Problem 20 · 2000 AMC 8
Stretch
Logic & Word Problems
caseworkcareful-counting
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?
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Answer: A — 1 dime.
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Hint 1 of 2
'At least one of each' is a gift: spend one of each coin up front (41¢) and the puzzle shrinks to placing the 5 leftover coins. Then look at the LAST digit of what's left.
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Hint 2 of 2
The total ends in 2 and only pennies change the units digit. So the number of pennies is forced by the final digit — pin that down before worrying about the bigger coins.
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Approach: pay one of each first, then let the units digit fix the pennies
- Take one of each coin off the top: 1 + 5 + 10 + 25 = 41¢, using 4 of the 9 coins. Remaining: 102 − 41 = 61¢ in 5 more coins.
- Units digit move: nickels, dimes, quarters all end in 0 or 5, so only pennies can produce the final '1' in 61. With 5 coins, you can't afford 6 pennies — so exactly 1 more penny. Now 60¢ in 4 coins.
- 60¢ in 4 coins needs a quarter (four dimes max out at 40¢). Two quarters = 50¢ leaves 10¢ in 2 coins = two nickels. That fills all four with no dime.
- So beyond the single starter dime, none are added: 1 dime.
- The reusable trick: in coin/total problems, read the *last digit* of the amount — only pennies (the 1¢ pieces) can change it, so the units digit alone often pins the smallest-coin count and collapses the casework.
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