AMC 10 · 2002 · #11

Grade 4 number-theory
prime-numbersparitydivisibility-rules casework ↑ Prerequisites: prime-numbersparity
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
Four positive integers are all prime at once: A, B, the difference A minus B, and the sum A plus B. We must decide what is always true about the total of these four primes.

Pick an answer.

(A)
$\ \mathrm{even}$
(B)
$\ \mathrm{divisible\ by\ }3$
(C)
$\ \mathrm{divisible\ by\ }5$
(D)
$\ \mathrm{divisible\ by\ }7$
(E)
$\ \mathrm{prime}$
How to solve
Strategy Eliminate Possibilities

Parity (odd/even) and the single even prime pin the numbers down so tightly that only one set of values survives. So the smart move is to rule out cases with even/odd reasoning and divisibility, until just one answer for A and B remains, then read off the sum.

1STEP 1

Name the four primes

Write the four plainly; since the difference must be a positive prime, A exceeds B.

A, B, A-B, A+B all prime, (A-B)+(A+B)=2A
2STEP 2

Both odd is impossible

If both were odd the sum and difference would both be even, impossible, so one is 2.

odd-odd=even, even prime=2 only
3STEP 3

Decide which one is 2

A cannot be 2 or the difference would not be positive, so B = 2.

A=2 → A-B=2-B ≤ 0 (not prime) → B=2
4STEP 4

Use divisibility by 3

Then A-2, A, A+2 cover all remainders mod 3, forcing the smallest to be 3 and A = 5.

A-2, A, A+2 (mod 3) = {0,1,2} → 3 ∣ (A-2) → A-2=3 → A=5
5STEP 5

Add them and pick the property

The four primes total 17, which is itself prime, choice (E).

5+2+3+7=17, 17 is prime
Answer
prime
Check the found values directly: 5, 2, 3, 7 are each prime, 5 - 2 = 3 is prime, and 5 + 2 = 7 is prime, so every requirement holds. The sum 17 is odd (so not choice A), leaves remainder 2 mod 3, remainder 2 mod 5, remainder 3 mod 7 (so not B, C, D), and 17 is prime, matching choice E.
💡Key takeaway

Since 2 is the only even prime, B has to be 2, and then divisibility by 3 forces A = 5, giving the primes 2, 3, 5, 7 whose sum 17 is itself prime.

  • Name the four primes
  • Both odd is impossible
  • Decide which one is 2
  • Use divisibility by 3
  • Add them and pick the property