AMC 10 · 2013 · #21

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

prime-factorizationfactorialdivisibility-rules convert-to-algebrapattern-recognition ↑ Prerequisites: prime-factorization

📏 Long solution 💡 3 insights

Problem

Twelve pirates split a chest of gold coins in order. The k-th pirate to go takes

\frac{k}{12}

of the coins that are in the chest at that moment. The chest starts with the smallest number of coins that lets every one of the 12 pirates walk away with a positive whole number of coins. Find how many coins the 12th (last) pirate takes.

Pick an answer.

(A)

720

(B)

1296

(C)

1728

(D)

1925

(E)

3850

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The starting count is unknown, so name it with a letter and write the 12th pirate's share as one fraction. The share is a whole number only when the starting count cancels the denominator, so the real work is a factoring subproblem: find the smallest starting count that supplies exactly the powers of 2 and 3 the denominator needs, then see what survives.

1STEP 1

Track the coins left each turn

Pirate k takes $\frac{k}{12}$ and leaves $\frac{12-k}{12}$ , so every turn is just one multiplication — and pirate 12 takes all that is left.

after pirate k: left = (12-k)/12×(previous pile)

2STEP 2

Write the last pirate's share

Call the start N. Chaining the eleven leftover fractions leaves the last pirate $\frac{N \cdot 11!}{12^{11}}$ — the whole scramble in one fraction.

S₁₂ = N·(11·10…1)/12¹¹ = (N · 11!)/12¹¹

3STEP 3

Break the pieces into primes

Prime-factoring $12^{11}$ and $11!$ cancels the shared 2s and 3s, shrinking the share to $\frac{N \cdot 5^{2} \cdot 7 \cdot 11}{2^{14} \cdot 3^{7}}$ .

S₁₂ = (N · 2⁸ · 3⁴ · 5² · 7 · 11)/(2²² · 3¹¹) = (N · 5² · 7 · 11)/(2¹⁴ · 3⁷)

4STEP 4

Find the smallest starting count

No earlier turn demands more, so the smallest start supplies exactly fourteen 2s and seven 3s: $N = 2^{14} \cdot 3^{7}$ , with no extra primes.

N_min = 2¹⁴ · 3⁷

5STEP 5

Cancel and read off the share

Substituting cancels every 2 and 3, leaving $5^{2} \cdot 7 \cdot 11$ = 1925 coins — the only odd choice, so the answer is (D).

S₁₂ = 2¹⁴ · 3⁷·(5² · 7 · 11)/(2¹⁴ · 3⁷) = 5² · 7 · 11 = 1925

Answer

1925

The starting count was chosen to hold exactly 2¹⁴ x 3⁷ and no more, so those primes cancel completely and the final share equals 5² x 7 x 11 = 1925. Since all factors of 2 are gone, the answer has to be odd; of the five choices only 1925 is odd, which confirms (D) and rules the others out at a glance.

💡Key takeaway

Turn a chain of 'fraction of what's left' into one expression, then let prime factors tell you the smallest start and exactly what survives.

Track the coins left each turn
Write the last pirate's share
Break the pieces into primes
Find the smallest starting count
Cancel and read off the share

Track the coins left each turn

Write the last pirate's share

Break the pieces into primes

Find the smallest starting count

Cancel and read off the share

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