListNode of this list
Throws:
NoSuchElementException - if this list is empty
• #### getNode

public DoublyLinkedList.ListNode<E> getNode(int index)
Returns the node at the specified position in this list.

This method has linear runtime complexity O(n).

Parameters:
index - index of the ListNode to return
Returns:
the ListNode at the specified position in this list
Throws:
IndexOutOfBoundsException - if the index is out of range (index < 0 || index >= size())
• #### indexOfNode

public int indexOfNode(DoublyLinkedList.ListNode<E> node)
Returns the index of the specified node in this list, or -1 if this list does not contain the node.

More formally, returns the index i such that node == getNode(i), or -1 if there is no such index. Because a ListNode is contained in at most one list exactly once, the returned index (if not -1) is the only occurrence of that node.

This method has linear runtime complexity O(n) to find node but returns in constant time O(1) if node is not contained in this list.

Parameters:
node - the node to search for
Returns:
the index of the specified node in this list, or -1 if this list does not contain node
Throws:
NullPointerException - if node is null
• #### removeNode

public boolean removeNode(DoublyLinkedList.ListNode<E> node)
Removes the node from this list. Returns true if node was in this list and is now removed. If node is not contained in this list, the list is left unchanged.

This method has constant runtime complexity O(1).

Parameters:
node - the node to remove from this list
Returns:
true if node was removed from this list
Throws:
NullPointerException - if node is null
• #### nodeOf

public DoublyLinkedList.ListNode<E> nodeOf(Object element)
Returns the first node holding the specified element in this list. More formally, returns the first ListNode such that Objects.equals(element, node.getValue()), or null if there is no such node.

This method has linear runtime complexity O(n).

Parameters:
element - the element whose ListNode is to return
Returns:
the first ListNode holding the element or null if no node was found
• #### lastNodeOf

public DoublyLinkedList.ListNode<E> lastNodeOf(Object element)
Returns the last node holding the specified element in this list. More formally, returns the last ListNode such that Objects.equals(element, node.getValue()), or null if there is no such node.

This method has linear runtime complexity O(n).

Parameters:
element - the element whose ListNode is to return
Returns:
the last ListNode holding the element or null if no node was found

Inserts the specified element at the front of this list. Returns the DoublyLinkedList.ListNode allocated to store the value. The returned ListNode is the new head of the list.

This method is equivalent to addFirst(Object) but returns the allocated ListNode.

Parameters:
element - the element to add
Returns:
the ListNode allocated to store the value

Inserts the specified element at the end of this list. Returns the DoublyLinkedList.ListNode allocated to store the value. The returned ListNode is the new tail of the list.

This method is equivalent to addLast(Object) but returns the allocated ListNode.

Parameters:
element - the element to add
Returns:
the ListNode allocated to store the value

E element)
Inserts the specified element before the specified successor in this list. Returns the ListNode allocated to store the value.
Parameters:
successor - ListNode before which the node holding value is inserted
element - the element to add
Returns:
the ListNode allocated to store the value
Throws:
IllegalArgumentException - if successor is not contained in this list
NullPointerException - if successor is null

public void addFirst(E e)
Specified by:
addFirst in interface Deque<E>

public void addLast(E e)
Specified by:
addLast in interface Deque<E>
• #### offerFirst

public boolean offerFirst(E e)
Specified by:
offerFirst in interface Deque<E>
• #### offerLast

public boolean offerLast(E e)
Specified by:
offerLast in interface Deque<E>
• #### removeFirst

public E removeFirst()
Specified by:
removeFirst in interface Deque<E>
• #### removeLast

public E removeLast()
Specified by:
removeLast in interface Deque<E>
• #### pollFirst

public E pollFirst()
Specified by:
pollFirst in interface Deque<E>
• #### pollLast

public E pollLast()
Specified by:
pollLast in interface Deque<E>
• #### getFirst

public E getFirst()
Specified by:
getFirst in interface Deque<E>
• #### getLast

public E getLast()
Specified by:
getLast in interface Deque<E>
• #### peekFirst

public E peekFirst()
Specified by:
peekFirst in interface Deque<E>
• #### peekLast

