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2010/11 AHSMC Part 1 Problem 5

Problem 5 of 16

In the sequence 1,2,3,4,6,7,8,9,…1, 2, 3, 4, 6, 7, 8, 9, \ldots, obtained by deleting the multiples of 55 from the sequence of the positive integers, the 20102010th term is

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Solution

Note that 2010×54=2512.52010 \times \frac{5}{4} = 2512.5. There are 502502 multiples of 55 from 55 to 25102510 inclusive. Hence 25122512 is the (2512−502)(2512 - 502)-th or 20102010-th number in the punctured sequence. The answer is (B).

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