AMC 8 · 2007 · #3

Grade 4 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

prime-factorizationprime-numbersfactors identify-subproblems ↑ Prerequisites: prime-numbersdivisibility-rules

📏 Short solution 💡 1 insight

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Problem

What is the sum of the two smallest prime factors of $250$ ?

Pick an answer.

(A)

$2$

(B)

$5$

(C)

$7$

(D)

$10$

(E)

$12$

AMC 8 2007 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Find the two smallest prime numbers that divide $250$ and add them.

Givens: The target number is $250$; We only need the two smallest distinct prime factors; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $10$, (E) $12$

Unknowns: The sum of the two smallest prime factors of $250$

Understand

Restated: Find the two smallest prime numbers that divide $250$ and add them.

Givens: The target number is $250$; We only need the two smallest distinct prime factors; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $10$, (E) $12$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

Tool #7 (Identify Subproblems) turns "find the two smallest prime factors" into two tiny questions: (1) what is the smallest prime that divides $250$? (2) once we divide that out, what is the smallest prime that divides what is left? Each step is a single divisibility check, which is much easier than factoring $250$ all at once. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net — once we have a candidate sum, we can verify it against the listed choices.

Execute — Answer: C

#7 Identify Subproblems 4.OA.B.4 Step 1

Find the smallest prime divisor of $250$.
The number ends in $0$, so it is even — divide by $2$.

$$250 \div 2 = 125$$

💡 The smallest prime is always $2$, and any even number is divisible by $2$. This is the Tool #7 "peel off one prime" move.

#7 Identify Subproblems 4.OA.B.4 Step 2

Find the smallest prime divisor of the quotient $125$.
It is odd, so $2$ does not work.
The digit sum is $1+2+5 = 8$, which is not a multiple of $3$, so $3$ does not work either.
It ends in $5$, so $5$ does work.

$$125 \div 5 = 25$$

💡 Checking divisibility in order ($2$, $3$, $5$, $7$, …) guarantees we catch the smallest prime first.

#3 Eliminate Possibilities 4.NBT.B.4 Step 3

The two smallest distinct prime factors of $250$ are $2$ and $5$.
Add them to get the answer.

$$2 + 5 = 7 \;\Rightarrow\; \textbf{(C)}$$

💡 $7$ matches choice (C). The other choices are quickly ruled out — (A) $2$ and (B) $5$ each forget one of the primes; (D) $10$ and (E) $12$ pull in composite numbers, but the problem asks for prime factors.

[1] #7 4.OA.B.4 Find the smallest prime divisor of $250$. The number ends in $0$, so it is even

[2] #7 4.OA.B.4 Find the smallest prime divisor of the quotient $125$. It is odd, so $2$ does no

[3] #3 4.NBT.B.4 The two smallest distinct prime factors of $250$ are $2$ and $5$. Add them to ge

Review

Reasonableness: Continue the factorization to double-check: $25 \div 5 = 5$ and $5 \div 5 = 1$, so $250 = 2 \times 5 \times 5 \times 5 = 2 \times 5^3$. The complete list of distinct primes is just $\{2, 5\}$, and those are also the two smallest, so the sum $2 + 5 = 7$ is correct. The answer (C) sits between (B) $5$ (only one prime) and (D) $10 = 2 \times 5$ (a composite distractor), exactly where a small-prime-sum should land.

Alternative: Tool #3 (Eliminate Possibilities) on its own: the smallest prime is always $2$, and $250$ is even, so $2$ must be one of the factors. That already rules out (A) $2$ (which would need both factors to be $2$, but the second factor must be a different prime). The next prime divisor must be at least $3$, so the sum is at least $2+3 = 5$ and at most a one-digit total. Only (B) $5$, (C) $7$, and (D) $10$ survive a rough size check. Testing $3$: digit sum $2+5+0 = 7$ is not a multiple of $3$, so $3$ is not a factor — eliminate (B). Testing $5$: $250$ ends in $0$, so $5$ works, giving $2 + 5 = 7$. Choice (C).

CCSS standards used (min grade 4)

4.OA.B.4 Find factor pairs, recognize multiples, and identify prime numbers within $1$-$100$ (Recognizing $2$ as a factor of $250$ (because it is even) and $5$ as a factor of $125$ (because it ends in $5$) — the Grade 4 prime-factor identification standard.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the two smallest prime factors $2 + 5 = 7$ to produce the final answer.)

⭐ To find the smallest prime factors, peel them off one at a time starting with $2$ — a Grade 4 divisibility-rules skill is all this AMC 8 problem asks for.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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