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2017 AMC 10B Problem 1

2017 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 1 of 5 · about 2 min · position P1-10

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?

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

Answer

Key insight

Un-switch the digits: the result 3x+11 must be one of 17, 27, 37, 47, 57, and only 47 = 3(12)+11 comes from a two-digit x.

Solution

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 $\boxed{\textbf{(B)}\ 12}$ .

Why this works

When a problem describes a chain of operations ending in a known range, reverse each operation in turn: the digit switch, then the $+11$ , then the $\times 3$ . Divisibility by $3$ filters the candidates immediately, so no trial and error over Mary's number is needed.

Alternative approach

Test the five choices directly: $3x + 11$ gives $44, 47, 50, 53, 56$ , whose digit reversals are $44, 74, 05, 35, 65$ . Only $74$ lands in $71$ to $75$ .

The trap

Solving 3x + 11 between 71 and 75 without reversing the digits first, which gives no integer and leads to guessing.

Common mistakes

Solving 3x + 11 between 71 and 75 without reversing the digits first, which gives no integer and leads to guessing.
Reversing the digits of the answer choices themselves (getting $21, 31, \dots$ ) instead of the digits of $3x + 11$ .

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Start from the end state / desired conclusion and reverse