largest integer

We know that : 33! = 1 × 2 × 3 × 4 ............. × 32 × 33

⇒ 33! = (2 × 4 × 6 ........ × 32) (1 × 3 × 5........ × 33)

⇒ 33! = 2¹⁶ (1 × 2 × 3 × 4 ...... × 15 × 16) (1 × 3 × 5 ...... × 33)

⇒ 33! = 2¹⁶ (2 × 4 ........ × 16) (1 × 3 ..... × 15) (1 × 3 ........×33)

⇒ 33! = 2¹⁶ (1 × 2 × ........ × 8) (1 × 3 × ....... × 15) (1 × 3 ...... × 33)

⇒ 33! = 2²⁴ (2 × 4 × 6 × 8) (1 × 3 × 5 × 7) (1 × 3 × ..... × 15) *1 × 3 × .......× 33)

⇒ 33! = 2²⁴ . 2⁴ (1 × 2 × 3 × 4) (1 × 3 × 5 × 7) (1 × 3 .......... × 15) (1 × 3 ........ × 33)

⇒ 33! = 2²⁸ (2 × 4) (1 × 3) (1 × 3 × 5 × 7) (1 × 3 × ........ × 15) (1 × 3 × ...... × 33)

⇒ 33! = 2³¹ (1 × 3) (1 × 3 × 5 × 7) (1 × 3 × ....... × 15) (1 × 3 × ....... × 33)

Thus the maximum value of n for which 33! is divisible by 2ⁿ is 31