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Find the last \(3\) digits of \(101^{101}\).

 Apr 12, 2021
 #1
avatar+31575 
+3

1012 = 10201

1013 = 1030301

1014 = 104060401  

 

continuing the pattern    101101    last 3 digits would be 101

 Apr 12, 2021
 #2
avatar+118626 
+1

To confirm EP's  answer....look at the last  two terms of the binomial expansion  of (101)^101  =   (100 + 1)^101  =

 

C(101, 100) * 100 * 1^100  +  C(101, 101) * 100^0  * 1^101  =

 

101 * 100  +  1 * 1 * 1  =

 

10100  +  1  =

 

10101

 

cool cool cool

 Apr 12, 2021

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