AMC 10 · 2003 · #14

Grade 4 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-numbersdigit-decompositionoptimization extremal-constructionsystematic-enumeration ↑ Prerequisites: prime-numbers
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Problem
Choose single-digit numbers d and e so that d, e, and the two-digit number 10d+e are all prime and all different. Among every such choice, take the one that makes the product n = d · e · (10d+e) as large as possible, then add up the digits of that n.

Pick an answer.

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

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The problem asks for the largest product, so Tool #14 (Extreme Principle) drives the search: push the biggest factor to its limit. The product n = d · e · (10d+e) is dominated by the two-digit factor 10d+e, and that factor grows fastest by making the tens digit d as large as possible. Because d and e can only be single-digit primes, Tool #3 (Eliminate Possibilities) shrinks the search to the four numbers 2,3,5,7, and Tool #2 (Make a Systematic List) checks each candidate for the second digit until 10d+e comes out prime. That turns an open-ended "largest" question into checking just a handful of cases.

1STEP 1

List the single-digit primes

Only 2, 3, 5, 7 are single-digit primes, so d and e both come from {2,3,5,7} and must differ.

d, e ∈ {2,3,5,7}, d ≠ e
2STEP 2

Make the tens digit as large as possible

n is ruled by its biggest factor 10d+e, which grows fastest when the tens digit is largest, so try d = 7 first.

d = 7 → 10d+e = 70+e
3STEP 3

Find e so that 70+e is prime

With d = 7 the units digit e is 2, 3, or 5: 72 and 75 are composite, but 70+3 = 73 is prime, so e = 3.

72 = 8 · 9, 75 = 3 · 25, 73 is prime → e = 3
4STEP 4

Multiply to get n, then add its digits

No smaller d can beat it, so n = 7 · 3 · 73 = 1533, and its digits give 1+5+3+3 = 12, choice (A).

n = 7 · 3 · 73 = 1533, 1+5+3+3 = 12 → (A)
Answer
12
The choice d=7, e=3 uses the largest single-digit prime as the tens digit, so its two-digit factor 73 is the largest achievable of the required form; any other start (d=5 gives at most 53, d=3 gives at most 37, d=2 gives at most 23) produces a smaller top factor and hence a smaller product. So n=1533 is genuinely the maximum. Its digit sum 12 is one of the listed choices, matching (A), and the smaller-d products would give different digit sums, so the answer is not ambiguous.
💡Key takeaway

To make the product biggest, make the two-digit prime biggest: start the tens digit at the largest single-digit prime 7, find the one units digit that keeps it prime (73), and 7 · 3 · 73 = 1533 has digit sum 12.

  • List the single-digit primes
  • Make the tens digit as large as possible
  • Find e so that 70+e is prime
  • Multiply to get n, then add its digits

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