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Jason counts up from 1 to 9 skipping the even numbers, and then immediately counts down without skipping any of the numbers, and then back up to 9 skipping the evens, and so on, alternately counting up and down

\((1, 3,5,7,9,8,7,6,5,4,3,2,1,3,5,\ldots )\)

What is the \(1000^{th}\) integer in his list?

ant101  Sep 1, 2018
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There is a cycle of 13 numbers that repeat, counting up from 1 and down to 1. So, we have:

1000 mod 13 = 12th number in the 13-number cycle. Therefore, the 12th number will be a 2 as the 1000th integer.

Guest Sep 2, 2018

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