Answers

2015-04-25T14:15:59+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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!.
1 5 1
2015-04-25T15:27:06+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

============================
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

1 5 1
click on Thanks (blue) button above pls