Question 1199903
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If 63^n is the greatest power of 63 that divides into 122!, find n.
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<pre>
{{{63^n}}} = {{{(9*7)^n}}} = {{{3^(2n)*7^n}}}.


The maximum degree  {{{3^k}}}  which divides  122!  is  

      [{{{122/3}}}] + [{{{122/9}}}] + [{{{122/27}}}] + [{{{122/81}}}] = 

    =  40    +   13    +   4    +   1     = 58.


  +---------------------------------------------------------------------------+
  | Here  [...]  is the floor-function (the maximum-integer-lesser function). |
  +---------------------------------------------------------------------------+


The maximum degree  {{{7^k}}}  which divides  122!  is  

      [{{{122/7}}}] + [{{{122/49}}}] = 17  + 2 = 19.


So,  n  must be the greatest integer such that 2n is lesser than or equal to 58
         and (at the same time) be the greatest integer such that n is lesser than or equal to 19.


From these conditions, n = 19.     <U>ANSWER</U>
</pre>

Solved.