Competition · AMC preparation · step 4 of 4
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.
Check whether 10001 is odd
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 PatternName the only even prime
Every 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 PossibilitiesSubtract to get the other addend
If 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.
If 10001 is written as a sum of two primes, those two primes have to be 2 and 9999.
▸ Why?
10001 is odd, and a sum comes out odd only when one addend is even and the other is odd, so one of the two primes must be the even one.
▸ Why?
Among the primes, 2 is the only even one, so the even prime in the pair can only be 2.
▸ Why?
Any even number past 2 equals 2 times some whole number bigger than 1, so it is that many equal groups of 2 and has a factor other than 1 and itself, which makes it composite rather than prime.
▸ Why?
With one prime fixed as 2, the partner is completely determined: it must be 10001 - 2 = 9999.
▸ Why?
The two primes add to 10001, and subtraction undoes addition, so the unknown partner equals 10001 take away 2.
Test 9999 with the digit-sum rule
The 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 PossibilitiesCount the valid ways
The 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!
- Check whether 10001 is odd
- Name the only even prime
- Subtract to get the other addend
- Test 9999 with the digit-sum rule
- Count the valid ways
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