AMC 8 · 2012 · #18
Grade 6 number-theoryProblem
What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
Pick an answer.
AMC 8 2012 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 positive integer $N$ that meets all four conditions at once: (i) $N$ is positive, (ii) $N$ is not prime, (iii) $N$ is not a perfect square, and (iv) every prime factor of $N$ is at least $50$.
Givens: $N$ must be a positive integer; $N$ is not prime (so $N$ is composite); $N$ is not a perfect square; Every prime factor of $N$ is greater than or equal to $50$; Answer choices: (A) $3127$, (B) $3133$, (C) $3137$, (D) $3139$, (E) $3149$
Unknowns: The smallest $N$ that satisfies all four conditions
Understand
Restated: Find the smallest positive integer $N$ that meets all four conditions at once: (i) $N$ is positive, (ii) $N$ is not prime, (iii) $N$ is not a perfect square, and (iv) every prime factor of $N$ is at least $50$.
Givens: $N$ must be a positive integer; $N$ is not prime (so $N$ is composite); $N$ is not a perfect square; Every prime factor of $N$ is greater than or equal to $50$; Answer choices: (A) $3127$, (B) $3133$, (C) $3137$, (D) $3139$, (E) $3149$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #3 Eliminate Possibilities
The phrase "smallest positive integer with no prime factor less than $50$" is a classic Tool #2 setup: list the allowed primes ($53, 59, 61, \ldots$) in order, then list the candidate products in order of size. Because $N$ must be composite, the smallest products to try are $53 \times 53$, $53 \times 59$, $53 \times 61$, $\ldots$ — easy to walk through in order. Tool #3 (Eliminate) then closes the deal: the problem is multiple choice, so once we find a valid candidate that matches a choice, the smaller-or-equal choices can be checked and eliminated.
Execute — Answer: A
4.OA.B.4 Step 1 - List the primes that are allowed as factors.
- Condition (iv) says every prime factor must be $\geq 50$.
- The primes just above $50$ are $53, 59, 61, 67, \ldots$ (note $51 = 3 \times 17$ and $57 = 3 \times 19$ are not prime).
💡 Recognizing primes versus composites in the $50$s is exactly the Grade $4$ "factors and primes" skill.
6.NS.B.4 Step 2 - Since $N$ is not prime, $N$ must be a product of at least two primes from that list (a prime can repeat).
- To make $N$ smallest, we want the smallest possible primes, so the candidates to test, in order, are $53 \times 53,\ 53 \times 59,\ 53 \times 61, \ldots$
💡 Building composite numbers from their smallest prime factors is the prime-factorization reasoning of Grade $6$ number sense.
6.EE.A.1 Step 3 - Apply condition (iii) to the first candidate.
- $53 \times 53 = 53^2 = 2809$ is a perfect square, so it is not allowed even though all its prime factors are $\geq 50$.
- Eliminate it.
💡 Spotting $p \times p = p^2$ as a perfect square uses the Grade $6$ exponent definition.
6.NS.B.4 Step 4 - Test the next candidate $N = 53 \times 59 = 3127$.
- Check all four conditions: it is positive, it is composite (a product of two primes), it is not a square (the two prime factors are different, so each exponent is $1$), and both prime factors $53$ and $59$ are $\geq 50$.
- All four pass.
💡 Reading the prime factorization $53^1 \cdot 59^1$ to confirm "composite but not a square" is core Grade $6$ factor reasoning.
6.NS.B.4 Step 5 - Confirm no smaller candidate is possible.
- Any product using $3$ or more primes from the allowed list is at least $53^3 = 148{,}877$, far larger than $3127$.
- Any product of $2$ allowed primes that is smaller than $53 \times 59$ would need both primes $\leq 53$, which forces $53 \times 53$ — already rejected.
- So $3127$ is the smallest.
- Matching the answer choices, this is option (A).
💡 Comparing factorizations to rule out smaller composites uses Grade $6$ factor and multiple reasoning.
4.OA.B.4 List the primes that are allowed as factors. Condition (iv) says every prime fac 6.NS.B.4 Since $N$ is not prime, $N$ must be a product of at least two primes from that l 6.EE.A.1 Apply condition (iii) to the first candidate. $53 \times 53 = 53^2 = 2809$ is a 6.NS.B.4 Test the next candidate $N = 53 \times 59 = 3127$. Check all four conditions: it 6.NS.B.4 Confirm no smaller candidate is possible. Any product using $3$ or more primes f Review
Reasonableness: Sanity-check by direct factoring of $3127$: $3127 \div 53 = 59$ exactly, and $59$ is prime, so $3127 = 53 \times 59$ — two distinct primes, both $\geq 50$, not a square, not prime. All four conditions hold. The size also feels right: $50 \times 50 = 2500$ is a rough lower bound for any product of two primes $\geq 50$, and $3127$ sits just above that, consistent with using the two smallest allowed primes.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices directly: try to factor each option. $3127 = 53 \times 59$ works. $3133 = 13 \times 241$ has factor $13 < 50$. $3137$ is prime. $3139 = 43 \times 73$ has factor $43 < 50$. $3149 = 47 \times 67$ has factor $47 < 50$. Only $3127$ satisfies all conditions — answer (A).
CCSS standards used (min grade 6)
4.OA.B.4Find factor pairs and identify prime vs. composite numbers (Listing the primes just above $50$ ($53, 59, 61, \ldots$) and recognizing that $51$ and $57$ are composite, so they are not allowed factors.)6.NS.B.4Find greatest common factors and least common multiples; use prime factorization (Building candidate composites $53 \times 53,\ 53 \times 59,\ \ldots$ from their prime factorizations and reading those factorizations to check the conditions.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recognizing $53 \times 53 = 53^2 = 2809$ as a perfect square and ruling it out via condition (iii).)
⭐ This AMC 8 problem only needs Grade 6 prime-factorization reasoning: list the allowed primes, multiply the two smallest, and check the conditions.
⭐ This AMC 8 problem only needs Grade 6 prime-factorization reasoning: list the allowed primes, multiply the two smallest, and check the conditions.
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