AMC 10 · 2013 · #17
Grade 6 arithmeticAlex has 75 red tokens and 75 blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Start: 75 red tokens and 75 blue tokens; Red booth: give 2 red, receive 1 silver and 1 blue; Blue booth: give 3 blue, receive 1 silver and 1 red
Unknowns: The number of silver tokens Alex has when no more trades are possible
Understand
Restated: 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.
Givens: Start: 75 red tokens and 75 blue tokens; Red booth: give 2 red, receive 1 silver and 1 blue; Blue booth: give 3 blue, receive 1 silver and 1 red
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #14 Extreme Principle, #4 Introduce a Variable, #11 Work Backwards
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.
Execute — Answer: E
4.OA.A.3 Step 1 Find each trade's net effect
- Write what each booth changes.
- The red booth removes 2 red and adds 1 blue and 1 silver.
- The blue booth removes 3 blue and adds 1 red and 1 silver.
- Every trade adds exactly one silver token.
💡 Boiling each booth down to a plus-and-minus list makes the two trades easy to compare.
6.EE.A.3 Step 2 Give each token a point value
- Pick values so that each trade is a fair swap of equal points.
- Let red be worth 4, blue worth 3, and silver worth 5.
- Then the red booth swaps 2 red (8 points) for 1 silver and 1 blue (5 + 3 = 8 points), and the blue booth swaps 3 blue (9 points) for 1 silver and 1 red (5 + 4 = 9 points).
💡 If both booths trade equal points for equal points, the total point value can never change.
4.NBT.B.5 Step 3 Compute the fixed total
- Because no trade changes the total points, the total at the end equals the total at the start.
- Add up the starting points from 75 red and 75 blue tokens.
💡 One fixed number, 525 points, ties the messy start to the messy end.
6.NS.C.7 Step 4 Describe the end state
- Trading stops when neither booth works.
- The red booth needs 2 red, so at the end there are fewer than 2 red tokens, meaning 0 or 1.
- The blue booth needs 3 blue, so at the end there are fewer than 3 blue tokens, meaning 0, 1, or 2.
- Silver tokens are never traded away.
💡 Only a tiny handful of red and blue tokens can survive to the end.
4.OA.B.4 Step 5 Split the 525 points at the end
- At the end the 525 points are spread over the leftover red and blue tokens plus the silver tokens, each worth 5.
- So 5 times the number of silvers equals 525 minus the leftover red and blue points.
- Since the number of silvers is a whole number, the leftover 4*red + 3*blue must be a multiple of 5.
- Checking the allowed leftovers, only red=0, blue=0 (leftover 0) and red=1, blue=2 (leftover 10) are multiples of 5.
💡 The silver count must be whole, so only leftovers that are multiples of 5 can actually happen.
4.OA.A.3 Step 6 Rule out zero-zero and finish
- Ending with 0 red and 0 blue is impossible: the very last trade always hands back a token, a blue at the red booth or a red at the blue booth, so at least one red or blue token must remain.
- That leaves 1 red and 2 blue, with leftover points 4 + 6 = 10.
- Then 5 times the number of silvers is 525 minus 10, which is 515, so there are 103 silver tokens.
- The answer is (E).
💡 The last trade always returns a coin, so you can never empty both colors at once.
4.OA.A.3 Write what each booth changes. The red booth removes 2 red and adds 1 blue and 1 6.EE.A.3 Pick values so that each trade is a fair swap of equal points. Let red be worth 4.NBT.B.5 Because no trade changes the total points, the total at the end equals the total 6.NS.C.7 Trading stops when neither booth works. The red booth needs 2 red, so at the end 4.OA.B.4 At the end the 525 points are spread over the leftover red and blue tokens plus 4.OA.A.3 Ending with 0 red and 0 blue is impossible: the very last trade always hands bac Review
Reasonableness: 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 4*1 + 3*2 = 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.
Alternative: Count the visits with variables. Let x be the number of red-booth visits and y the number of blue-booth visits. Ending with 1 red and 2 blue gives 75 - 2x + y = 1 and 75 + x - 3y = 2. Solving this pair yields x = 59 and y = 44. Each visit produces one silver, so the total is x + y = 59 + 44 = 103, the same answer.
CCSS standards used (min grade 6)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Writing each booth's net change in tokens and doing the final subtract-and-divide to reach 103.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Assigning red=4, blue=3, silver=5 so each trade balances equal point values on both sides.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing the fixed starting total 75*4 + 75*3 = 525 points.)6.NS.C.7Understand ordering and absolute value of rational numbers (Turning 'fewer than 2 red' and 'fewer than 3 blue' into the end-state ranges for red and blue tokens.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Requiring the leftover 4*red + 3*blue to be a multiple of 5 so the silver count stays whole.)
⭐ 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.
⭐ 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.
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