1990 AJHSME Problems/Problem 1

Problem

What is the smallest sum of two $3$ -digit numbers that can be obtained by placing each of the six digits $4,5,6,7,8,9$ in one of the six boxes in this addition problem?

$[asy] unitsize(12); draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle); draw((4,1)--(6,1)--(6,3)--(4,3)--cycle); draw((4,4)--(6,4)--(6,6)--(4,6)--cycle); draw((7,1)--(9,1)--(9,3)--(7,3)--cycle); draw((7,4)--(9,4)--(9,6)--(7,6)--cycle); [/asy]$

$\text{(A)}\ 947 \qquad \text{(B)}\ 1037 \qquad \text{(C)}\ 1047 \qquad \text{(D)}\ 1056 \qquad \text{(E)}\ 1245$

Solution

Let the two three-digit numbers be $\overline{abc}$ and $\overline{def}$ . Their sum is equal to $100(a+d)+10(b+e)+(c+f)$ .

To minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ . Similarly, we let $b=6$ , $e=7$ , and then $c=8$ and $f=9$ . The sum is $100(9)+10(13)+(17)=1047 \rightarrow \boxed{\text{C}}$

1990 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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1990 AJHSME Problems/Problem 1

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