Question

what is the largest integer n such that 33! is divisible by 2n

Answer 1

Let me help you with this question.
We know that,
33! = 33 x 32 x 31 x 30 x ......... x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
33! = 33 x (2)⁵ x 31 x 30  x  ......... x (2)³ x 6 x 5 x (2²) x 3 x (2)¹ x 1
Taking all 2s out, we get:
33! = 2⁵ x 2⁴ x 2³ x 2² x 2¹ ( 33 x 31 x 30 x ...... x 6 x 5 x 3 x 1 )
33! = 2⁵⁺⁴⁺³⁺²⁺¹  ( 33 x 31 x 30 x ...... x 6 x 5 x 3 x 1 )
33! = 2¹⁵  ( 33 x 31 x 30 x ...... x 6 x 5 x 3 x 1 )  .......... (1)
Now,
( 33 x 31 x 30 x ...... x 6 x 5 x 3 x 1 ) contains numbers 
6 , 10 , 12 , 14 , 18 , 20 , 22 , 24 , 26 , 28 , 30 which have 2 as their common factor.
So, (6,10,12,14,18,20,22,24,26,28,30) = (2x3,2x5,2x2x3,2x7,2x3x3,2x2x5,2x11,2x2x2x3,2x13,2x2x7,2x3x5)
(6,10,12,14,18,20,22,24,26,28,30) = (2,2,2²,2,2,2²,2,2³,2,2²,2)
The product of these numbers is
(2x2x2²x2x2x2²x2x2³x2x2²x2) = 2¹⁶
Put this value in equation (1), we get:
33! = 2¹⁵  ( 33 x 31 x 30 x ...... x 6 x 5 x 3 x 1 )
33! = 2¹⁵ . 2¹⁶
33! = 2³¹
which shows that 31 is the largest number such that 33! is divisible by 2ⁿ