AMC 10 · 2013 · #17

Grade 6 logicnumber-theory

Archetype Branching State Enumeration └ Sub-archetype Rule-Governed Process Simulation

invariant-monovariantsystems-of-equationsmultiples invariant-monovariantwork-backwards ↑ Prerequisites: systems-of-equations

📏 Long solution 💡 3 insights

Problem

Alex starts with 75 red tokens and 75 blue tokens. At the red booth he trades 2 red tokens for 1 silver and 1 blue token. At the blue booth he trades 3 blue tokens for 1 silver and 1 red token. He keeps trading until neither booth can be used. Find how many silver tokens he ends with.

Pick an answer.

(A)

(B)

(C)

(D)

102

(E)

103

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

How to solve

Strategy Organize Information in More Ways

Following every trade one by one would take dozens of steps. Instead, re-describe the tokens by giving each color a point value chosen so that every trade keeps the total number of points the same. Then the total is fixed from start to finish, and the only thing left to pin down is how the leftover points split among the few tokens that remain at the end.

1STEP 1

Find each trade's net effect

Red booth: lose 2 red, gain 1 blue and 1 silver. Blue booth: lose 3 blue, gain 1 red and 1 silver. Each trade adds 1 silver.

Red booth: -2 red, +1 blue, +1 silver; Blue booth: +1 red, -3 blue, +1 silver

2STEP 2

Give each token a point value

Set red 4, blue 3, silver 5. Then 2 red = 8 = 1 silver + 1 blue, and 3 blue = 9 = 1 silver + 1 red: every trade is an even swap.

red=4, blue=3, silver=5: 2 · 4 = 5 + 3, 3 · 3 = 5 + 4

3STEP 3

Compute the fixed total

No trade changes the total, so the ending total equals the start: 75 red and 75 blue are worth 525 points.

75 · 4 + 75 · 3 = 300 + 225 = 525

4STEP 4

Describe the end state

Trading stops with fewer than 2 red and fewer than 3 blue, so red is 0 or 1 and blue is 0, 1, or 2. Silver is never spent.

red ∈ {0, 1}, blue ∈ {0, 1, 2}

5STEP 5

Split the 525 points at the end

Silver is 5 points each, so the leftover 4·red + 3·blue must be a multiple of 5. Of the allowed leftovers only 0 and 10 qualify.

5·(silver) = 525 - (4·red + 3·blue), 4·red+3·blue∈{0, 10}

6STEP 6

Rule out zero-zero and finish

The last trade always hands a token back, so 0 red and 0 blue is impossible. Leftover 1 red and 2 blue give 515 / 5 = 103 silvers.

5·(silver) = 525 - 10 = 515 → silver = 103

Answer

103

Check the leftover directly: 1 red is 4 points and 2 blue is 6 points, so leftover points are 10 and silver points are 515, giving 515 / 5 = 103 silvers. Also 41 + 32 = 10 is indeed a multiple of 5, matching the whole-number requirement, and the total 4 + 6 + 515 = 525 equals the starting value. Everything lines up, so 103, choice (E), is correct.

💡Key takeaway

Give each token a point value so every trade is an even swap, then the unchanging total tells you the answer no matter how the trades happen.

Find each trade's net effect
Give each token a point value
Compute the fixed total
Describe the end state
Split the 525 points at the end
Rule out zero-zero and finish