Trailing zeroes in factorial sum
Number Theory
Level
3
Let . How many trailing zeroes does have?
The answer is 6.
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1 solution
Notice that:
2 ! − 1 ! = 1 ! ( 2 − 1 ) = 1 ⋅ 1 !
3 ! − 2 ! = 2 ! ( 3 − 1 ) = 2 ⋅ 2 !
4 ! − 3 ! = 3 ! ( 4 − 1 ) = 3 ⋅ 3 !
...
2 5 ! − 2 4 ! = 2 4 ! ( 2 5 − 1 ) = 2 4 ⋅ 2 4 !
Adding all of these up, we get S = 0 ! + ( 2 ! − 1 ! ) + ( 3 ! − 2 ! ) + ( 4 ! − 3 ! ) + . . . + ( 2 5 ! − 2 4 ! ) .
Most terms cancel, and we are left with S = 0 ! − 1 ! + 2 5 ! , which is equal to 2 5 ! . The number of trailing zeroes in 2 5 ! is simply ⌊ 5 2 5 ⌋ + ⌊ 5 2 2 5 ⌋ = 6