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If n is an integer, which of the following must also be an i

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by Max@Math Revolution » Wed Jul 18, 2018 3:50 am

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[GMAT math practice question]

If n is an integer, which of the following must also be an integer?

I. n(n+1)/2
II. n(n+1)(n+2)/6
III. n(n+1)(n+2)(n+3)/8

A. I only
B. II only
C. III only
D. I and II
E. I, II and III
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by Max@Math Revolution » Fri Jul 20, 2018 12:16 am
=>

Statement I
Since n and n + 1 are consecutive integers, n(n+1) is a multiple of 2.
Thus, n(n+1)/2 is an integer.

Statement II
Since n and n + 1 are consecutive integers, n(n+1) and n(n+1)(n+2) are multiples of 2.
Since n, n + 1 and n + 2 are three consecutive integers, n(n+1)(n+2) is a multiple of 3.
Thus, n(n+1)(n+2) is a multiple of 6, and n(n+1)(n+2) / 6 is an integer.

Statement III
Since n and n + 1 are two consecutive integers, n(n+1) is a multiple of 2.
Similarly, (n+2)(n+3) is a multiple of 2.
Also, either n and n + 2 or n + 1 and n + 3 are consecutive even integers. Thus, either (n + 1)(n+3) is a multiple of 8 or n(n+2) is a multiple of 8 since one of them is a multiple of 4.
It follows that n(n+1)(n+2)(n+3) is a multiple of 8, and n(n+1)(n+2)(n+3)/8 is an integer.

Therefore, the answer is E.

Answer: E
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by Scott@TargetTestPrep » Tue Apr 23, 2019 6:41 pm
Max@Math Revolution wrote:[GMAT math practice question]

If n is an integer, which of the following must also be an integer?

I. n(n+1)/2
II. n(n+1)(n+2)/6
III. n(n+1)(n+2)(n+3)/8

A. I only
B. II only
C. III only
D. I and II
E. I, II and III

We can use the rule that the product of n consecutive integers will always be divisible by n!.

Thus, n(n+1), or the product of 2 consecutive integers, will always be divisible by 2! = 2.

n(n+1)(n+2), or the product of 3 consecutive integers, will always be divisible by 3! = 6.

n(n+1)(n+2)(n+3), or the product of 4 consecutive integers, will always be divisible by 4! = 24, which is divisible by 8.

Thus I, II, and III are all true.

Answer: E

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