AMC 10 · 2022 · #4
Grade 6 number-theoryPick an answer.
The two conditions each force a small, listable shape on n. Tool #2 (Systematic List): the GCD condition says 15 ∣ n, so write the multiples of 15 up to 180 in order. That short list is the entire candidate pool. Tool #3 (Eliminate Possibilities): now run each candidate through the two original conditions and keep the one survivor. This avoids the prime-exponent algebra of the reference solution while still being rigorous, and shows the student exactly why exactly one n works.
Narrow to multiples of fifteen
The gcd forces a multiple of fifteen.
Listing every multiple of 15 in order guarantees no candidate is missed.
4.OA.B.4Make A Systematic ListNarrow to divisors of 180
The lcm forces a divisor of 180.
Any number whose LCM with anything equals 180 must itself be a divisor of 180.
Any number whose least common multiple with something equals a given number must itself divide that number.
▸ Why?
A common multiple contains every prime at least as often as each number does, so nothing can overflow it.
▸ Why?
Every number has exactly one prime recipe, so a prime not in the target cannot hide in a candidate.
Filter by the lcm
Check the lcm directly.
Computing lcm for a handful of small pairs is quick once the candidate pool is shrunk.
6.NS.B.4Eliminate PossibilitiesFilter by the gcd
Check the gcd too.
If 45 already divides n, then gcd(n,45)=45, not 15 — so n may be a multiple of 15 but not of 45.
6.NS.B.4Eliminate PossibilitiesAdd the digits
Adding the digits gives 6.
Reading digits and adding them is the same skill as place value in second-grade arithmetic.
2.NBT.A.1Make A Systematic ListThis AMC 12 problem only needs Grade 6 LCM and GCD reasoning you already know — list multiples of 15, keep the ones that divide 180, and check which one gives LCM 180 and GCD 15; that is n=60, with digit sum 6.