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![X and y are integers such that when [m]xy^3[/m](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "X and y are integers such that when [m]xy^3[/m"){kind=icon}
23 May 2026, 22:14
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\(x\) and \(y\) are integers such that when \(xy^3\) is divided by \(24\), the remainder is \(12\). Which of the following must be true?
I. \(x + y\) is odd
II. \(\frac{x}{2}\) is even
III. \(x + y\) is a multiple of \(3\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
I. \(x + y\) is odd
II. \(\frac{x}{2}\) is even
III. \(x + y\) is a multiple of \(3\)
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
25 May 2026, 04:56
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\(x\) and \(y\) are integers such that when \(xy^3\) is divided by \(24\), the remainder is \(12\). Which of the following must be true?
From the prompt, we know the following:
\(xy^3 = 24d + 12 = 12\) \(×\) an odd number
I. \(x + y\) is odd
\(xy^3 = 12\) \(×\) an odd number.
Thus, either \(x\) or \(y^3\) must be odd,
and either \(x\) or \(y^3\) must be even.
odd \(+\) even \(=\) odd
Must be true.
II. \(\frac{x}{2}\) is even
For \(\frac{x}{2}\) to be even, \(x\) has to be a multiple of \(4\).
So, let's check to see whether \(x\) must be a multiple of \(4\).
We know from our analysis of statement I that \(x\) or \(y^3\) is odd and \(x\) or \(y^3\) is even.
Let's see whether \(x\) has to be even by making \(x\) odd and seeing whether the original expression works.
If \(x\) is odd, \(y\) must be even. The cube of an even \(y\) is always a multiple of \(2^3 = 8\).
Any multiple of \(12\) that is a multiple of \(8\) is a multiple of \(24\).
If \(xy^3\) is is multiple of \(24\), there will be no remainder when \(xy^3\) is divided by \(24\).
Thus, \(x\) cannot be the odd one. It must be that \(x\) is even and \(y^3\) is odd.
Then, for \(xy^ 3\) to be a multiple of \(12\), if \(y\) is odd, \(x\) must be a multiple of \(4\).
\(x = 4d\)
\(\frac{4d}{2} =\) even
Must be true.
III. \(x + y\) is a multiple of \(3\)
From our analysis of II, we know that \(y^3\) is odd.
We can also see that, since \(xy^3\) is a multiple of \(12\), either \(x\) or \(y\) is a multiple of \(3\), but not necessarily both.
If \(x\) is a multiple of \(3\) and \(y\) is not, \(x + y\) is not a multiple of \(3\).
So, this does not have to be true.
Eliminate.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Correct answer: D
From the prompt, we know the following:
\(xy^3 = 24d + 12 = 12\) \(×\) an odd number
I. \(x + y\) is odd
\(xy^3 = 12\) \(×\) an odd number.
Thus, either \(x\) or \(y^3\) must be odd,
and either \(x\) or \(y^3\) must be even.
odd \(+\) even \(=\) odd
Must be true.
II. \(\frac{x}{2}\) is even
For \(\frac{x}{2}\) to be even, \(x\) has to be a multiple of \(4\).
So, let's check to see whether \(x\) must be a multiple of \(4\).
We know from our analysis of statement I that \(x\) or \(y^3\) is odd and \(x\) or \(y^3\) is even.
Let's see whether \(x\) has to be even by making \(x\) odd and seeing whether the original expression works.
If \(x\) is odd, \(y\) must be even. The cube of an even \(y\) is always a multiple of \(2^3 = 8\).
Any multiple of \(12\) that is a multiple of \(8\) is a multiple of \(24\).
If \(xy^3\) is is multiple of \(24\), there will be no remainder when \(xy^3\) is divided by \(24\).
Thus, \(x\) cannot be the odd one. It must be that \(x\) is even and \(y^3\) is odd.
Then, for \(xy^ 3\) to be a multiple of \(12\), if \(y\) is odd, \(x\) must be a multiple of \(4\).
\(x = 4d\)
\(\frac{4d}{2} =\) even
Must be true.
III. \(x + y\) is a multiple of \(3\)
From our analysis of II, we know that \(y^3\) is odd.
We can also see that, since \(xy^3\) is a multiple of \(12\), either \(x\) or \(y\) is a multiple of \(3\), but not necessarily both.
If \(x\) is a multiple of \(3\) and \(y\) is not, \(x + y\) is not a multiple of \(3\).
So, this does not have to be true.
Eliminate.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Correct answer: D
