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1990 AJHSME Problems/Problem 1

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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?

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);...

\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}}

See Also

1990 AJHSME (ProblemsResources)
Preceded by
First
Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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