1967 AHSME Problems/Problem 1

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Problem

The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$. If $5b9$ is divisible by 9, then $a+b$ equals

$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution

If $5b9$ is divisible by $9$, this must mean that $5 + b + 9$ is a multiple of $9$. So, \[5 + b + 9 = 9, 18, 27, 36...\]

Because $5 + 9 = 14$ and $b$ is in between 0 and 9,

\[5 + b + 9 = 18\] \[b = 4\]

The question states that \[2a3 + 326 = 549\] so \[2a3 = 549 - 326\] \[2a3 = 223\] \[a = 2\]

\[a + b = 6\] which is answer choice $\boxed{C}$.

See Also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AHSME Problems and Solutions

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