AMC 8 · 2014 · #4
Grade 4 number-theoryPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sum 85 is odd, and odd + odd = even while even + even = even. So one of the two primes must be even and the other odd — that is a powerful elimination (Tool #3). The only even prime is 2, so one prime is forced to be 2 and the other is 85 - 2 = 83. Tool #2 (Systematic List) is the backup: we can also list small primes 2, 3, 5, 7, 11, …, pair each with 85 - p, and check primality — both routes land on the same pair.
The sum 85 is odd, but odd+odd and even+even are both even — so one prime is even and the other odd.
Knowing odd + even = odd is a Grade 2 fact, and we use it to rule out every pair of two odd primes in one shot.
2.OA.C.3Eliminate PossibilitiesThe only even prime is 2 — every even number past 2 is divisible by 2 — so one prime must be 2.
Identifying 2 as the lone even prime is a Grade 4 prime/composite fact and finishes the elimination.
4.OA.B.4Eliminate PossibilitiesSubtract to get the other prime: 85 - 2 = 83.
A single subtraction pins down the second number; the only remaining job is to check it really is prime.
4.OA.B.4Eliminate PossibilitiesCheck 83 is prime: √83 ＜ 10, so test 2, 3, 5, 7 — none divide it, so 83 is prime.
Listing the primes up to √(83) in order is a tiny Tool #2 list — only four numbers to test.
4.OA.B.4Make A Systematic ListMultiply the two primes: 2 × 83 = 166, choice (E).
2 × 83 is a Grade 3 multiplication; doubling 83 gives 166 and matches choice (E).
3.OA.C.7Eliminate PossibilitiesWhenever two primes add to an odd number, one of them has to be 2 — that single Grade 4 parity trick cracks the whole problem.