Question

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 $1000^{\text{th}}$ integer in his list?

Answer 1

The 1000th integer in Jason's list is 7.

Jason's pattern. He starts counting up from 1, skipping even numbers, until he reaches 9. Then, he counts down without skipping any numbers until he reaches 1 again. This pattern repeats. First, let's find the length of each cycle. In the counting-up phase, he covers five numbers (1, 3, 5, 7, 9), and in the counting-down phase, he covers seven numbers (9, 8, 7, 6, 5, 4, 3, 2, 1). So, each cycle has a length of 5 + 7 = 12 integers. Now, to find the 1000th integer in his list, we can divide 1000 by the length of each cycle: 1000 ÷ 12 = 83 R 4

This means that after completing 83 cycles, there are 4 integers left in the sequence. Since the sequence repeats every 12 integers, the 1000th integer will be the same as the 4th integer in the cycle. Counting the first few numbers in the cycle (1, 3, 5, 7, 9, 8, 7, 6, 5, 4, 3, 2), we see that the 4th integer is 7.

Therefore, the 1000th integer in Jason's list is 7.