Question 1207676
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Find the remainder when  {{{40^13}}}  is divided by  81.
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        It is clear that the mathematical meaning of this problem is not to follow literally 

        the written formula.


        Its meaning is to decrease / (to reduce)  the degrees and values of participating numbers 

        to make calculations easier using standard properties of operations of modular arithmetic.



<pre>
Following this idea, I write  40 = 36 + 4,

    {{{40^13}}} = {{{(36+4)^13}}}.


Next we should apply the Newtonian binomial formula.


It will give the sum of the terms  {{{C[13]^k*36^k*4^(13-k)}}},  k = 0, 1, 2, 3, . . . , 13.


All the terms with k >= 2  will be zero by modulo 81, since 36 = 9*4.


Therefore, we can exclude all these terms from our consideration.


So, the terms under our consideration are the terms with k= 0 and k= 1, or

    {{{C[13]^0*4^13}}} + {{{C[13]^1*36*4^12}}} = {{{4^13}}} + {{{13*36*4^12}}}.


This expression is easy to calculate using a regular calculator or Excel spreadsheet; 
its value is 7918845952.


Finally,  7918845952 mod 81 is 22  (use long division or Excel function mod)


So, the  <U>ANSWER</U>  is 22.
</pre>

Solved.



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