How many positive integers
such that the decimal expansion of
ends with exactly 20152015 zeros.
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Using the formula for the number of trailing zeroes:
f ( n ) = ∑ i = 1 ⌊ lo g 5 n ⌋ ⌊ 5 i n ⌋
we can see that
f ( 5 k − 1 ) < f ( 5 k ) = f ( 5 k + 1 ) = f ( 5 k + 2 ) = f ( 5 k + 3 ) = f ( 5 k + 4 ) < f ( 5 k + 5 ) .
So there are 5 numbers n for which f ( n ) = x or in this case 2 0 1 5 2 0 1 5 .