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If $2 a+b$ and $\frac{a}{b}$ are even, $a$ and $b$ are integers, then
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22 Jun 2025, 22:05
22 Jun 2025, 22:05
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If $\(2 a+b\)$ and $\(\frac{a}{b}\)$ are even, $a$ and $b$ are integers, then which of the following must be true?
(A) $\(a\)$ is even
(B) $\(b\)$ is even
(C) $\(a+b\)$ is even
(D) $\(a-b\)$ is even
(E) $\(a\)$ is a multiple of 4
(F) $\(b\)$ is a multiple of 4
(A) $\(a\)$ is even
(B) $\(b\)$ is even
(C) $\(a+b\)$ is even
(D) $\(a-b\)$ is even
(E) $\(a\)$ is a multiple of 4
(F) $\(b\)$ is a multiple of 4
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Official Answer
A,B,C,D,E
Re: If $2 a+b$ and $\frac{a}{b}$ are even, $a$ and $b$ are integers, then
[#permalink]
24 Jun 2025, 04:15
24 Jun 2025, 04:15
Expert Reply
Let $b=2$ (even, not multiple of 4 ).
Then $\(\frac{a}{2}\)$ is even, so $a=4 k$. Let $k=1$, so $a=4$.
Now, check $\(2 a+b=8+2=10\)$, which is even.
Check options:
A: $a=4$ is even. True.
B: $b=2$ is even. True.
C: $a+b=6$ is even. True.
$\(\mathrm{D}: a-b=2\)$ is even. True.
E: $a=4$ is multiple of 4. True.
$\(\mathrm{F}: b=2\)$ is not multiple of 4 . False.
Example 2:
Let $b=4$ (multiple of 4 ).
Then $\(\frac{a}{4}\)$ is even, so $a=8 k$. Let $k=1$, so $a=8$.
Now, $\(2 a+b=16+4=20\)$, which is even.
Check options:
All options A-E are true, and F is also true here ( $b=4$ is multiple of 4 ).
But since in the first example $F$ was false, $F$ doesn't have to always be true.
Then $\(\frac{a}{2}\)$ is even, so $a=4 k$. Let $k=1$, so $a=4$.
Now, check $\(2 a+b=8+2=10\)$, which is even.
Check options:
A: $a=4$ is even. True.
B: $b=2$ is even. True.
C: $a+b=6$ is even. True.
$\(\mathrm{D}: a-b=2\)$ is even. True.
E: $a=4$ is multiple of 4. True.
$\(\mathrm{F}: b=2\)$ is not multiple of 4 . False.
Example 2:
Let $b=4$ (multiple of 4 ).
Then $\(\frac{a}{4}\)$ is even, so $a=8 k$. Let $k=1$, so $a=8$.
Now, $\(2 a+b=16+4=20\)$, which is even.
Check options:
All options A-E are true, and F is also true here ( $b=4$ is multiple of 4 ).
But since in the first example $F$ was false, $F$ doesn't have to always be true.
gmatclubot
Re: If $2 a+b$ and $\frac{a}{b}$ are even, $a$ and $b$ are integers, then [#permalink]
24 Jun 2025, 04:15
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