1996 AHSME Problems/Problem 1

Problem

The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?

$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5 6}\end{tabular}$

$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$

Solution

Doing the addition as is, we get $641 + 852 + 973 = 2466$ . This number is $10$ larger than the desired sum of $2456$ . Therefore, we must make one of the three numbers $10$ smaller.

We may either change $641 \rightarrow 631$ , $852 \rightarrow 842$ , or $973 \rightarrow 963$ . Either change results in a valid sum. The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\boxed{D}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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.

Page

Toolbox

Search

1996 AHSME Problems/Problem 1

Problem

Solution

See also