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 $12$.

Work backwards from the final number. Switching the digits of $3x + 11$ produced $71, 72, 73, 74$ or $75$, so $3x + 11$ itself was $17, 27, 37, 47$ or $57$.

Subtract $11$ from each candidate: $6, 16, 26, 36, 46$. Divide by $3$: only $36$ is a multiple of $3$, giving $x = 12$. Check: $3 \cdot 12 + 11 = 47$, and switching the digits gives $74$, which is in range. (The candidate $6$ would give $x = 2$, not two-digit.)

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