AMC 8 · 2011 · #24
Grade 4 number-theoryPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Listing every prime under 10001 is hopeless, so we use the parity pattern (Tool #5) to slash the candidates. Odd + odd = even and even + even = even, but odd + odd-or-even rules out most cases — to land on an odd sum like 10001, exactly one of the two primes must be even. The only even prime is 2, which uses Tool #3 (Eliminate Possibilities) to crush infinitely many cases down to a single candidate: the pair must be (2, 9999). Then one quick divisibility check on 9999 decides whether even that lone candidate works.
10001 ends in 1, so it is odd; an odd sum needs exactly one even addend and one odd addend.
Recognizing odd/even and the parity of sums is a Grade 2 standard, and it does the heavy lifting here.
2.OA.C.3Look For A PatternEvery even number above 2 has 2 as a factor, so the only even prime is 2 — that must be the even addend.
Grade 4 students learn to test small numbers for prime/composite by checking factors, which is exactly what eliminates every even number above 2.
4.OA.B.4Eliminate PossibilitiesIf one prime is 2, the other is 10001 - 2 = 9999, so the only candidate pair is (2, 9999).
A single multi-digit subtraction at the Grade 4 fluency level locks in the only candidate.
4.NBT.B.4Eliminate PossibilitiesThe digit sum of 9999 is 9+9+9+9 = 36, a multiple of 3, so 9999 is divisible by 3 — composite, not prime.
Recognizing multiples of 3 (and using factor pairs to declare a number composite) lives in the Grade 4 prime/composite standard.
4.OA.B.4Eliminate PossibilitiesThe lone candidate (2, 9999) fails since 9999 is composite, so 10001 is a sum of two primes in 0 ways.
The final count is the number of surviving candidates after every elimination — a direct application of the Grade 4 prime/composite reasoning.
4.OA.B.4Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 prime-and-composite reasoning you already know!