AMC 10 · 2024 · #3
Grade 4 number-theoryProblem
What is the sum of the digits of the smallest prime that can be written as a sum of 5 distinct primes?
Pick an answer.
AMC 10 2024 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: Find the smallest prime number that can be written as a sum of $5$ distinct primes, then add the digits of that prime.
Givens: We need a prime $P$ that equals $p_1 + p_2 + p_3 + p_4 + p_5$ where all five $p_i$ are distinct primes; Among all such $P$, we want the smallest; Answer choices: (A) $5$, (B) $7$, (C) $8$, (D) $10$, (E) $13$
Unknowns: The digit sum of the smallest qualifying prime $P$
Understand
Restated: Find the smallest prime number that can be written as a sum of $5$ distinct primes, then add the digits of that prime.
Givens: We need a prime $P$ that equals $p_1 + p_2 + p_3 + p_4 + p_5$ where all five $p_i$ are distinct primes; Among all such $P$, we want the smallest; Answer choices: (A) $5$, (B) $7$, (C) $8$, (D) $10$, (E) $13$
Plan
Primary tool: #6 Guess and Check
Secondary: #3 Eliminate Possibilities
To make $P$ as small as possible, we want the five primes themselves to be as small as possible — that points straight at Tool #6 (Guess and Check). We try the smallest candidate set first, check whether the sum is prime, and if not, swap in the next prime up. Before guessing, Tool #3 (Eliminate Possibilities) rules out using $2$: if $2$ is one of the five, the other four are odd, the sum is even, and the only even prime is $2$ itself — way too small. So all five must be odd primes, which narrows the search to a tiny list.
Execute — Answer: B
2.OA.C.3 Step 1 - Eliminate the prime $2$ from the candidate set.
- If $2$ were one of the five primes, the other four would be distinct odd primes, so the sum would be $2 + (\text{odd} + \text{odd} + \text{odd} + \text{odd}) = 2 + \text{even} = \text{even}$.
- The smallest such sum is $2+3+5+7+11 = 28$, which is even and bigger than $2$, so it cannot be prime.
- The same argument kills every sum that uses $2$.
💡 Adding four odd numbers gives an even number, and adding $2$ keeps it even. Even numbers above $2$ are never prime — Grade 2 odd/even rules eliminate the case.
4.NBT.B.4 Step 2 - Make the first guess: use the five smallest odd primes $3, 5, 7, 11, 13$.
- This gives the smallest possible sum, so if it works, we are done.
💡 Starting from the smallest possible numbers guarantees the smallest possible sum — there is nowhere lower to look.
4.OA.B.4 Step 3 - Check whether $39$ is prime.
- Test small divisors: $39 = 3 \times 13$, so $39$ is composite.
- We need to try the next sum up.
💡 Divisibility by $3$ is fast: digits $3 + 9 = 12$ is a multiple of $3$, so $39$ is too.
4.NBT.B.4 Step 4 - Adjust the guess.
- To get the next smallest sum, replace the largest prime $13$ with the next prime $17$ — this is the cheapest change we can make.
- Any other swap would raise the sum more.
💡 Trading $13$ for $17$ adds $4$ to the sum: $39 + 4 = 43$. That is the smallest possible next sum.
4.OA.B.4 Step 5 - Check whether $43$ is prime.
- The only primes to test are those with $p^2 \leq 43$, i.e.
- $2, 3, 5$: $43$ is odd, $4+3=7$ is not a multiple of $3$, and $43$ does not end in $0$ or $5$.
- So $43$ is prime.
💡 Only need to check primes up to $\sqrt{43} \approx 6.6$, so $2, 3, 5$ suffices.
4.OA.B.4 Step 6 - Confirm no smaller prime works.
- The only prime strictly between $39$ and $43$ is $41$.
- The four smallest odd primes already sum to $3+5+7+11 = 26$, so the fifth prime would need to be $41 - 26 = 15$, which is not prime.
- Any other five distinct odd primes give a sum of at least $43$.
- So $43$ is the smallest.
💡 Only $41$ sits between $39$ and $43$, and it cannot be hit by any allowed combination — so $43$ is locked in.
2.NBT.A.1 Step 7 Add the digits of $43$ as the problem asks.
💡 Reading the tens digit ($4$) and the ones digit ($3$) and adding them is a Grade 2 place-value move.
2.OA.C.3 Eliminate the prime $2$ from the candidate set. If $2$ were one of the five prim 4.NBT.B.4 Make the first guess: use the five smallest odd primes $3, 5, 7, 11, 13$. This g 4.OA.B.4 Check whether $39$ is prime. Test small divisors: $39 = 3 \times 13$, so $39$ is 4.NBT.B.4 Adjust the guess. To get the next smallest sum, replace the largest prime $13$ w 4.OA.B.4 Check whether $43$ is prime. The only primes to test are those with $p^2 \leq 43 4.OA.B.4 Confirm no smaller prime works. The only prime strictly between $39$ and $43$ is 2.NBT.A.1 Add the digits of $43$ as the problem asks. Review
Reasonableness: Double-check the qualifying sum: $3 + 5 + 7 + 11 + 17 = 8 + 7 + 11 + 17 = 15 + 11 + 17 = 26 + 17 = 43$. All five primes are distinct and odd, and $43$ is prime. The digit sum $4 + 3 = 7$ appears as choice (B). The other choices are easy to rule out: $5$ would come from a $2$-digit prime like $14$ (not prime) or $23, 32, 41, 50$ (only $23, 41$ are prime, both too small to be sums of $5$ distinct primes); $8$ would point at $17$ or $53$ (both too small); $10$ at $19, 37, 73$ (too small); $13$ at $67$ or larger (bigger than $43$).
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: the candidate prime $P$ has digit sum equal to one of $5, 7, 8, 10, 13$. Combined with the parity argument that $P$ is a sum of five odd primes (so $P$ is odd) and $P \geq 3+5+7+11+13 = 39$, the smallest odd primes with those digit sums above $39$ are $41$ (sum $5$), $43$ (sum $7$), $53$ (sum $8$), $73$ (sum $10$), $67$ (sum $13$). Check $41$ first: it forces fifth prime $15$, not prime; eliminate. Check $43$: hits with $\{3,5,7,11,17\}$. Done — (B).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs for a whole number; determine whether a given whole number is prime or composite (Recognising that $39 = 3 \times 13$ is composite, that $43$ is prime (no divisors $\leq \sqrt{43}$), and that $15$ is composite — the divisibility checks that drive the search.)2.OA.C.3Determine whether a group of objects has an odd or even number of members (Showing that $2 + \text{odd} + \text{odd} + \text{odd} + \text{odd}$ is even, so any sum that includes the prime $2$ cannot itself be a prime larger than $2$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Computing the candidate sums $3+5+7+11+13 = 39$ and $3+5+7+11+17 = 43$.)2.NBT.A.1Understand place value: a two-digit number represents amounts of tens and ones (Reading the tens and ones digits of $43$ and adding them to get the digit sum $4 + 3 = 7$.)
⭐ Throw out the prime $2$ on parity grounds, add the five smallest odd primes, and bump the largest one up until the sum is prime. The answer $43$ falls out on the second try, and its digits add to $7$.
⭐ Throw out the prime $2$ on parity grounds, add the five smallest odd primes, and bump the largest one up until the sum is prime. The answer $43$ falls out on the second try, and its digits add to $7$.
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