# Difference between revisions of "1986 AJHSME Problems/Problem 8"

## Problem

In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is $$\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}$$ $\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

## Solution

Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.

So, $2 \times \text{B}$ has a units digit of $6$, so $\text{B}$ is either $3$ or $8$. If $\text{B}=3$, then the product is $32\times 73$, which is clearly too small, so $\text{B}=8$ $\boxed{\text{E}}$

## See Also

 1986 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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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