1998 AHSME Problems/Problem 3

Problem 3

If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which

$\begin{tabular}{rr}&\ \texttt{7 a 2}\\ -& \texttt{4 8 b} \\ \hline &\ \texttt{c 7 3} \end{tabular}$

then $\texttt{a+b+c =}$

$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$

Solution

Working from right to left, we see that $2 - b = 3$ . Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$ . Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$ .

Since $1$ was borrowed from $a$ , we have from the tens column $(a-1) - 8 = 7$ . Again for single digit integers this will not work. Again, borrow $1$ from $7$ , giving $10 + (a-1) - 8 = 7$ . Solving for $a$ :

$10 + a - 1 - 8 = 7$

$1 + a = 7$

$a = 6$

Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$ , giving $c = 2$ .

As a check, the problem is $762 - 489 = 273$ , which is a true sentence.

The desired quantity is $a + b + c = 6 + 9 + 2 = 17$ , and the answer is $\boxed{D}$ .

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1998 AHSME Problems/Problem 3

Problem 3

Solution

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