Again it is dancing!
Find the number of positive integers which satisfy the condition
Notation : denotes the floor function .
The answer is 2499.
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1 solution
We can test the integers that work for the 1st few multiples of 99 and 101. We can start with an integer n = 0, and, by substitution, find that 101n≤x<99(n+1). By substituting n=0, then n=1, then n=2, etc, we can find that the number of solutions in the interval starts at 99 and decreases by 2 each time. Thus, the total number of solutions is the sum 99+97+95....+3+1, or the 1st 49 odd numbers. We can calculate this sum as 2500, but since 0 is nonpositive and thus doesn't count, the total number is 2499.