Question

product if any four consecutive integers is divisible by
plz give ans as fast as possible

Answer 1

Statement: (∀n∈N)[(∏n+3n)|24](∀n∈N)[(∏nn+3)|24]

Proof:

I will induct on n.Base Case: (n=1):1∗2∗3∗4=24 (which is divisible by 24)We can also see that for n=1:(24∣∣(∏n+3n+1))I will induct on n.Base Case: (n=1):1∗2∗3∗4=24 (which is divisible by 24)We can also see that for n=1:(24|(∏n+1n+3))

Inductive Hypothesis:Inductive Step: [ Assume 24∣∣(∏n+3n) and 24∣∣(∏n+3n+1)] For some n∈N[I must show 24∣∣(∏n+4n+1)]Inductive Hypothesis:[ Assume 24|(∏nn+3) and 24|(∏n+1n+3)] For some n∈NInductive Step: [I must show 24|(∏n+1n+4)]

1:2:3:4:∏n+4n+1=(n+4)∗∏n+3n+1=(n∗∏n+3n+1)+(4∗∏n+3n+1)=(∏n+3n)+4∗(∏n+3n+1)24∣∣(∏n+3n)24∣∣(∏n+3n+1)∣∣(4∗∏n+3n+1)24∣∣((n∗∏n+3n+1)+(4∗∏n+3n+1))[By def of ∏][By I.H.][By I.H.][By 2,3 and modular arithmetic]1:∏n+1n+4=(n+4)∗∏n+1n+3[By def of ∏]=(n∗∏n+1n+3)+(4∗∏n+1n+3)=(∏nn+3)+4∗(∏n+1n+3)2:24|(∏nn+3)[By I.H.]3:24|(∏n+1n+3)|(4∗∏n+1n+3)[By I.H.]4:24|((n∗∏n+1n+3)+(4∗∏n+1n+3))[By 2,3 and modular arithmetic]

∴24∣∣(∏n+4n+1)