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

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.

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