Numbers Problem 2

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aditiniyer: Senior | Next Rank: 100 Posts; Posts: 45; Joined: Fri Aug 19, 2016 1:42 am

by aditiniyer » Sat Jan 28, 2017 4:58 am

If the integer a & n are greater than 1, and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

1) a^n = 64
2) n=6

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Source: Beat The GMAT — Data Sufficiency | Sat Jan 28, 2017 4:58 am

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by GMATGuruNY » Sat Jan 28, 2017 5:11 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Mon Jan 30, 2017 10:10 pm

aditiniyer wrote:If the integer a & n are greater than 1, and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

1) a^n = 64
2) n=6

We have 1*2*3*4*5*6*7*8 = K*a^n; where K is a positive integer

=> 2â�·*3*5*7 = K*a^n

S1: a^n = 64

If a^n = 64 = 2^6 => a = 2; however,

if a^n = 64 = 4^3 => a = 4. No unique value of 'a.' Insufficient.

S2: n=6

We have 2â�·*3*5*7 = K*a^n

2â�·*3*5*7 = K*a^6

Since a > 1, the minimum value of a = 2. This is an important information. Had this been not there, a^6 = 1^6 = 1, rendering a =1.

If a =2, K=2*3*5*7--possible.

'a' cannot be greater than 2 since K cannot be a be a negative exponent of 3, 5, or 7; K is a positive integer.

Thus, a = 2. Sufficient.

The correct answer: B

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

Hope this helps!

-Jay
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8149; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Thu Feb 02, 2017 11:16 am

aditiniyer wrote:If the integer a & n are greater than 1, and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

1) a^n = 64
2) n=6

We are given that integers a and n are greater than 1, and the product of the first 8 positive integers is a multiple of a^n. Thus:

8!/a^n = integer

Recall that 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = (2^7) x (3^2) x 5 x 7.

We must determine the value of a.

Statement One Alone:

a^n = 64

There are multiple possible values of a. For instance, a = 2 and n = 6 (since 2^6 = 64), or a = 4 and n = 3 (since 4^3 = 64). Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

n = 6

Since n = 6, we know the following:

8!/a^6 = integer

[(2^7) x (3^2) x 5 x 7]/a^6 = integer

The only factor in the prime factorization of 8! that has a power greater than or equal to 6 is 2^7. Thus, the only possible value of a that will allow for 8!/a^6 to be an integer is 2.

Answer: B

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