Problem 2 · AMC 8 Stretch
Core
Logic & Word Problems
Number Theory
work-backwardpattern-recognition
Same game as before: take \(1\), \(2\), \(3\), or \(4\) counters from a pile of \(27\). But now the player who takes the LAST counter LOSES. 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 1 (leaving 26)
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Hint 1 of 4
Now losing means being forced to take the final counter. So you want to hand your opponent exactly \(1\) counter at the very end — then they must take it and lose.
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Hint 2 of 4
Work backward from leaving \(1\). What is the next safe number to leave below that?
Still stuck? Show hint 3 →
Hint 3 of 4
Just like the last game used multiples of \(5\), this game uses numbers that are \(1\) more than a multiple of \(5\): \(1, 6, 11, 16, 21, 26\).
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Approach: Working backward to force the opponent to take the last counter
- You win by forcing your opponent to take the last counter, so on your final turn you want to leave exactly \(1\).
- Leaving \(1\) is safe (they must take it and lose). Working backward, the safe numbers to leave are one more than a multiple of \(5\): \(1, 6, 11, 16, 21, 26\).
- From any of these, whatever your opponent takes (\(k\)), you take \(5 - k\) to get back to the next safe number.
- Since \(27 = 26 + 1\), you go FIRST and take \(1\), leaving \(26\). Then always leave a number that is \(1\) more than a multiple of \(5\); your last move leaves \(1\), your opponent must take it, and they lose.
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