AMC 8 · 2003 · #2
Easy mode Grade 4Problem
Look at the five numbers below. Each one can be broken into prime factors.
For each number, find its smallest prime factor. Which number has the smallest one of all?
Pick an answer.
AMC 8 2003 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: From the five numbers $55, 57, 58, 59, 61$, find the one whose smallest prime factor is the smallest. In other words, look at each number's smallest prime divisor and pick whichever number has the smallest such divisor.
Givens: Five candidate numbers: $55, 57, 58, 59, 61$; We compare them by their smallest prime factor, not by their size; Answer choices: (A) $55$, (B) $57$, (C) $58$, (D) $59$, (E) $61$
Unknowns: Which of the five numbers has the smallest prime factor
Understand
Restated: From the five numbers $55, 57, 58, 59, 61$, find the one whose smallest prime factor is the smallest. In other words, look at each number's smallest prime divisor and pick whichever number has the smallest such divisor.
Givens: Five candidate numbers: $55, 57, 58, 59, 61$; We compare them by their smallest prime factor, not by their size; Answer choices: (A) $55$, (B) $57$, (C) $58$, (D) $59$, (E) $61$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #5 Look for a Pattern
This is a multiple-choice problem with only five candidates, so Tool #3 (Eliminate Possibilities) is the natural fit: test each choice against the smallest primes in order. Tool #5 (Look for a Pattern) sharpens the test — the smallest prime is $2$, and the pattern "divisible by $2$ $=$ even" lets us scan the list at a glance. If any candidate is even, it must be the winner, because no number can have a prime factor below $2$.
Execute — Answer: C
4.OA.B.4 Step 1 - List the primes from smallest to largest.
- The smallest prime factor any number can possibly have is $2$.
- After that comes $3$, then $5$, and so on.
- So the search for "smallest prime factor" starts at $2$.
💡 Knowing the order of primes turns the problem into a quick checklist: try $2$ first, then $3$, then $5$.
3.OA.D.9 Step 2 - Check each candidate for divisibility by $2$.
- A number is divisible by $2$ exactly when it is even (ones digit is $0, 2, 4, 6,$ or $8$).
- Scan the ones digits: $55, 57, 58, 59, 61$ end in $5, 7, 8, 9, 1$.
- Only $58$ ends in an even digit.
💡 The even/odd pattern is the fastest divisibility test there is — one glance at the ones digit decides it.
4.OA.B.4 Step 3 - Conclude with the smallest-prime rule.
- Since $58$ is divisible by $2$, its smallest prime factor is $2$.
- The other four numbers are odd, so their smallest prime factor is at least $3$.
- Nothing can beat $2$, so $58$ wins.
💡 Once one candidate hits the smallest possible prime, the search is over — no later check can produce a smaller answer.
4.OA.B.4 List the primes from smallest to largest. The smallest prime factor any number c 3.OA.D.9 Check each candidate for divisibility by $2$. A number is divisible by $2$ exact 4.OA.B.4 Conclude with the smallest-prime rule. Since $58$ is divisible by $2$, its small Review
Reasonableness: Confirm by finding the smallest prime factor of each candidate. $55 = 5 \times 11$, smallest prime $5$. $57 = 3 \times 19$, smallest prime $3$. $58 = 2 \times 29$, smallest prime $2$. $59$ is prime, smallest prime $59$. $61$ is prime, smallest prime $61$. The smallest of $\{5, 3, 2, 59, 61\}$ is $2$, achieved by $58$. The answer (C) holds.
Alternative: Tool #2 (Make a Systematic List): for each number $55, 57, 58, 59, 61$, list factor pairs and read off the smallest prime in each. $55: 5 \cdot 11$; $57: 3 \cdot 19$; $58: 2 \cdot 29$; $59: \text{prime}$; $61: \text{prime}$. Compare the smallest primes: $5, 3, 2, 59, 61$. The minimum is $2$, from $58$, giving (C).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs for a whole number in the range 1-100; determine whether a given whole number is prime or composite (Knowing what "prime factor" means, and that the primes in order start $2, 3, 5, \ldots$, so the search for a smallest prime factor begins at $2$.)3.OA.D.9Identify arithmetic patterns, including patterns in the addition table or multiplication table, and explain them using properties of operations (Using the even/odd pattern (divisibility by $2$) to scan the candidates at a glance and find the one with $2$ as a factor.)
⭐ The smallest prime is $2$, so on a "smallest prime factor" question, first ask: is any choice even? If yes, that choice wins immediately — here $58$ is the only even number, so the answer is (C).
⭐ The smallest prime is $2$, so on a "smallest prime factor" question, first ask: is any choice even? If yes, that choice wins immediately — here $58$ is the only even number, so the answer is (C).
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