2003 AMC 10A Problem 14

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

1. The single-digit primes are $2, 3, 5, 7$. So $d$ and $e$ are two different numbers from this list.
2. The number $10d + e$ is the two-digit number with tens digit $d$ and units digit $e$. For example, $d = 7, e = 3$ gives $73$.
3. 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 $7$.
4. Check every case:

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

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.

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