Question

The maximum value of n such that 2^n divides 21!

Answer 1

In order to find the maximum value of $n$ such that $2^n$ divides $21!$, it is enough to find the exponent of $2$ in the prime factorization of $21!$ :

$\left\lfloor\dfrac{21}{2}\right\rfloor+\left\lfloor\dfrac{21}{2^2}\right\rfloor+\left\lfloor\dfrac{21}{2^3}\right\rfloor+\left\lfloor\dfrac{21}{2^4}\right\rfloor$

$=10+5+2+1\\=18$

That means $18$ is the maximum value of $n$ such that $2^n$ divides $21!$.

Answer 2

21 !  =  2 * 3 * 4 * 5 * ... 18 * 19 * 20 * 21

The factors of 21 !  which are multiples of 2 are :  (ignore odd integers)
  = 2 * 4 * 6 * 8 ...  * 18 * 20 
  =  2¹°  *  ( 1 * 2 * 3 * .....  * 8 * 9 * 10 ),    now we ignore the odd integers.
  = 2¹° * 2⁵ * ( 1 * 2 * 3 * 4 * 5 )      ,  ignore the odd integers
  = 2¹° * 2⁵ * 2²  * ( 1 * 2 )
  = 2¹°⁺⁵⁺²⁺¹ 
  =2¹⁸
maximum value of n such that 2^n divides 21!  is  18

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Another way is:

 The power of 2 in the value of  N! is given by the sum:

          [ N / 2 ]  + [ N / 4 ]  + [ N / 8 ]  + ...  + 2  + 1
                   where  [ X ]   is the greatest integer function with value <= X.

     = [21/2 ]  +  [ 21/4 ] + [ 21 / 8 ] + [21/16 ]
     = 10 + 5 + 2 + 1
     = 18