AMC 8 · 2011 · #24
Easy mode Grade 4Problem
A prime number is a whole number greater than 1 that can only be divided evenly by 1 and itself. For example, 2,3,5,7,11,13,… are primes.
Suppose you want to write 10001 as the sum of two prime numbers, like this: 10001=p+q, where both p and q are primes.
How many ways can this be done?
Pick an answer.
AMC 8 2011 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: Count the number of ways the integer $10001$ can be written as a sum of two prime numbers. Order of the two primes does not matter (so $p + q$ and $q + p$ count as the same way).
Givens: Target sum is $10001$; Both addends must be prime numbers; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: How many unordered prime pairs $(p, q)$ satisfy $p + q = 10001$
Understand
Restated: Count the number of ways the integer $10001$ can be written as a sum of two prime numbers. Order of the two primes does not matter (so $p + q$ and $q + p$ count as the same way).
Givens: Target sum is $10001$; Both addends must be prime numbers; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #5 Look for a Pattern
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.
Execute — Answer: A
2.OA.C.3 Step 1 - Notice the parity of $10001$: it ends in $1$, so it is odd.
- Use the parity pattern for sums — odd $+$ odd $=$ even, even $+$ even $=$ even, and odd $+$ even $=$ odd.
- To get an odd sum, exactly one addend must be even and the other must be odd.
💡 Recognizing odd/even and the parity of sums is a Grade 2 standard, and it does the heavy lifting here.
4.OA.B.4 Step 2 - Identify the only even prime.
- Every even number greater than $2$ has $2$ as a factor besides $1$ and itself, so it is composite.
- That leaves $2$ as the only even prime.
- So the even prime in our pair must be $2$.
💡 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.NBT.B.4 Step 3 - Pin down the other addend.
- If one prime is $2$, the other must be $10001 - 2 = 9999$.
- So the only possible prime pair is $(2,\, 9999)$.
- Either $9999$ is prime — and we have $1$ way — or it is composite — and we have $0$ ways.
💡 A single multi-digit subtraction at the Grade 4 fluency level locks in the only candidate.
4.OA.B.4 Step 4 - Test $9999$ for primality with the divisibility-by-$3$ rule (a number is divisible by $3$ iff the sum of its digits is divisible by $3$).
- The digit sum of $9999$ is $9+9+9+9 = 36$, and $36 = 3 \times 12$.
- So $3$ divides $9999$, and since $9999 \neq 3$, it is 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.4 Step 5 - Combine the findings.
- The only candidate pair $(2,\, 9999)$ fails because $9999$ is not prime.
- No other pair survives the parity filter.
- Therefore $10001$ cannot be written as a sum of two primes in any way.
💡 The final count is the number of surviving candidates after every elimination — a direct application of the Grade 4 prime/composite reasoning.
2.OA.C.3 Notice the parity of $10001$: it ends in $1$, so it is odd. Use the parity patte 4.OA.B.4 Identify the only even prime. Every even number greater than $2$ has $2$ as a fa 4.NBT.B.4 Pin down the other addend. If one prime is $2$, the other must be $10001 - 2 = 9 4.OA.B.4 Test $9999$ for primality with the divisibility-by-$3$ rule (a number is divisib 4.OA.B.4 Combine the findings. The only candidate pair $(2,\, 9999)$ fails because $9999$ Review
Reasonableness: Sanity-check the parity argument on a known case: $10 = 3 + 7 = 5 + 5$ (even target $\rightarrow$ odd $+$ odd works fine), but $9 = 2 + 7$ is the only split (odd target forces one prime to be $2$). The pattern matches our argument. For our problem, $10001 - 2 = 9999$, and the digit-sum test correctly flags $9999$ as a multiple of $3$. Answer $0$ is consistent with both the parity logic and the divisibility check.
Alternative: Tool #2 (Systematic List): start listing prime pairs $(p, 10001 - p)$ with the smaller prime $p$ first. $p = 2 \Rightarrow 9999 = 3 \times 3333$ (composite). $p = 3 \Rightarrow 9998 = 2 \times 4999$ (composite, even). $p = 5 \Rightarrow 9996$ (even, composite). Every odd $p \geq 3$ gives an even $10001 - p > 2$, which is automatically composite. So no $p$ works, confirming $0$ ways. This is slower but reaches the same conclusion.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Establishing that $10001$ is odd and that an odd sum requires exactly one odd and one even addend.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Concluding that $2$ is the only even prime and that $9999 = 3 \times 3333$ is composite, which closes out the candidate pair $(2, 9999)$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $10001 - 2 = 9999$ to pin down the only candidate partner for $2$.)
⭐ This AMC 8 problem only needs Grade 4 prime-and-composite reasoning you already know!
⭐ This AMC 8 problem only needs Grade 4 prime-and-composite reasoning you already know!
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