2003 AMC 10A Problem 14

P11-15~3 minPrint

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

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 $n$ ?

Show solution without answering

Key insight

d and e must be among 2, 3, 5, 7, and a two-digit prime cannot end in 2 or 5. That leaves only a few cases to list and compare.

Solution

There are very few single-digit primes, so list every possibility and pick the biggest.

The single-digit primes are $2, 3, 5, 7$ . So $d$ and $e$ are two different numbers from this list.
The number $10d + e$ is the two-digit number with tens digit $d$ and units digit $e$ . For example, $d = 7, e = 3$ gives $73$ .
A two-digit number ending in $2$ is even, and one ending in $5$ is a multiple of $5$ , so neither is prime. So $e$ must be $3$ or $7$ .
Check every case:

$e$	two-digit number	prime?
$3$	$23$	yes
$3$	$53$	yes
$3$	$73$	yes
$7$	$27 = 3 \times 9$	no
$7$	$37$	yes
$7$	$57 = 3 \times 19$	no

5. Work out the product $d \times e \times (10d + e)$ for the four that work:
$2 \times 3 \times 23 = 138, \quad 5 \times 3 \times 53 = 795, \quad 3 \times 7 \times 37 = 777, \quad 7 \times 3 \times 73 = 1533.$
6. The largest is $n = 1533$ . Its digits add to $1 + 5 + 3 + 3 = 12$ .

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

Why this works

When the choices are limited to a handful of small primes, a neat table of every case is fast and safe. The rule "a prime bigger than $5$ cannot end in $2$ , $4$ , $5$ , $6$ , $8$ or $0$ " cuts the list down before you start testing. Guessing "use the biggest $d$ " happens to work here, but only the table shows it for sure.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 2: Primes and Composites
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.2, §4.4)