how to find number of digits in a factorial
We know,
log(a*b) = log(a) + log(b)
Therefore
log( n! ) = log(1*2*3....... * n)
= log(1) + log(2) + ........ +log(n)
Now, observe that the floor value of log base
10 increased by 1, of any number, gives the
number of digits present in that number.
Hence, output would be : floor(log(n!)) + 1.Copyright © 2021 Codeinu
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