AMC 10 · 2015 · #13

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
linear-equations-one-varmultiplessystematic-enumeration logical-deduction ↑ Prerequisites: linear-equations-one-var
📏 Medium solution 💡 3 insights
Problem
Claudia has 12 coins, each worth either 5 cents or 10 cents. Counting every total she can make from one or more of her coins, exactly 17 different values appear. Figure out how many of her coins are 10-cent coins.

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
7

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The puzzle hides a single unknown — how many 10-cent coins there are — so Tool #4 (Introduce a Variable) names it y and pins the 5-cent count to 12-y. The phrase "17 different values" feels slippery until Tool #5 (Look for a Pattern) reveals the key regularity: because there is at least one 5-cent coin, the obtainable totals are exactly the multiples of 5 marching from the smallest coin up to the whole pile, with no gaps. Tool #8 (Analyze the Units) then cashes that in: counting in 5-cent units turns "17 different values" into "the pile is worth 17 fives," a clean number that ties straight back to y.

1STEP 1

Name the unknown coins

Let y be the number of 10-cent coins; the rest, 12-y, are 5-cent coins.

y=number of 10-cent coins, 12-y=number of 5-cent coins
2STEP 2

Totals are gap-free multiples of 5

With a 5-cent coin on hand, totals climb in unbroken 5-cent steps from 5 up to the whole pile.

different totals = (value of whole pile)/5
3STEP 3

Turn 17 values into a total

17 different totals means 17 fives: 17× 5 = 85 cents total.

17 different totals → whole pile=17× 5=85 cents
4STEP 4

Write the value equation and solve

Solving 5(12-y)+10y=85 gives y = 5 dimes — answer (C).

5(12-y)+10y=85 → 60+5y=85 → y=5 → (C)
Answer
5
Check the count directly: 5 dimes and 12-5=7 nickels are worth 5· 10 + 7· 5 = 50+35 = 85 cents, which is 17 fives, matching the required 17 different totals (the multiples 5,10,…,85). The answer is also forced rather than lucky: each time you trade a nickel for a dime the pile gains 5 cents and the value-count rises by exactly 1, so y=3 would give 15 totals, y=4 gives 16, y=5 gives 17, y=6 gives 18, y=7 gives 19. Only y=5 lands on 17, ruling out (A), (B), (D), and (E).
💡Key takeaway

Counting in 5-cent steps, the totals equal the pile's value in fives, so 17 totals means 85 cents and 5 dimes.

  • Name the unknown coins
  • Totals are gap-free multiples of 5
  • Turn 17 values into a total
  • Write the value equation and solve

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