1987 AJHSME Problems/Problem 8

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

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

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

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 $9876+932+91=10899$

Since all the possible sums are between $10019$ and $10899$ , they must have $5$ digits.

$\boxed{\text{B}}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions