SOLUTION: Using binomial theorem what is the Last three digits of the number 27^27?

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Question 1095973: Using binomial theorem what is the Last three digits of the number 27^27?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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1.  27%5E27 = %28%2827%5E3%29%5E3%29%5E3.

    Let us go up on this upstairs from the bottom to the top step by step.



2.  First consider the number  27%5E3.  It is  19683:

    27%5E3 = 19683.

     Write it in the form  27%5E3 = 19%2A10%5E3+%2B+683.


     It is clear that for  %2827%5E3%29%5E3  and  for  %28%2827%5E3%29%5E3%29%5E3  the last three digits are determined by last 
     
     three digits "683" of the number  19%2A10%5E3+%2B+683.

     The part  19%2A10%5E3  does not affect the last three digits of the number  %2827%5E3%29%5E3.

     It is exactly what the binomial theorem says and provides in this situation.


     Therefore, in finding the three last digits of the number  %2827%5E3%29%5E3  we can track only  for  683%5E3  and do not concern about other terms.


     It implies that the last three digits of the number  %2827%5E3%29%5E3  are exactly the same as the last three digits of the number  683%5E3.


     The number  683%5E3 = 318611987,  as easy to calculate (I used Excel in my computer),  so its last three digits are  987.



3.  Now we can make the next (and the last) step up on this upstairs in the same way.

    The last three digits of the number  %28%2827%5E3%29%5E3%29%5E3  are the same as the last three digits of the number  987%5E3,  and it is easy to calculate.

    987%5E3 = 961504803,    and its last three digits are  803.


    Therefore, the last three digits of the number  27%5E27  are 803.

Solved.