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

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

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

Givens: There are 12 pirates who take shares one after another.; The k-th pirate takes k/12 of whatever coins remain in the chest when it is his turn.; The starting number of coins is the smallest one for which every share is a positive whole number.; The 12th pirate is last, and 12/12 = 1, so he takes everything still in the chest.

Unknowns: The number of coins the 12th pirate receives.

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #5 Look for a Pattern, #7 Identify Subproblems, #3 Eliminate Possibilities

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.

Execute — Answer: D

#5 Look for a Pattern 5.NF.B.4 Step 1

Track the coins left each turn

When the k-th pirate takes k/12 of the chest, the part he leaves behind is 1 - k/12 = (12-k)/12 of what was there.
So each turn multiplies the pile by a simple fraction: the 1st pirate leaves 11/12, the 2nd leaves 10/12, and so on down to the 11th, who leaves 1/12.
The 12th pirate then takes 12/12 of that, meaning he scoops up everything still in the chest.

$$\text{after pirate } k: \ \text{left} = \frac{12-k}{12}\times(\text{previous pile})$$

💡 Taking a fraction of what remains is the same as multiplying the pile by the leftover fraction.

#4 Introduce a Variable 6.EE.B.6 Step 2

Write the last pirate's share

Let N be the starting number of coins.
After pirates 1 through 11 each leave their fraction, the pile has been multiplied by 11/12, 10/12, ..., 1/12.
So the coins left for the 12th pirate, which he takes entirely, are N times the product of those numerators over twelve copies of 12.
The numerators multiply to 11 x 10 x 9 x ...
x 1 = 11!
(11 factorial), and there are eleven factors of 12 in the bottom.

$$S_{12} = N\cdot\frac{11\cdot10\cdots1}{12^{11}} = \frac{N\cdot 11!}{12^{11}}$$

💡 Chaining the leftover fractions turns the whole process into one clean expression for the final share.

#7 Identify Subproblems 6.EE.A.1 Step 3

Break the pieces into primes

To see when this is a whole number, factor the top and bottom into primes.
The denominator 12^{11} = (2^2 x 3)^{11} = 2^{22} x 3^{11}.
The numerator 11!
= 2^8 x 3^4 x 5^2 x 7 x 11 (count the factors: 11!
has eight 2s, four 3s, two 5s, one 7, one 11).
Cancelling the shared 2s and 3s leaves the 12th pirate's share as N times 5^2 x 7 x 11, divided by whatever powers of 2 and 3 are still stuck in the bottom.

$$S_{12} = \frac{N\cdot 2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7\cdot 11}{2^{22}\cdot 3^{11}} = \frac{N\cdot 5^{2}\cdot 7\cdot 11}{2^{14}\cdot 3^{7}}$$

💡 Prime factoring shows exactly which factors must be supplied and which cancel.

#7 Identify Subproblems 6.NS.B.4 Step 4

Find the smallest starting count

The bottom of the fraction now has 2^{14} x 3^{7} left over.
For the share to be a whole number, N must supply at least fourteen 2s and seven 3s.
Checking each earlier pirate the same way, no earlier turn ever demands more 2s or 3s than this last one does, so once the last share is whole, all the earlier shares are whole too.
The smallest N that works is therefore exactly N = 2^{14} x 3^{7}, adding no extra primes.

$$N_{\min} = 2^{14}\cdot 3^{7}$$

💡 The smallest safe starting count carries just enough 2s and 3s to cancel the toughest step.

#3 Eliminate Possibilities 4.OA.B.4 Step 5

Cancel and read off the share

Substitute N = 2^{14} x 3^{7} into the share.
The 2^{14} and 3^{7} in N cancel the 2^{14} and 3^{7} in the denominator perfectly, leaving only 5^2 x 7 x 11 = 25 x 77 = 1925.
Notice every factor of 2 has cancelled, so the answer must be odd, and 1925 is the only odd choice.
The 12th pirate receives 1925 coins, answer (D).

$$S_{12} = 2^{14}\cdot 3^{7}\cdot\frac{5^{2}\cdot 7\cdot 11}{2^{14}\cdot 3^{7}} = 5^{2}\cdot 7\cdot 11 = 1925$$

💡 Because N was built to cancel every 2 and 3, only the odd factors 5, 5, 7, 11 survive.

[1] #5 5.NF.B.4 When the k-th pirate takes k/12 of the chest, the part he leaves behind is 1 - k

[2] #4 6.EE.B.6 Let N be the starting number of coins. After pirates 1 through 11 each leave the

[3] #7 6.EE.A.1 To see when this is a whole number, factor the top and bottom into primes. The d

[4] #7 6.NS.B.4 The bottom of the fraction now has 2^{14} x 3^{7} left over. For the share to be

[5] #3 4.OA.B.4 Substitute N = 2^{14} x 3^{7} into the share. The 2^{14} and 3^{7} in N cancel t

Review

Reasonableness: The starting count was chosen to hold exactly 2^{14} x 3^{7} and no more, so those primes cancel completely and the final share equals 5^2 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.

Alternative: Work backwards from the last pirate. He takes everything left, which is N x 11!/12^{11}. Rather than minimizing N first, cancel the numerator 11 x 10 x ... x 1 against 12^{11} = 2^{22} x 3^{11}; the smallest N just fills the remaining 2^{14} x 3^{7}, and the odd survivors 5^2 x 7 x 11 = 1925 are exactly what the last pirate collects. Faster still for a timed test: the powers of two must all cancel, so the answer is odd, and only (D) 1925 is odd.

CCSS standards used (min grade 6)

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Turning 'take k/12 of the remaining coins' into multiplying the pile by the leftover fraction (12-k)/12)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the starting count N and writing the 12th pirate's share as N x 11!/12^{11})
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Factoring 12^{11} = 2^{22} x 3^{11} and 11! = 2^8 x 3^4 x 5^2 x 7 x 11 into prime powers)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Finding the smallest N = 2^{14} x 3^{7} that supplies just enough factors to make every share a whole number)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Reading off the surviving odd factors 5 x 5 x 7 x 11 = 1925 and using parity to confirm the choice)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

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

Review

CCSS standards used (min grade 6)

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