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2021 AMC 10B Problem 5

2021 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 1 of 5 · about 2 min · position P1-10

The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?

A)
$21$
B)
$22$
C)
$23$
D)
$24$
E)
$25$

Answer

Key insight

30 has only one single-digit factor pair, 5 × 6, so 6 is taken and 24 must be 3 × 8.

Solution

Write each product as a pair of distinct single-digit factors.

$30 = 1 \cdot 30 = 2 \cdot 15 = 3 \cdot 10 = 5 \cdot 6$ ; only $5 \cdot 6$ uses single digits. So two cousins are $5$ and $6$ .

$24 = 3 \cdot 8 = 4 \cdot 6$ (the pairs $1 \cdot 24$ and $2 \cdot 12$ are out). Since $6$ is already used and ages are distinct, the other two cousins are $3$ and $8$ .

Sum: $3 + 8 + 5 + 6 = 22$ .

The answer is $\boxed{\textbf{(B)}\ 22}$ .

Why this works

Start with the more constrained product. The digit limit leaves $30$ exactly one factorization, and distinctness then eliminates one of the two factorizations of $24$ . Small enumeration plus one constraint is faster than any algebra here.

The trap

Using 4 × 6 for 24 together with 5 × 6 for 30, forgetting that the four ages must be distinct (which gives 21).

Common mistakes

Using 4 × 6 for 24 together with 5 × 6 for 30, forgetting that the four ages must be distinct (which gives 21).
Overlooking $3 \cdot 8$ as a factorization of $24$ and concluding there is no valid set.

Techniques

Organized listing / direct enumeration