AMC 8 · 2011 · #24
Easy mode Grade 4Problem
A prime number is a whole number greater than that can only be divided evenly by and itself. For example, are primes.
Suppose you want to write as the sum of two prime numbers, like this: , where both and are primes.
How many ways can this be done?
Pick an answer.
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!