Problem 1 · AMC 8 Stretch
Core
Logic & Word Problems
Number Theory
work-backwardpattern-recognition
Two players take turns removing \(1\), \(2\), \(3\), or \(4\) counters from a pile that starts with \(27\) counters. The player who takes the very last counter wins. Should you go first or second, and what is your winning plan? (Give the number of counters to take on your first move.)
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Answer: Go first and take 2 (leaving 25)
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Hint 1 of 4
There are lots of ways to start, but only one way to finish (grab the last few counters). When the END is clearer than the start, work backward from the end.
Still stuck? Show hint 2 →
Hint 2 of 4
Ask: how many counters can I leave for my opponent so that no matter what they take (\(1\), \(2\), \(3\), or \(4\)), I can take the rest and grab the last one?
Still stuck? Show hint 3 →
Hint 3 of 4
If you leave exactly \(5\), you win: whatever they take, you take the rest of those \(5\). Now work back one more step. To be able to leave \(5\) next time, what should you leave now?
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Approach: Working backward from the finish to find the safe positions
- To win you want to take the last counter, so after your final move there are \(0\) left. Work backward from there.
- The magic numbers to leave for your opponent are the multiples of \(5\). Leaving \(5\) works: if they take \(k\) (one of \(1, 2, 3, 4\)), you take the remaining \(5 - k\) and grab the last counter.
- So every turn you hand them a multiple of \(5\): \(25, 20, 15, 10, 5, 0\). Whatever they remove, you remove enough to land on the next multiple of \(5\).
- Since \(27 = 25 + 2\), you go FIRST and take \(2\), leaving \(25\). After that always leave a multiple of \(5\); your last move leaves \(0\), so you took the last counter and win.
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