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If \(x\) is an odd integer and \(y=x(x+1)(x-1),\) which of the following must also be an integer?

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If \(x\) is an odd integer and \(y=x(x+1)(x-1),\) which of the following must also be an integer?

by VJesus12 » Fri Sep 04, 2020 6:08 am

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If \(x\) is an odd integer and \(y=x(x+1)(x-1),\) which of the following must also be an integer?

I. \(\dfrac{y}{2^3\cdot 3}\)

II. \(\dfrac{y}{2^2\cdot 3^2}\)

III. \(\dfrac{y}{2\cdot 3}\)

(A) I only
(B) III only
(C) I and III
(D) I, II, and III
(E) none of the above

Answer: C

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Re: If \(x\) is an odd integer and \(y=x(x+1)(x-1),\) which of the following must also be an integer?

by Scott@TargetTestPrep » Wed Sep 09, 2020 1:38 pm
VJesus12 wrote: ↑
Fri Sep 04, 2020 6:08 am
 If \(x\) is an odd integer and \(y=x(x+1)(x-1),\) which of the following must also be an integer?

I. \(\dfrac{y}{2^3\cdot 3}\)

II. \(\dfrac{y}{2^2\cdot 3^2}\)

III. \(\dfrac{y}{2\cdot 3}\)

(A) I only
(B) III only
(C) I and III
(D) I, II, and III
(E) none of the above

Answer: C

Solution:

Since y is a product of 3 consecutive integers (regardless if x is odd or even), it’s divisible by 3! = 6. Therefore, III will result in an integer.

Furthermore, since we are given that x is odd, we can write x = 2k + 1 for some integer k. Therefore, in terms of k, y = (2k + 1)(2k + 2)(2k) = 2k(2(k + 1))(2k + 1) = 4k(k + 1)(2k + 1). Since k(k + 1) is a product of 2 consecutive integers, it’s divisible by 2! = 2. Along with the factor 4, y is divisible by 4 * 2 = 8. Since y is divisible by 6 (mentioned above), it’s divisible by 3 and therefore, y is divisible by 8 x 3 = 24. We see that I will result in an integer.

However, II might not result in an integer. For example, if x = 3, y = 3(4)(2) = 24 is not divisible by 2^2 * 3^2 = 36.

Answer: C

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