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avatar+282 

In this expression, what are the last 2 digits? $\displaystyle \prod_{n=0}^{99} 99^n$ 

 Jun 17, 2022
 #1
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0

The last two digits are 99.

 Jun 17, 2022
 #2
avatar+282 
+1

Explain.

hipie  Jun 17, 2022
 #3
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+1

productfor(n, 0, 99, (99^n)) mod 10^10==7087255001 - these are the last 10 digits.

 Jun 17, 2022
 #4
avatar+578 
+2

\(\displaystyle \prod_{n=0}^{99} 99^n\)

 

\(99^{4950}\)

 

As the guest before mentioned the last 2 digits are 01

 

-Vinculum

 Jun 18, 2022
 #5
avatar+578 
+1

So the guest in 3rd post was right

Vinculum  Jun 18, 2022
 #6
avatar+117746 
+1

\(99^0*99^1*99^2* .....*99^{99}\\ =99^{0+1+2+3+4+...99}\\ =99^{\text{an even number}}\)

 

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

 Jun 18, 2022

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