Question 1137877: 2017 students arrange themselves in a single file.
In the first round of counting, they number themselves 1, 2, 3, 1, 2, 3, 1, 2, … and so on from left to right.
In the second round of counting, they number themselves 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1,2, … and so on from right to left.
Find the number of students whose difference between their numbers in the first and second rounds of counting is 1.
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!
The period of the first sequence of numbers is 3; the period of the second sequence is 5. That means the period of the absolute difference between the two sequences is 15.
So make a diagram to show how many students in each group of 15 have numbers in the two sequences that differ by 1:
1st sequence: 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
2nd sequence: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
|diff| = 1? * * * * *
5 students in each group of 15 have numbers in the two sequences that differ by 1.
2017/15 = 134 remainder 7
Only 1 of the first 7 students in each group has numbers in the two sequences that differ by 1. So the number of students in the group of 2017 students who have numbers in the two sequence that differ by 1 is
134*5 + 1 = 671
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