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# If m is the product of all the integers from 2 to 11, inclusive, and n

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Math Expert
Joined: 02 Sep 2009
Posts: 56266
If m is the product of all the integers from 2 to 11, inclusive, and n  [#permalink]

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21 Dec 2018, 02:50
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Difficulty:

15% (low)

Question Stats:

85% (00:54) correct 15% (01:44) wrong based on 50 sessions

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If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of n/m?

A. 1/12
B. 1/8
C. 1/6
D. 1/3
E. 1/2

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Re: If m is the product of all the integers from 2 to 11, inclusive, and n  [#permalink]

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21 Dec 2018, 05:00
Bunuel wrote:
If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of n/m?

A. 1/12
B. 1/8
C. 1/6
D. 1/3
E. 1/2

m = 2*3*4*5*6*7*8*9*10*11
n=4*5*6*7*8*9*10*11

So, whenever we are going to divide n by m we will left with only 2 and 3 in the denominator . Rest numbers will be eliminated due to division.

n/m= 1/2*3 = 1/6.

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2942
If m is the product of all the integers from 2 to 11, inclusive, and n  [#permalink]

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21 Dec 2018, 05:29

Solution

Given:
• m is the product of all the integers from 2 to 11, inclusive
• n is the product of all the integers from 4 to 11, inclusive

To find:
• The value of $$\frac{n}{m}$$

Approach and Working:
• m = 2 * 3 * 4 * .... * 10 * 11
• n = 4 * 5 * 6 * ...… * 10 * 11
• Therefore, $$\frac{n}{m} = \frac{1}{(2*3)} = \frac{1}{6}$$

Hence, the correct answer is Option C

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Re: If m is the product of all the integers from 2 to 11, inclusive, and n  [#permalink]

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22 Dec 2018, 02:16
Bunuel wrote:
If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of n/m?

A. 1/12
B. 1/8
C. 1/6
D. 1/3
E. 1/2

I found that if we write the 2 numbers in terms of factorials, then it can be solved slightly quicker.

Setting up the statement in mathematical form:
m = 11! (Notice that 11! = 1 x 2 x 3 .... x 11 which is same as 2 x 3 x .... x 11)

n = $$\frac{11!}{3!}$$ (n is product of all integers from 1 to 11, except 1x2x3, which is 3!)

$$\frac{n}{m}$$ = $$\frac{11!}{3!}$$ divided by 11!
i.e $$\frac{11!}{3!}$$*$$\frac{1}{11!}$$
=$$\frac{1}{3!}$$ =$$\frac{1}{6}$$

This method helped train my mind to simplify multiplication of consecutive integers in terms of factorials (in case required in other more complex problems).

Cheers!
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Re: If m is the product of all the integers from 2 to 11, inclusive, and n   [#permalink] 22 Dec 2018, 02:16
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