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2011 AMC 10B Problem 5

2011 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

In multiplying two positive integers $a$ and $b$ , Ron reversed the digits of the two-digit number $a$ . His erroneous product was $161$ . What is the correct value of the product of $a$ and $b$ ?

A)
$116$
B)
$161$
C)
$204$
D)
$214$
E)
$224$

Answer

Key insight

161 = 7 x 23 has only one two-digit factor, 23, so the reversed number was 23, the true a is 32, and b = 7.

Solution

Ron's product is (reversed $a$ ) $\times\ b = 161$ . Factor $161$ : it is odd, not divisible by $3$ or $5$ , and $161 = 7 \cdot 23$ . Its factorizations into two positive integers are $1 \cdot 161$ and $7 \cdot 23$ .

The reversed $a$ is a two-digit number, so it must be $23$ (the only two-digit factor), and $b = 7$ . Reversing $23$ back gives $a = 32$ .

The correct product is $32 \cdot 7 = 224$ , so the answer is $\boxed{\textbf{(E)}\ 224}$ .

Why this works

Knowing the erroneous product pins down the reversed number through factorization, because $161$ has so few factors. Whenever a puzzle gives a product of unknown integers, factor it first; the constraint "two-digit" then selects the factor.

The trap

Failing to factor 161 (it is not prime) or reversing the wrong number, such as treating 161 itself as the two-digit number.

Common mistakes

Failing to factor $161$ (it is not prime) or reversing the wrong number, such as treating $161$ itself as the two-digit number.
Answering $161$ (the erroneous product) or computing $23 \cdot 7$ again instead of $32 \cdot 7$ .

Techniques

Set up the equation/formula and compute; no special trick needed