AMC 10 · 2002 · #15

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

prime-numbersparitydivisibility-rules casework ↑ Prerequisites: prime-numbersparity

📏 Medium solution 💡 2 insights

Problem

The four numbers A, B, the difference A minus B, and the sum A plus B are all positive integers, and they are all prime numbers at the same time. Which property must the sum of these four primes always have?

Pick an answer.

(A)

$\ \mathrm{even}$

(B)

$\ \mathrm{divisible\ by\ }3$

(C)

$\ \mathrm{divisible\ by\ }5$

(D)

$\ \mathrm{divisible\ by\ }7$

(E)

$\ \mathrm{prime}$

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

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 primes down: A, B, A - B, A + B. Since A - B is positive, A is bigger than B, and A - B and A + B differ by 2B.

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

2STEP 2

Both odd is impossible

If A and B were both odd, A - B and A + B would both be even — but 2 is the only even prime, so one of A, B is 2.

odd-odd=even, even prime=2 only

3STEP 3

Decide which one is 2

It cannot be A: if A = 2 then A - B is zero or negative for every prime B, never a positive prime. So B = 2.

A=2 → A-B=2-B ≤ 0 (not prime) → B=2

4STEP 4

Use divisibility by 3

A - 2, A, A + 2 sit 2 apart, so they cover every remainder mod 3; the multiple of 3 must be the prime 3 itself, so 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 are 5, 2, 3, 7, totalling 17: odd and not divisible by 3, 5 or 7, so (A) to (D) die and (E) is left.

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

Name the four primes

Both odd is impossible

Decide which one is 2

Use divisibility by 3

Add them and pick the property

More like this