Lesson Find the last three digits of these numbers

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Find the last three digits of these numbers


Problem 1

Using binomial theorem,  find the Last three digits of the number  27%5E27.

Solution

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.

Problem 2

Find the last three digits of the number  126%5E2018.

Solution

      Intermediate statement.   The last three digits of the number  126%5En  are  376,  for any  n >= 3.
    First consider the number  126%5E2.  It is  15876.

    Next                       126%5E3.  It is  2000376.

    Next                       126%5E4.  It is  252047376.


    Notice that  126%5E3  and  126%5E4  have the last 3 digits  376.


    Now it is clear that for  126%5E%28n%2B1%29  the last three digits are determined by last three digits  of the number  126%5En.


    To prove it, write  126%5En  in the form

       126%5En = 1000*N + 376,


    and notice that  376*126 = 47376  has the three last digits  376.


    It implies, by the method of Mathematical induction, that ALL  the numbers  126%5En  have the three lest numbers 376, starting from n = 3.

Thus the intermediate statement is proved,  and the solution to the problem is completed.


My other lessons in this site on miscellaneous problems on divisibility of integer numbers are
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    - Introductory problems on divisibility of integer numbers
    - Finding Greatest Common Divisor of integer numbers
    - Relatively prime numbers help to solve the problem
    - Solving equations in integer numbers
    - Quadratic polynomial with odd integer coefficients can not have a rational root
    - Proving an equation has no integer solutions
    - Composite number of the form (4n+3) must have a prime divisor of the form (4n+3)
    - Problems on divisors of a given number
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    - How many 3-digit numbers are not divisible by 2; not divisible by 3; not divisible by either 2 or 3
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    - Find the sum of digits of integer numbers
    - Two-digit numbers with digit "9"
    - Find a triangle with integer side lengths and integer area
    - Math circle level problem on the hundred-handed monster Briareus
    - Math Circle level problem on lockers and divisors of integer numbers
    - Nice entertainment problems related to divisibility property
    - Solving problems on modular arithmetic
    - Using the little Fermat's theorem to solve a problem on modular arithmetic
    - OVERVIEW of miscellaneous solved problems on divisibility of integer numbers


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