AMC 10 · 2013 · #17
Grade 6 number-theoryPick an answer.
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.
Track the coins left each turn
Each turn scales what is left by a fraction.
Taking a fraction of what remains is the same as multiplying the pile by the leftover fraction.
Taking a fraction of what remains is the same as multiplying the pile by the leftover fraction.
▸ Why?
A fraction of a pile is that pile cut into equal shares, so what is left is the remaining shares.
▸ Why?
Each turn multiplies by its own fixed factor, so the whole process is one chain of multiplications.
Write the last pirate's share
The last share is one long product.
Chaining the leftover fractions turns the whole process into one clean expression for the final share.
6.EE.B.6Introduce A VariableBreak the pieces into primes
Breaking into primes exposes what must divide.
Prime factoring shows exactly which factors must be supplied and which cancel.
6.EE.A.1Identify SubproblemsFind the smallest starting count
That names the smallest possible start.
The smallest safe starting count carries just enough 2s and 3s to cancel the toughest step.
6.NS.B.4Identify SubproblemsCancel and read off the share
Cancelling leaves 1925, choice (D).
Because N was built to cancel every 2 and 3, only the odd factors 5, 5, 7, 11 survive.
4.OA.B.4Eliminate PossibilitiesTurn 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