Competition · AMC preparation · step 4 of 4

AMC 10 · 2023B · #18

Grade 8 number-theory
gcdprime-factorizationdivisibility-ruleslogical-deduction caseworkconvert-to-algebralogical-deduction ↑ Prerequisites: gcdprime-factorization
📏 Long solution 💡 3 insights
Problem
Positive integers a, b, c satisfy a/14 + b/15 = c/210. Decide which of the three statements I, II, III about gcd(a,14), gcd(b,15), gcd(c,210) are necessarily true, and pick the choice naming exactly that set.

Pick an answer.

(A)
I, II, and III
(B)
I only
(C)
I and II only
(D)
III only
(E)
II and III only

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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".

1STEP 1

Clear the denominators

Clear denominators: the LCM of 14, 15, 210 is 210, so multiplying through gives c = 15a + 14b.

210·a/14 + 210·b/15 = 210·c/210 ⟹ c = 15a + 14b
2STEP 2

Break Statement I with an example

Counterexample to I: a=1 (so gcd(a,14)=1), b=3 gives c=57=3·19, so gcd(57,210)=3≠1 — Statement I is false.

a=1, b=3 → c=57, gcd(57,210)=3
3STEP 3

Prove Statement III forward

Forward III: if gcd(a,14)=1 and gcd(b,15)=1, then mod 2,7 give c≡a, mod 3,5 give c≡2b,4b — never 0, so gcd(c,210)=1.

c ≢ 0 (mod 2,3,5,7) ⟹ gcd(c,210)=1
4STEP 4

Prove Statement III backward

Reverse III: using gcd(x+ky,y)=gcd(x,y), gcd(c,14)=gcd(a,14) and gcd(c,15)=gcd(b,15), so gcd(c,210)=1 forces both to be 1.

gcd(c,14)=gcd(a,14), gcd(c,15)=gcd(b,15)
5STEP 5

Settle Statement II

Statement II wants only the weaker 'or', which III's reverse 'and' already gives — so II is true; with I false, II and III are the answer.

(II AND III true, I false) → (E) II and III only
Answer
II and III only
Cross-check Statement III on a clean example: a=1, b=1 gives c = 15 + 14 = 29 — prime, so gcd(29, 210) = 1, matching gcd(1, 14) = gcd(1, 15) = 1. Cross-check the reverse: any c coprime to 210 comes from (a, b) where a is coprime to 14 and b to 15. Counterexample to I: a=1, b=3 gives c = 57, gcd(57, 210) = 3 ≠ 1 despite gcd(1, 14)=1. So I is false, III is true (both directions), II is true (weaker than III's reverse). The right choice is (E) II and III only.
💡Key takeaway

Multiplying 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.

  • Clear the denominators
  • Break Statement I with an example
  • Prove Statement III forward
  • Prove Statement III backward
  • Settle Statement II

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