2013 AMC 8 Problems/Problem 13

Problem

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 47 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 49$

Video Solution

https://www.youtube.com/watch?v=9FkjSCcdTqY ~David

https://youtu.be/KBM2YN4kKGA ~savannahsolver

Solution

Let the two digits be $a$ and $b$ .

The correct score was $10a+b$ . Clara misinterpreted it as $10b+a$ . The difference between the two is $|9a-9b|$ which factors into $|9(a-b)|$ . Therefore, since the difference is a multiple of 9, the only answer choice that is a multiple of 9 is $\boxed{\textbf{(A)}\ 45}$ .

Solution 2

We can simply test out numbers to see which one works. We can see that Clara’s score can’t be a multiple of ten because the reverse of the score is a one digit number, to small for the answer choices. After testing multiples, the answer should be $\textbf{A}$ .

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AJHSME/AMC 8 Problems and Solutions

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

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2013 AMC 8 Problems/Problem 13

Contents

Problem

Video Solution

Solution

Solution 2

See Also