2017 AMC 10B Problem 1

Mary did three things in order: multiply by $3$, add $11$, switch the digits. We know only the end result, so undo the steps starting from the last one.

1. Undo the switch. The final number was $71$, $72$, $73$, $74$ or $75$. Switching the digits back, the number before the switch was $17$, $27$, $37$, $47$ or $57$.
2. Undo the "add 11". Subtract $11$ from each: $6$, $16$, $26$, $36$, $46$.
3. Undo the "multiply by 3". Mary's number times $3$ gave one of these, so the right one must be a multiple of $3$. Only $6 = 3 \cdot 2$ and $36 = 3 \cdot 12$ are.
4. The number $2$ is not two-digit, so Mary's number is $12$.
5. Check: $3 \cdot 12 + 11 = 47$, and switching the digits gives $74$, which is between $71$ and $75$.

The answer is $\boxed{\textbf{(B)}\ 12}$.

When a problem tells you a chain of steps and only the final result, run the chain backwards: undo the last step first. Here the "multiple of $3$" test threw out four of the five candidates at once, so no guessing was needed.

Try each answer choice forward.

Only $74$ is between $71$ and $75$, so the answer is $12$.