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2000 AMC 10 Problem 17

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2000 AMC 10 Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Modular Arithmetic · difficulty 3 of 5 · about 3 min · position P16-20

Boris has an incredible coin-changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?

  1. A)

    $3.63

  2. B)

    $5.13

  3. C)

    $6.30

  4. D)

    $7.45

  5. E)

    $9.07

Answer

D

Key insight

Quarter-to-nickels and nickel-to-pennies preserve value; penny-to-quarters adds exactly 124 cents, so every reachable total is 1 more than a multiple of 124.

Solution

Track the total value in cents through each kind of exchange.

  • Quarter in, five nickels out: 25→2525 \to 25, no change.
  • Nickel in, five pennies out: 5→55 \to 5, no change.
  • Penny in, five quarters out: 1→1251 \to 125, a gain of 124124.

So no matter what Boris does, his total is 1+124k1 + 124k cents, where kk counts how many pennies he has fed in. Subtract 11 from each choice and test divisibility by 124124:
362=2⋅124+114,512=4⋅124+16,629=5⋅124+9,744=6⋅124,906=7⋅124+38.362 = 2 \cdot 124 + 114,\quad 512 = 4 \cdot 124 + 16,\quad 629 = 5 \cdot 124 + 9,\quad 744 = 6 \cdot 124,\quad 906 = 7 \cdot 124 + 38.
Only $7.45\$7.45 fits, with k=6k = 6. (It is also reachable: each time, break a quarter into nickels and a nickel into pennies, then feed in a penny.)

The answer is (D) $7.45\boxed{\textbf{(D)}\ \$7.45}.

Why this works

When a process has a few repeatable moves, ask what each move does to a single summary quantity. Two moves leave the total alone and the third shifts it by a fixed amount, so the total mod 124124 is invariant. Multiple choice then reduces to five remainder checks, and the invariant alone eliminates four options without any construction.

The trap

Simulating the machine step by step, or testing whether each choice is a multiple of 124 instead of whether it leaves remainder 1.

Common mistakes

  • Simulating the machine step by step, or testing whether each choice is a multiple of 124 instead of whether it leaves remainder 1.
  • Miscounting the gain as 125125 (forgetting the penny that was surrendered), which makes no choice work.

Techniques

Use an invariant, parity, or coloring argument