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

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


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

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 \[9876+932+91=10899\]

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

$\boxed{\text{B}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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