AMC 8 · 2007 · #3
Easy mode Grade 4Problem
Think about the number .
A prime factor of is a prime number that divides evenly. Find the two smallest ones.
What do you get when you add those two prime factors together?
Pick an answer.
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
4.OA.B.4 Step 1 - Find the smallest prime divisor of $250$.
- The number ends in $0$, so it is even — divide by $2$.
💡 The smallest prime is always $2$, and any even number is divisible by $2$. This is the Tool #7 "peel off one prime" move.
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.
💡 Checking divisibility in order ($2$, $3$, $5$, $7$, …) guarantees we catch the smallest prime first.
4.NBT.B.4 Step 3 - The two smallest distinct prime factors of $250$ are $2$ and $5$.
- Add them to get the answer.
💡 $7$ 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.
4.OA.B.4 Find the smallest prime divisor of $250$. The number ends in $0$, so it is even 4.OA.B.4 Find the smallest prime divisor of the quotient $125$. It is odd, so $2$ does no 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.4Find 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.4Fluently 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.
⭐ 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.
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