Problem 25 · 1998 AJHSME
Stretch
Algebra & Patterns
work-backwardinvariant
Three generous friends redistribute their money as follows: Amy gives Jan and Toy enough to double each of their amounts; then Jan gives Amy and Toy enough to double theirs; finally Toy gives Amy and Jan enough to double theirs. Toy had $36 at the beginning and $36 at the end. What is the total amount the three friends have?
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Answer: D — $252.
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Hint 1 of 2
Money is only handed around, never created or destroyed — so the grand total is the SAME at every moment. You just need to catch it at one convenient instant.
Still stuck? Show hint 2 →
Hint 2 of 2
Watch Toy. In the first two rounds he's a receiver, and each time his amount is doubled. So follow Toy's pile from $36 up to the moment right before his own turn.
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Approach: use the unchanging total; track Toy through his two doublings
- Key idea: the total never changes, since every step just moves dollars between people. So if we can pin the total at any single moment, we're done.
- Toy receives (and doubles) in rounds 1 and 2: $36 → $72 → $144 just before his own turn.
- On his turn Toy gives away enough to double Amy and Jan, and he ends at $36 — so he handed out 144 − 36 = $108. That $108 was exactly what it took to double Amy and Jan together, meaning they held $108 between them just before.
- At that instant the total is Toy's 144 + the others' 108 = $252 — and since the total is constant, that's the answer.
- Why this transfers: in any 'pass things around' puzzle, first ask what stays fixed (here, the total). An invariant lets you ignore the messy middle and read the answer off one clean snapshot.
Another way — work backward from the end:
- Suppose the total is T. At the very end all three have whole amounts and Toy has $36. The last move (Toy doubling Amy and Jan) means just before it, Amy and Jan each had half their final amount, and Toy had everything else.
- Peeling back each doubling step in reverse keeps the total T fixed at every stage. Using Toy's $36 start, the backward chain pins T = $252, matching the invariant method — a good check that working forward, backward, or by invariant must all agree.
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