AMC 8 · 2010 · #7

Grade 3 arithmetic
systematic-enumerationmulti-digit-arithmeticpattern-recognition systematic-enumerationoptimization-counting ↑ Prerequisites: multi-digit-arithmetic
📏 Medium solution 💡 3 insights
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Problem
Freddie keeps a fixed collection of U.S. coins — only pennies (1¢), nickels (5¢), dimes (10¢), and quarters (25¢). He wants the smallest such collection so that for any whole-cent amount from 1¢ to 99¢, he can hand over exact change from his collection. What is the smallest total number of coins?

Pick an answer.

(A)
6
(B)
10
(C)
15
(D)
25
(E)
99

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The full range 1 to 99 is intimidating, so use Tool #9: solve the easier problem of paying every amount from 1¢ to 24¢ first — that locks in the small coins. Tool #2 (Systematic List) is the workhorse inside that easier case — list the multiples of 5 up to 20 and check which {nickel, dime} combos hit each one. Then Tool #7 (Subproblems) handles the upper amounts: once we can cover 0–24¢, adding one quarter at a time extends the range by 25¢, so 3 quarters reach all the way to 99¢. Tool #6 (Guess and Check) is the fast sanity test against the answer choices — (A) 6 is clearly too few, (C) 15 and (D) 25 are wasteful, so the smallest plausible number is around 10.

1STEP 1

Solve an easier case first: pennies alone must cover the ones digit 0–4, so you need 4 pennies.

Pennies needed = 4 (to cover 1¢, 2¢, 3¢, 4¢ on top of any multiple of 5)
2STEP 2

For the multiples of 5 up to 20¢, 1 nickel and 2 dimes hit every one (5, 10, 15, 20); a single dime would leave 20¢ short.

Nickels = 1, Dimes = 2 → covers {5, 10, 15, 20} ¢
3STEP 3

Running total: 4 + 1 + 2 = 7 coins, enough to build every amount from 0¢ to 24¢.

4 + 1 + 2 = 7 coins → covers 0–24 ¢
4STEP 4

Each quarter pushes the reach up by 25¢, so 3 quarters stretch the collection all the way to 99¢.

Quarters needed = 3 → covers 0–99 ¢
5STEP 5

Add them all up: 4 + 1 + 2 + 3 = 10 coins, matching choice (B).

4 pennies + 1 nickel + 2 dimes + 3 quarters = 4 + 1 + 2 + 3 = 10 coins → (B)
Answer
10
Spot-check a few amounts. 7¢ = 1 nickel + 2 pennies. 38¢ = 1 quarter + 1 dime + 3 pennies. 99¢ = 3 quarters + 2 dimes + 4 pennies — uses every coin in the collection, so we cannot drop any of them. Trying to remove one penny breaks 4¢; removing the nickel breaks 5¢ (we have no other way to make 5 without using 5 pennies); removing a dime breaks 20¢; removing a quarter breaks 75¢. So 10 is tight, and the answer is (B).
💡Key takeaway

This AMC 8 problem only needs Grade 3 reasoning: solve the easier 0–24¢ case with pennies, nickels, and dimes, then let quarters do the rest.