# 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

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

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

## 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.

$\boxed{\text{B}}$