Question 1179135
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Prove 4•10^(2n)+9•10^(2n-1)+5 is divisible by 99 for n is in N.
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<pre>
The given number is


    {{{4*10^(2n) + 9*10^(2n-1) + 5}}}.    (1)



    +---------------------------------------------------------------+
    |    Notice that the number  {{{10^m}}}  gives the remainder 1,      |
    |    when is divided by 9, for ANY positive integer m.          |
    +---------------------------------------------------------------+



It implies that the first addend in  (1)  gives the remainder 4, when is divided by 9.

The second addend in  (1)  is divisible by 9 without the remainder (OBVIOUSLY).


    It makes it OBVIOUS that the number (1) is divisible by 9 without the remainder.



Next, the number (1) has the form  49000 . . .005, where 

    - the leading digit 4 is located in some <U>O D D</U> position (2n+1), counting from the most right "ones" position;

    - the digit 9 is placed in the next (<U>E V E N</U>) position (2n);

    - and the last, "ones" digit 5 is located in the most right position number ONE (counting from the right);

    - while all the other digits are zeros.


Applying the "divisibility by 11 rule", we see that the number (1) has alternate sum of digits  4 - 9 + 5 = 0.


        Since the alternate sum of digits equal 0 (i.e. is divisible by 11),
        it means that the number (1) itself is divisible by 11,
        according to the "divisibility by 11 rule.



Thus the number (1) is multiple of 9 and 11 --- HENCE, it is multiple of 99.
</pre>

Solved (mentally), explained and completed.


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On the "divisibility by 11 rule" see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-11-rule.lesson>Divisibility by 11 rule</A>  

in this site.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;*******************************************************

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;An integer number is divisible by &nbsp;<B>11</B>&nbsp; if and only if 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the alternate sum of its digits is divisible by &nbsp;<B>11</B>.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;*******************************************************