AMC 10 · 2024 · #12
Grade 6 algebranumber-theoryPick an answer.
The words "geometric sequence" cry out for Tool #13 (Convert to Algebra): name the common ratio r, write r = p/q in lowest terms, and the integer constraints on a and b collapse into a clean number-theory statement — both p and q must be divisors of 720. To minimize b = 720 · p/q with p ＞ q, we need p/q as close to 1 as possible, i.e. consecutive integers. Tool #2 (Systematic List) then sweeps through pairs of consecutive divisors of 720 — small list, only need the largest pair. No need for full algebra of inequalities; the divisor structure does the work.
Write the ratio as a fraction
Take the reduced ratio p over q.
Naming the unknown ratio with a letter — and pinning it down as a reduced fraction — is the Grade 6 "letters stand for numbers" move that turns a vague "some sequence" into two clean formulas.
6.EE.A.2Convert To AlgebraTranslate the integer condition
Both p and q must be divisors of 720.
Coprime plus divides-the-product is the GCF rule from Grade 6: if q shares no factor with p, then q has to take all its factors from 720 alone.
If a number shares no factor with another, then it must take all its factors from what remains.
▸ Why?
Every number has exactly one prime recipe, so a prime not shared cannot be supplied by that partner.
▸ Why?
The quotient is a whole number only when every needed factor is present, with nothing left over.
What makes b smallest
Making the ratio as close to 1 as possible minimizes b.
Consecutive integers are automatically coprime (gcd(n, n+1) = 1), so the coprime condition comes for free — we just need both to divide 720.
6.RP.A.1Convert To AlgebraFind consecutive divisor pairs
The largest such pair is 15 and 16.
After (15, 16) the next candidates would need q ≥ 16 with q+1 also dividing 720. The divisors of 720 jump from 16 to 18 to 20 to 24 — no more consecutive pairs. So (15, 16) is the winner.
4.OA.B.4Make A Systematic ListGet b and add its digits
b is 768 and its digits add to 21.
Plugging the optimal (p, q) = (16, 15) back into b = 720p/q is just division and multiplication — divide 720 by the small factor 15 first to keep the arithmetic light.
6.NS.B.2Convert To AlgebraWhen a geometric sequence has to land on integers, the common ratio is a fraction p/q in lowest terms where both p and q divide the middle term. To squeeze b as close to 720 as possible, find the largest pair of consecutive divisors of 720 — that pair is (15, 16), so b = 720 · 16/15 = 768.