AMC 10 · 2002 · #15

Grade 4 number-theory
prime-numbersplace-value systematic-enumeration ↑ Prerequisites: prime-numbers
📏 Medium solution 💡 2 insights
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Problem
Use each of the digits 1, 2, 3, 4, 5, 6, 7, 9 exactly once to build four two-digit prime numbers, then find the sum of those four primes.

Pick an answer.

(A)
150
(B)
160
(C)
170
(D)
180
(E)
190

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The tempting plan is to hunt for four actual primes, but that is slow and unnecessary. Tool #16 (Change Focus) says look at what the question really asks: only the sum, and the sum of four two-digit numbers is 10×(sum of tens digits) + (sum of units digits). So the sum is fixed the moment we know which four digits land in the units place. To find those, Tool #3 (Eliminate Possibilities) rules out any digit that cannot end a prime: a two-digit prime is never even and never ends in 5. That forces the split of the eight digits into a tens group and a units group. Tool #7 (Identify Subproblems) then finishes with two easy sub-sums.

1STEP 1

Which digits can end a prime

A two-digit prime is never even and never ends in 5, so its units digit must be one of 1, 3, 7, 9.

units digit ∈ {1,3,7,9}, ruled out: 2,4,6 (even), 5
2STEP 2

Force the tens and units groups

Only four digits can end a prime and there are exactly four units slots, so the leftovers 2, 4, 5, 6 are forced into the tens place.

tens digits = {2,4,5,6}, units digits = {1,3,7,9}
3STEP 3

Add the two place-value groups

Sum = 10 × (2+4+5+6) + (1+3+7+9) = 170 + 20 = 190, which is choice (E).

10 (2+4+5+6) + (1+3+7+9) = 10 · 17 + 20 = 170 + 20 = 190 → (E)
Answer
190
The argument never built a single prime, so check that a valid set actually exists: 23, 41, 59, 67 are all prime and use the digits 2,3,4,1,5,9,6,7 — each of the eight digits once. Their sum is 23+41+59+67 = 190, matching (E). A size check also fits: four two-digit numbers between 10 and 99 average about 47.5, so a sum near 4 × 47.5 = 190 is exactly the right ballpark. The trap answers 150–180 come from mis-sorting a digit (for example letting a 5 or a 6 slip into a units place), which the primality rule forbids.
💡Key takeaway

You never have to find the primes: even digits and 5 can't end a prime, so 1,3,7,9 are the units and 2,4,5,6 are the tens, and the sum is 10 × 17 + 20 = 190.

  • Which digits can end a prime
  • Force the tens and units groups
  • Add the two place-value groups