AMC 10 · 2002 · #15
Grade 4 number-theoryThe positive integers A,B,A−B, and A+B are all prime numbers. The sum of these four primes is
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: A and B are prime numbers; A - B is a prime number (so A > B); A + B is a prime number
Unknowns: A shared property of the sum A + B + (A - B) + (A + B) = 3A + B that must hold for every valid A and B
Understand
Restated: 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.
Givens: A and B are prime numbers; A - B is a prime number (so A > B); A + B is a prime number
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #4 Introduce a Variable, #14 Extreme Principle
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.
Execute — Answer: E
2.OA.C.3 Step 1 Name the four primes
- Write the four primes plainly: A, B, A - B, and A + B.
- Since A - B must be a positive prime, A is bigger than B.
- Notice that A - B and A + B differ from each other by 2B, and they add up to 2A.
💡 Getting the four objects and their relationships on paper is what makes the hidden even/odd clue visible.
2.OA.C.3 Step 2 Both odd is impossible
- Suppose A and B were both odd.
- Then A - B and A + B would both be even.
- But the only even prime is 2, and A - B and A + B cannot both be 2 because they differ by 2B, which is at least 4.
- So A and B are not both odd.
- One of them must be the even prime 2.
💡 Odd minus odd is even, and 2 is the lone even prime, so two big even primes simply cannot both exist here.
4.OA.B.4 Step 3 Decide which one is 2
- One of A, B equals 2.
- It cannot be A: if A = 2 then A - B = 2 - B is zero or negative for any prime B, so it is not a positive prime.
- Therefore B = 2.
💡 Only the smaller number can be 2, because subtracting from 2 would drop below the primes.
4.OA.B.4 Step 4 Use divisibility by 3
- With B = 2, the primes A - 2, A, and A + 2 are three numbers spaced 2 apart.
- Looking at remainders when divided by 3, these three cover all of remainder 0, 1, and 2.
- So exactly one of them is a multiple of 3.
- A prime that is a multiple of 3 must be 3 itself, and the smallest of the three is A - 2, giving A - 2 = 3, so A = 5.
💡 Three numbers two apart always hit every remainder mod 3, so one is forced to be the prime 3.
4.OA.B.4 Step 5 Add them and pick the property
- So A = 5, B = 2, A - B = 3, A + B = 7, all prime.
- Their sum is 5 + 2 + 3 + 7 = 17.
- Now test the choices: 17 is odd (rules out A), not divisible by 3, 5, or 7 (rules out B, C, D).
- But 17 itself is prime.
- So the sum is prime, and the answer is (E).
💡 Once the only surviving primes are 2, 3, 5, 7, their total 17 is itself prime.
2.OA.C.3 Write the four primes plainly: A, B, A - B, and A + B. Since A - B must be a pos 2.OA.C.3 Suppose A and B were both odd. Then A - B and A + B would both be even. But the 4.OA.B.4 One of A, B equals 2. It cannot be A: if A = 2 then A - B = 2 - B is zero or neg 4.OA.B.4 With B = 2, the primes A - 2, A, and A + 2 are three numbers spaced 2 apart. Loo 4.OA.B.4 So A = 5, B = 2, A - B = 3, A + B = 7, all prime. Their sum is 5 + 2 + 3 + 7 = 1 Review
Reasonableness: 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.
Alternative: Instead of the general argument, just search small primes: the only way A - B, A, A + B, B are all prime forces B = 2 and a prime pair A - 2, A + 2 both prime with A prime, and A = 5 pops out almost immediately. Summing 2 + 3 + 5 + 7 = 17 again shows the total is prime.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Reasoning that odd minus odd is even and that only 2 is an even prime, forcing one value to be 2)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using that a prime multiple of 3 must equal 3, and checking that 17 is prime)
⭐ 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.
⭐ 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.
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