Difference between revisions of "1998 AHSME Problems/Problem 3"

Revision as of 13:28, 5 July 2013

Problem 3

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

$\begin{tabular}{r}&\ \texttt{7 a 2}\\ &- \texttt{4 8 b} \\

\hline

&\ \texttt{c 7 3} \end{tabular}$ (Error compiling LaTeX. Unknown error_msg)

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
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 30: / Line 30: @@
 ==See Also==
 {{AHSME box|year=1998|num-b=2|num-a=4}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "1998 AHSME Problems/Problem 3"

Revision as of 13:28, 5 July 2013

Problem 3

Solution

See Also