In this expression, what are the last 2 digits? $\displaystyle \prod_{n=0}^{99} 99^n$
The last two digits are 99.
Explain.
productfor(n, 0, 99, (99^n)) mod 10^10==7087255001 - these are the last 10 digits.
\(\displaystyle \prod_{n=0}^{99} 99^n\)
\(99^{4950}\)
As the guest before mentioned the last 2 digits are 01
-Vinculum
So the guest in 3rd post was right
\(99^0*99^1*99^2* .....*99^{99}\\ =99^{0+1+2+3+4+...99}\\ =99^{\text{an even number}}\)
99^1=99
99^2=9801
99^3=ends in 99
99^4 ... ends in 01
etc
99 ^ (any even positive integer) will end in 01
+344 
