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# If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100

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Math Expert
Joined: 02 Sep 2009
Posts: 45305
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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29 Jan 2016, 03:18
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

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Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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29 Jan 2016, 06:56
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Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

hi,
bunuel a good Q..

what does three factor means?
it means the number is a square of a prime number.

THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?

one number has to be 1 and other the perfect square itself..

lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...

total ways = 100*99..

prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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29 Jan 2016, 07:32
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I like to add to chetan post

the clue to the sum is a number containing 3 factor

to get this we should have a square power to a prime number in the canonical form

For ex :2^2 , 3^2 , 4^2
All the above numbers will have 3 factors , since their powers are 2

We have constrain in this sum , it should be less than 100.
So perfect squares of prime numbers less than 100 will fit
2^2 = 4
3^2 – 9
5^2 -25
7^2 = 49
11^2 -121 >100 not ok
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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29 Jan 2016, 21:43
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Expert's post
Aside: Chetan noticed that, since the product ab has an ODD number of factors, we can conclude that ab is a perfect square.
For more on this concept, see our free video: http://www.gmatprepnow.com/module/gmat- ... /video/829
Here's a practice question to reinforce your learning: http://www.gmatprepnow.com/module/gmat- ... /video/830

Cheers,
Brent
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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31 Jan 2016, 18:56
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.

Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.

Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.

Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.
Math Expert
Joined: 02 Aug 2009
Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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31 Jan 2016, 20:45
Nez wrote:
Hei folks,
I don't get where it said "the product is less than 100".
The numbers being multiplied are specified as <=100.

Cmon guys,
where did we get that from?
chetan2u and EMPOWERgmatRichC please throw light.

Hi Nez,

the Q is
Quote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

why should be the product be <100..
the condition the resulting number has EXACTLY 3 factors? narrows down the answer to perfect square of a prime number..
so factor of that number= 1*p*p, where p is the prime number..
the other condition is a and b are different number, so 1*p*p can have two different factors as 1*p^2, where a=1 and b=p^2 or vice versa...
p*p would mean both a and b are same, which means a=b=p.. BUT this is not true as per the restrictions..
that is why a*b will have one of a or b as 1, and the other number perfect square <100.. so its product will always be <100..
hope it helps
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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10 Feb 2016, 16:10
chetan2u wrote:
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

hi,
bunuel a good Q..

what does three factor means?
it means the number is a square of a prime number.

THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?

one number has to be 1 and other the perfect square itself..

lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...

total ways = 100*99..

prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B

You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?

100C2 = 50*99

Also, video solving the problem for those who need it.
http://gmatpractice.q-51.com/arithmetic ... ty-4.shtml
Math Expert
Joined: 02 Aug 2009
Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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10 Feb 2016, 19:14
ZaydenBond wrote:
chetan2u wrote:
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

hi,
bunuel a good Q..

what does three factor means?
it means the number is a square of a prime number.

THE CATCH IS :- How do we get two different integers a and b such that the multiple of these numbers is a perfect square?

one number has to be 1 and other the perfect square itself..

lets see how many perfect square of which all prime numbers are <100...
2,3,5,7.. next 11 will lead to 121, which is >100..
if say a is one in every case b will be 4,9,25,49.... so 4 ways
vice versa.. b is 1 and a is 4,9,25,49..
so total ways =2*4=8...

total ways = 100*99..

prob= $$\frac{8}{{100*99}}$$..
prob= $$\frac{2}{{25*99}}$$
B

You list the total # of ways of selecting 2 numbers as 100*99, but isn't that incorrect?

100C2 = 50*99

Also, video solving the problem for those who need it.
http://gmatpractice.q-51.com/arithmetic ... ty-4.shtml

HI,
100C2 means combination of two numbers in 100 numbers..
But here you have a then b and b, then a are two different selections..
first number can be selected in 100 ways..
when the first one is selected, the remaining are 99, out of which you have to select the 2nd one..
so ways = 100*99
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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11 Feb 2016, 14:25
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

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Joined: 02 Aug 2009
Posts: 5779
Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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11 Feb 2016, 18:26
livio04 wrote:
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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12 Feb 2016, 01:17
if a number has exactly 3 factors means, it is a square of a prime number.
take example prime 2 square number is 4. factors are 1 ,2 and4.
same for 9 is 1,3 and 9.

among numbers from 1to 100. we are selecting two distinct integers a and b and multiply it.

so two distinct integers with 3 factors one number must be 1 and other numbers are 4,9,25, and 49.

so total 4 pairs of combination available.
number of selctions of 2 numbers from 1 to 100 is 100c2

so probability= favourable events/number of events

=4/100*99/2=4/50*99
=2/25*99. so option B is correct.
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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12 Feb 2016, 02:47
chetan2u wrote:
livio04 wrote:
Guys, can anyone explain me why numbers such as 22 (1*2*11) or 51 (3*17*1) are not in your list of possible outcomes? If we make a matrix of prime numbers from 2 to 47 both in horizontal and vertical we'll find out 31 numbers between 1 and 100 that would fit this rule of having 3 factors and with products below 100:

2*2, 2*3, 2*5, 2*7, 2*11.... 2*47
3*3, 3*5, 3*7,.... 3*31
5*5, 5*7.... 5*17
7*7, 7*11, 7*13

HI,
why we do not take numbers like 22 or 51 is bcause they have more than 3 factors..
22- 1,2,11,22- 4 factors
51-1,3,17,52- 4 factors..
as explained above only square of prime number fit into this category..
hope it helped

Thank you! It helped indeed!
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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16 Mar 2016, 00:47
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

Lets consider 121 (perfect square of 11) -> 3 factors of 121 = 1, 11, 121 (which would be the possible values of a and b)

So a and b = (121 and 1) or (11 and 11) - but a, b <100 and also distinct. Hence the perfect squares need to be <100.

Hope this helps.
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Re: If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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31 Jan 2018, 16:32
Hello,

following up on the answers above, once you realize that one of the numbers has to be 1 and the others have to be squares of prime numbers (in this case, 4, 9, 25, 49), you can use the hypergeometric distribution as a quick way to figure out the probability:

In this case, you're finding the probability of choosing 1 from 1 number (1 - 1C1) AND 1 from 4 numbers (4,9,25,49 - 4C1), from a total 2 out of a total of 100 numbers (100C2).

numerator: The possible number of event outcomes - in this case, the number of outcomes will be 1C1 * 4C1 (just 4)
denominator: The total possible outcomes - 100C2 ((99 * 100) / 2 or 99*50)

So the total probability would be (1C1 * 4C1) / (100C2) = 4 / (99 * 50) = = 2 / (99 * 25)

hope this helps
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100 [#permalink]

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31 Jan 2018, 20:53
Bunuel wrote:
If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100} and multiplied, what is the probability that the resulting number has EXACTLY 3 factors?

A. 4/(25*99)
B. 2/(25*99)
C. 8/(25*99)
D. 16/(25*99)
E. 32/(25*99)

I feel here that the resultant of multiplication is important rather than the order.
Total ways of selecting number with EXACTLY 3 factors (1,4) (1,9) (1,25) (1,49) = 4
Total ways of choosing the number = $$100C2$$= $$50*99$$
Hence, Probability = $$4/50*99$$ = $$2/25*99$$
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If two distinct integers a and b are picked from {1, 2, 3, 4, .... 100   [#permalink] 31 Jan 2018, 20:53
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