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