AMC 8 · 2012 · #18

Grade 6 number-theory
prime-numbersprime-factorizationprimality-testperfect-squares caseworksystematic-enumeration ↑ Prerequisites: prime-numbersprime-factorization
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

(A)
$hspace{.05in}3127$
(B)
$hspace{.05in}3133$
(C)
$hspace{.05in}3137$
(D)
$hspace{.05in}3139$
(E)
$hspace{.05in}3149$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

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, …) in order, then list the candidate products in order of size. Because N must be composite, the smallest products to try are 53 × 53, 53 × 59, 53 × 61, … — 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.

1STEP 1

Condition (iv) forces every prime factor to be at least 50, so the usable primes start at 53, 59, 61, 67 (51 and 57 aren't prime).

Allowed primes = 53, 59, 61, 67, …
2STEP 2

N is composite, so multiply the smallest allowed primes in order — the candidates to test are 53 × 53, 53 × 59, 53 × 61, …

53 × 53 = 2809, 53 × 59 = 3127, 53 × 61 = 3233, …
3STEP 3

The first candidate 53 × 53 = 2809 is a perfect square, so condition (iii) eliminates it despite its big prime factor.

53² = 2809 is a perfect square → rejected
4STEP 4

The next candidate 53 × 59 gives 3127: positive, composite, factors 53 and 59 distinct (not a square), both at least 50 — all four hold.

N = 53 × 59 = 3127 ✓
5STEP 5

Three primes give at least 53³ = 148,877, and the only smaller two-prime product is the banned 53 × 53, so 3127 is smallest — choice (A).

53³ = 148,877 ≫ 3127, 53 × 59 = 3127 → (A)
Answer
hspace{.05in}3127
Sanity-check by direct factoring of 3127: 3127 ÷ 53 = 59 exactly, and 59 is prime, so 3127 = 53 × 59 — two distinct primes, both ≥ 50, not a square, not prime. All four conditions hold. The size also feels right: 50 × 50 = 2500 is a rough lower bound for any product of two primes ≥ 50, and 3127 sits just above that, consistent with using the two smallest allowed primes.
💡Key takeaway

This AMC 8 problem only needs Grade 6 prime-factorization reasoning: list the allowed primes, multiply the two smallest, and check the conditions.