Problem of the Day


If a and b are positive integers, is 4a+3b divisible by 12?

(1) b! is a multiple of 3

(2) 2a+5b=26

Reveal Answer

Answer

C. Both statements together are sufficient, but neither statement alone is sufficient.

See the Solution

Solution

[latexpage]

If $a$ and $b$ are positive integers, is $4a+3b$ divisible by 12?

(1) $b!$ is a multiple of 3.  This implies only that $b \geq 3$, and tells us nothing about the value of $a$. INSUFFICIENT.

(2) $2a+5b=26$.  The only possible solutions of this equation where $a$ and $b$ are positive integers are $a=8, b=2$ or $a=3, b=4$. In the former case, $4a+3b=38$, which is not divisible by 12. In the latter case, $4a+3b=24$, which is divisible by 12. INSUFFICIENT.

Statements 1&2: Combining the two statements, we can eliminate $a=8, b=2$ as a possibility. Thus, it must be true that $a=3,b=4$, which means $4a+3b$ is divisible by 12. SUFFICIENT.


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