2013 AMC 8 Problems/Problem 13

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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(educated guess)

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(brute force elimination)

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 $\textbf{A}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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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