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2003 AMC 10A Problem 14

2003 AMC 10A Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Primes · difficulty 3 of 5 · about 3 min · position P11-15

Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ , $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$ ?

A)
$12$
B)
$15$
C)
$18$
D)
$21$
E)
$24$

Answer

Key insight

Only e = 3 or 7 lets 10d + e be prime; the four survivors 23, 37, 53, 73 make 7 * 3 * 73 = 1533 the largest product.

Solution

The single-digit primes are $2, 3, 5, 7$ , so $d$ and $e$ come from this set and are distinct.

For the two-digit number $10d + e$ to be prime, its units digit $e$ cannot be $2$ or $5$ , so $e \in \{3, 7\}$ . Check the candidates:

$e = 3$ : $23$ (prime), $53$ (prime), $73$ (prime).
$e = 7$ : $27$ (no), $37$ (prime), $57 = 3 \cdot 19$ (no).

The possible products $d \cdot e \cdot (10d + e)$ are
$2 \cdot 3 \cdot 23 = 138,\quad 5 \cdot 3 \cdot 53 = 795,\quad 7 \cdot 3 \cdot 73 = 1533,\quad 3 \cdot 7 \cdot 37 = 777.$
The largest is $n = 1533$ , whose digit sum is $1 + 5 + 3 + 3 = 12$ .

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

Why this works

The search space is tiny once you use the structure: only four single-digit primes exist, and a two-digit prime's last digit is restricted. A short organized list beats guessing that "the biggest $d$ wins," which happens to be true here but needs the check that $73$ is prime while $75$ and $77$ are not.

The trap

Allowing e = 5 or e = 2 (75 or 72 are not prime) or picking d = e = 7, which violates 'distinct' and gives the non-prime 77.

Common mistakes

Allowing e = 5 or e = 2 (75 or 72 are not prime) or picking d = e = 7, which violates 'distinct' and gives the non-prime 77.
Reporting the digit sum of the primes ( $7 + 3 + 7 + 3$ ) or of $73$ rather than of the product $n$ .

Techniques

Bound the quantity above/below or estimate to pin it down · Organized listing / direct enumeration