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Find the remainder when $30^{99} + 61^{100}$ is divided by 31.

 Jan 26, 2021
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30^99  =  ( 31   - 1 )*99

61^100  =  ( 62  -1 )^100

 

The binomial expansion  of the  first   will   have all terms divisible by 31  in every  term except the last  which    =  (-1)^99 = -1

 

The binomial expansion of the  second  will   have some  power of  62  in every term  (each divisible by 31)  except the last term which is  (1)^100  =   1

 

So....adding the  expansions   will  give us  all terms  that will be  divisible  by  31  plus  -1 + 1  =  0

 

So  the remainder of this sum divided  by 31 =    0

 

 

 

cool cool cool

 Jan 26, 2021

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