33! = 33 × 32 × 31 × 30 × 29 … × 16 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 33 × (2)5× 31 × 30 × 29 × … × (2)4… × (2)3× 7 × 6 × 5 × (2)2× 3 × (2)1× 1

Now take all 2’s out we get

33! = (2)5∙ (2)4∙ (2)3 ∙ (2)2∙ (2)1[33 × 31 × 30 × 29 × … × 1] [33 × 31 × 30 × 29 × … × 1]

contains numbers 6, 10, 12, 14, 18, 20, 22, 24, 26, 28, 30,

which have 2 as one of their factors.

So, the product of 2 present in these numbers

= 2 × 2 × 22× 2 × 2 × 22× 2 × 23× 2 × 22× 2

= 216

Therefore, we have 215× 216= 231

Thus, 31 is the largest integer such that 33! Is divisible by 2n.