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1987 AJHSME Problems/Problem 8

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Problem

If \text{A} and \text{B} are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is

\begin{tabular}[t]{cccc}9 & 8 & 7 & 6 \\& A & 3 & 2 \\& & B & 1 \\ \hline \end{tabular}


\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ \text{depends on the values o...

Solution

The minimum possible value of this sum is when A=B=1, which is 9876+132+11=10019

The largest possible value of the sum is when A=B=9, making the sum 9876+999+91=10966

Since all the possible sums are between 10019 and 10966, they must have 5 digits.

\boxed{\text{B}}

See Also

1987 AJHSME (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
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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