find 0's at end of 1*2^2 *3^3 *......100^100??
Let be the maximum such that
First note that so the number of 0s at the end of will be exactly
For example for we have , note that the coefficients of the multiples of 5 are the maximum powers of 5 dividing these numbers. This comes from - (k,5)=1- So the maximum power of 5 dividing the number is itself multiplied by the maximum such that
Now since it follows that the maximum such that is
Thus:
It may also be written as: ( note that appears in as many sums as the maximum such that )
PS- That's not a factorial
Hello, chandni!
Opalg pointed out a big error in my calculatitons . . . *blush*
Find the number of 0's at the end of: .
Every factor of 5 (combined with an even number) produces a final zero.
How many 5's are in that product?
Zeros arise from: .
We are concerned with the factors of 5 (only).
The sum of the series is
. . Hence, the product contains a factor of . . . . no
I overlooked these factors: .
. .
My total was short by: . fives.
The product contains a factor of: .
Therefore, the product ends in 1300 zeros.
Thank you, Opalg!
.