AMC 10 · 2002 · #17

Grade 5 number-theory
prime-numbersplace-valueoptimization extremal-construction ↑ Prerequisites: prime-numbersplace-value
📏 Medium solution 💡 3 insights
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Problem
Split the nine nonzero digits 1,2,…,9 into a set of prime numbers so that each digit is used exactly once — the example {7,83,421,659} shows one way to do it. Find the smallest sum such a set can have.

Pick an answer.

(A)
193
(B)
207
(C)
225
(D)
252
(E)
447
How to solve
Strategy Extreme Principle

"Smallest possible" is exactly the signal for Tool #14 (Extreme Principle), and a minimum question always splits into two halves via Tool #7 (Identify Subproblems): prove a floor that no set can go under, then exhibit one set that sits exactly on the floor. Tool #15 (Organize Information in More Ways) supplies the floor — stop reading the set as a list of numbers and read it as nine digits, each charged according to its place, so the question becomes which digits are forbidden from ending a prime. Tool #6 (Guess and Check) closes the argument by producing an actual set of primes that pays exactly that price; without it, a floor is only a bound, not an answer.

1STEP 1

Write the sum by place value

The total is each digit weighted by its place value, so push every digit as far right as possible.

sum = Σ_d d × (place value of d), place value ∈ {1, 10, 100, …}
2STEP 2

Three digits cannot end a prime

A prime never ends in 4, 6 or 8, and none is prime alone, so those three cost at least the tens place.

last digit ∈ {4,6,8} → number is even and > 2 → not prime
3STEP 3

Add up the floor: 207

Charging the minimum everywhere gives 180 + 27 = 207 as a floor no set can beat.

(40+60+80) + (1+2+3+5+7+9) = 180 + 27 = 207
4STEP 4

Land on the floor with an example

The set {41, 67, 89, 2, 3, 5} is all primes, uses each digit once, and sums to 207, choice (B).

41+67+89+2+3+5 = 207 → (B)
Answer
207
The floor argument bites hard enough to be checkable: choice (A) 193 is below 207, so it is unreachable no matter how the digits are arranged — a genuine impossibility, not a near miss. On the other side, 207 is not a fluke of one lucky arrangement: {43, 61, 89, 2, 5, 7} and {47, 61, 89, 2, 3, 5} both consist of primes, both use each nonzero digit once, and both also sum to 207. The problem's own example {7,83,421,659} totals 1170, far above the floor, exactly as expected once a digit is pushed into the hundreds place.
💡Key takeaway

To find a smallest, first prove a floor that nothing can go under, then build one example that lands exactly on it.

  • Write the sum by place value
  • Three digits cannot end a prime
  • Add up the floor: 207
  • Land on the floor with an example