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

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

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New post 21 Dec 2018, 02:50
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

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

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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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New post 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.

C is the correct answer.
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If m is the product of all the integers from 2 to 11, inclusive, and n  [#permalink]

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New post 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

Answer: 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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New post 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!)

Solving to obtain the answer:
\(\frac{n}{m}\) = \(\frac{11!}{3!}\) 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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