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2016 AMC 12B Problem 7

Problem 7 of 25EasierNumber TheoryProblem-Solving Techniques

Josh writes the numbers 1,1, 2,2, 3,3, …,\ldots, 99,99, 100.100. He marks out 1,1, skips the next number (2),(2), marks out 3,3, and continues skipping and marking out the next number to the end of his list. Then he goes back to the start of his list, marks out the first remaining number (2),(2), skips the next number (4),(4), marks out 6,6, skips 8,8, marks out 10,10, and so on to the end. Josh continues in this manner until only one number remains. What is that number?

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Solution

The first pass removes the odd numbers, leaving the multiples of 2.2. The second pass removes 2,6,10,…,2,6,10,\ldots, leaving the multiples of 4.4. In general, after the nnth pass only the multiples of 2n2^n remain. The surviving number is the highest power of 22 not exceeding 100,100, which is 26=64.2^6=64. Thus, the correct answer is D.
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Tagged: power of 2 · pattern recognition

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