Prove that 33! Is divisible by 215, what is the largest integer n suc

1.	Prove that 33! Is divisible by 215, what is the largest integer n such that 33! Is divisible by 2n?
Answer» 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.

Prove that 33! Is divisible by 215, what is the largest integer n such that 33! Is divisible by 2n?

Answer»

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.