Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #23
Grade 5 number-theoryPick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Combining conditions (I) and (II) is a small subproblem: N ends in 99 means N + 1 = k² ends in 00, which forces k to be a multiple of 10. That collapses the search from thousands of four-digit numbers down to a tiny finite list of k values (Tool #7 splits the work cleanly). Tool #2 (Systematic List) then enumerates every candidate k = 40, 50, …, 100 in order. For each, condition (III) is checked by noting k²-1 = (k-1)(k+1) — both factors must be prime — which is Tool #3 (Eliminate Possibilities) crossing off any k where k-1 or k+1 is composite. No algebra needed beyond grade-school multiplication and primality checks.
Combine the first two clues
Ends in 99 means N + 1 ends in 00 — a square ending in 00 forces its root to be a multiple of 10, so k = 10m.
Patterns in trailing zeros of place value are a Grade 5 idea: a square ending in 00 has to come from a number ending in 0.
5.NBT.A.2Identify SubproblemsNarrow the possible squares
With 1001 ≤ k² ≤ 10000 and k a multiple of 10, the only candidates are k = 40, 50, 60, 70, 80, 90, 100.
Multi-digit multiplication (40², 50², …) is a Grade 5 fluency standard, plenty to range-check the candidates.
5.NBT.B.5Make A Systematic ListFactor as a difference of squares
Factor N = k² - 1 = (k-1)(k+1): N is a product of two primes exactly when both k-1 and k+1 are prime.
Verifying (k-1)(k+1) = k²-1 is just multiplication, and "is it prime?" is the Grade 4 prime/composite skill.
The candidate number, being 1 less than a perfect square, equals (k-1)(k+1), so it is a product of exactly two primes precisely when both k-1 and k+1 are prime.
▸ Why?
The number is 1 less than a perfect square, so it has the form k²-1, and k²-1 can be rewritten as the product (k-1)(k+1).
▸ Why?
Multiplying (k-1)(k+1) out gives k · k + k · 1 - 1 · k - 1 · 1 = k²-1, because multiplying a sum by a sum lets you multiply each part separately and add the results.
▸ Why?
Because the number equals the two whole numbers k-1 and k+1 multiplied together, its full list of prime factors is exactly the prime factors of k-1 combined with the prime factors of k+1.
▸ Why?
Since every whole number above 1 splits into primes in only one way, that combined list is the number's one and only prime factorization; so the number is a product of exactly two primes precisely when k-1 and k+1 each contribute exactly one prime — that is, when both are already prime, because any composite factor would contribute more than one prime and push the count past two.
Eliminate the composite cases
Only k = 60 survives — 59 and 61 are both prime, giving N = 59 × 61 = 3599; every other k has a composite neighbor.
Testing each k-1 and k+1 for primality by trial division is exactly the Grade 4 "determine prime or composite" task.
4.OA.B.4Eliminate PossibilitiesCount the survivors
Exactly one four-digit number works: N = 3599 = 59 × 61 — the count is 1, choice (B).
Tallying the surviving candidates from the systematic list is Grade 4 counting.
4.OA.B.4Make A Systematic ListThis AMC 8 problem only needs Grade 5 place-value patterns plus the Grade 4 "is this number prime?" check you already know!
- Combine the first two clues
- Narrow the possible squares
- Factor as a difference of squares
- Eliminate the composite cases
- Count the survivors
A parent dashboard for the family lives at sensimlab.com.