1967 AHSME Problems/Problem 1

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

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

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,

$5 + b + 9 = 18$ and $b = 4$

$2a3 + 326 = 549$ , so $2a3 = 549 - 326$ $a = 2$

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

1967 AHSC (Problems • Answer Key • Resources)
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