prove that 33! is divisible by 2^15 what is the largest integer n such that 33! is divisible by 2^n - Maths - Permutations and Combinations

Question

prove that 33! is divisible by 2^15 what is the largest integer
n such that 33! is divisible by 2^n?

Shipra D. · Accepted Answer

33!=33×32×31×30×29× ×16
×8×7×6×5×4×3×2×1=33×25×31×30×29×
×24
×23×7×6×5×22×3×21×1Now take all 2s out we get33!=25
24 23 22 2133××31×30×29×
×1=21533××31×30×29×
×1 [Since, xa
xb=xa+b]This shows that 33! is divisible by 215 
Consider,33!=21533××31×30×29×
×133××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=216Therefore, we have 215×216=231Thus, 31 is the largest integer such that 33! is divisible by 2n

prove that 33! is divisible by 2^15.what is the largest integer n such that 33! is divisible by 2^n?