public E peekLast()
Specified by:
peekLast in interface Deque<E>
• #### offer

public boolean offer(E e)
Specified by:
offer in interface Deque<E>
Specified by:
offer in interface Queue<E>
• #### remove

public E remove()
Specified by:
remove in interface Deque<E>
Specified by:
remove in interface Queue<E>
• #### poll

public E poll()
Specified by:
poll in interface Deque<E>
Specified by:
poll in interface Queue<E>
• #### element

public E element()
Specified by:
element in interface Deque<E>
Specified by:
element in interface Queue<E>
• #### peek

public E peek()
Specified by:
peek in interface Deque<E>
Specified by:
peek in interface Queue<E>
• #### push

public void push(E e)
Specified by:
push in interface Deque<E>
• #### pop

public E pop()
Specified by:
pop in interface Deque<E>
• #### invert

public void invert()
Inverts the list. For instance, calling this method on the list $(a,b,c,\dots,x,y,z)$ will result in the list $(z,y,x,\dots,c,b,a)$. This method does only pointer manipulation, meaning that all the list nodes allocated for the previously added elements are valid after this method finishes.
• #### moveFrom

public void moveFrom(int index,
Moves all ListNodes of the given sourceList to this list and inserts them all before the node previously at the given position. All the nodes of movedList are moved to this list. When this method terminates this list contains all nodes of movedList and movedList is empty.
Parameters:
index - index of the first element of list in this list after it was added
movedList - the DoublyLinkedList to move to this one
Throws:
NullPointerException - if movedList is null
• #### append

public void append(DoublyLinkedList<E> movedList)
Appends the movedList to the end of this list. All the elements from movedList are transferred to this list, i.e. the list is empty after calling this method.
Parameters:
movedList - the DoublyLinkedList to append to this one
Throws:
NullPointerException - if movedList is null
• #### prepend

public void prepend(DoublyLinkedList<E> movedList)
Prepends the movedList to the beginning of this list. All the elements from movedList are transferred to this list, i.e. the movedList is empty after calling this method.
Parameters:
movedList - the DoublyLinkedList to prepend to this one
Throws:
NullPointerException - if movedList is null
• #### circularIterator

public DoublyLinkedList.NodeIterator<E> circularIterator(E firstElement)
Returns a DoublyLinkedList.NodeIterator that starts at the first DoublyLinkedList.ListNode of this list that is equal to the specified firstElement, iterates in forward direction over the end of this list until the first node.

The first call to DoublyLinkedList.NodeIterator.nextNode() returns the first node that holds a value such that Objects.equals(node.getValue, firstElement) returns true. The returned NodeIterator iterates in forward direction returning the respective next element in subsequent calls to next(Node). The returned iterator ignores the actual bounds of this DoublyLinkedList and iterates until the node before the first one is reached. Its hasNext() returns false if the next node would be the first one.

Parameters:
firstElement - the element equal to the first next()
Returns:
a circular NodeIterator iterating forward from firstElement
• #### reverseCircularIterator

public DoublyLinkedList.NodeIterator<E> reverseCircularIterator(E firstElement)
Returns a DoublyLinkedList.NodeIterator that starts at the first DoublyLinkedList.ListNode of this list that is equal to the specified firstElement, iterates in reverse direction over the end of this list until the first node.

The first call to DoublyLinkedList.NodeIterator.nextNode() returns the first node that holds a value such that Objects.equals(node.getValue, firstElement) returns true. The returned NodeIterator iterates in reverse direction returning the respective previous element in subsequent calls to next(Node). The returned iterator ignores the actual bounds of this DoublyLinkedList and iterates until the node before the first one is reached. Its hasNext() returns false if the next node would be the first one.

Parameters:
firstElement - the element equal to the first next()
Returns:
a circular NodeIterator iterating backwards from firstElement


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# Question - Shortest Distance Between Two Lines

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#### Question

Show that the following two lines are coplanar:

(x−a+d)/(α−δ)= (y−a)/α=(z−a−d)/(α+δ) and (x−b+c)/(β−γ)=(y−b)/β=(z−b−c)/(β+γ)

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