AMC 10 · 2022 · #7
Grade 6 number-theoryPick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
The GCD condition forces 15 to divide n, and the LCM condition forces n to divide 180 — so n is a multiple of 15 up to 180.
Listing every multiple of 15 in order guarantees no candidate is missed.
4.OA.B.4Make A Systematic ListKeep only the listed numbers that themselves divide 180: 15, 30, 45, 60, 90, 180 survive; the rest are cut.
Any number whose LCM with anything equals 180 must itself be a divisor of 180.
4.OA.B.4Eliminate PossibilitiesTest each survivor's LCM with 18: only 60 and 180 give 180; the others give 90 and fail.
Computing lcm for a handful of small pairs is quick once the candidate pool is shrunk.
6.NS.B.4Eliminate PossibilitiesNow the GCD test: gcd(60,45)=15 passes, but gcd(180,45)=45 fails since 45 divides 180 — only n = 60 remains.
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 PossibilitiesThe digits of 60 are 6 and 0, so the digit sum is 6 + 0 = 6 — choice (B).
Reading digits and adding them is the same skill as place value in second-grade arithmetic.
2.NBT.A.1Make A Systematic ListThis AMC 10 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.