AMC 10 · 2017 · #1

Grade 3 arithmetic
place-valuemulti-digit-arithmeticdivisibility-rules work-backwardsguess-and-check ↑ Prerequisites: place-value
📏 Short solution 💡 2 insights
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Problem
Mary picked a positive two-digit number, multiplied it by 3, and added 11. She then switched the two digits of that result and got a number from 71 to 75, inclusive. Find Mary's original number.

Pick an answer.

(A)
11
(B)
12
(C)
13
(D)
14
(E)
15

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Work Backwards

We are told the END of the chain (the switched number lands in 71–75) and asked for the START (Mary's number), so Tool #11 (Work Backwards) fits exactly: un-switch the digits, undo the +11, then undo the × 3. Tool #6 (Guess and Check) closes the problem by running the chain forward on the survivor to confirm it lands in the target range.

1STEP 1

Switching digits undoes itself, so flip 71–75 back — those flips are the possible values of 3 × (number) + 11.

71{→}17, 72{→}27, 73{→}37, 74{→}47, 75{→}57
2STEP 2

Undo each candidate — subtract 11, then divide by 3 — and only 47 gives a whole two-digit start: 47 − 11 = 36, 36 ÷ 3 = 12.

47 - 11 = 36, 36 ÷ 3 = 12
3STEP 3

Run 12 forward — 12 × 3 = 36, + 11 = 47, switch to 74 — 74 lands in 71–75, confirming (B).

12 × 3 = 36, 36 + 11 = 47, 47 → 74 → (B)
Answer
12
The forward check is airtight: 12 → 36 → 47 → 74, and 71 ≤ 74 ≤ 75. Every other choice fails the divisibility-by-3 test in the backward step, so 12 is the unique answer, matching (B).
💡Key takeaway

When a problem hands you the ending and asks for the start, undo each step in reverse order — subtract before you divide.