AMC 10 · 2023 · #15
Grade 8 number-theoryPick an answer.
Three independent yes/no questions live in one problem — perfect setup for Tool #7 (Identify Subproblems). Clear denominators first: multiplying by 210 gives c = 15a + 14b. Then test each statement separately. Tool #3 (Eliminate Possibilities) supplies a single counterexample to kill Statement I. Tool #13 (Convert to Algebra) — specifically modular reduction mod 2, 3, 5, 7 — settles Statement III, and Statement II follows immediately because "and" implies "or".
Clear the denominators
Clearing gives an integer relation.
Working with c = 15a + 14b is much easier than fractions — and the coprime split 15 = 3 · 5, 14 = 2 · 7 already hints that the primes will separate cleanly.
6.NS.B.4Convert To AlgebraFind a counterexample
A single counterexample kills one statement.
One counterexample is enough to kill a universal claim — choose a to satisfy the hypothesis, then b to sneak a forbidden prime into c.
6.NS.B.4Eliminate PossibilitiesCheck prime by prime
Check the remainder at each prime.
Reducing c = 15a + 14b mod each prime kills one of the two terms, so divisibility of c by that prime is forced by just one variable — making each case a one-line check.
Reducing the expression modulo each prime kills one of the two terms, so one variable carries the whole condition.
▸ Why?
Only the remainder after dividing survives, so a term that is a multiple of the prime disappears.
▸ Why?
Every number has exactly one prime recipe, so the primes can be tested one at a time with no interference.
Prove the other two
The relation guarantees the other two.
Modding out the 14b piece leaves 15a mod 14, and since 15 ≡ 1 mod 14 the gcd with 14 is just gcd(a,14) — same trick the other way.
6.NS.B.4Identify SubproblemsPick the choice
The answer is the second and third only.
A logical "and" is stronger than "or" — once we proved the harder "and" claim of III, the weaker "or" claim of II rides along for free.
6.NS.B.4Eliminate PossibilitiesMultiplying through by 210 turns the equation into c = 15a + 14b. Mod each prime in {2, 3, 5, 7}, one of the two terms vanishes, so gcd(c, 210) = 1 is equivalent to gcd(a, 14) = gcd(b, 15) = 1 (Statement III). That "and" automatically implies the "or" in Statement II, but Statement I fails because a = 1, b = 3 gives c = 57 — divisible by 3. Answer: (E) II and III only.