Prove that the product of three consecutive no.s is divisible by 6.

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Prove that the product of three consecutive no.s is divisible by 6.

Mathematics
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X+(X+1)+(X+2):6 X=1--->1+1+1+1+2:6 6:6 X=k---->k+k+1+k+2:6 3k+3:6 3(k+1):6 X=K+1------>k+1+K+2+K+3:6
3k+6:6 w8 i stuck here...:S
I think u added every term. Are we not to multiply x(x+1)(x+2)? Plz explain in detail.

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no u add them....as far as i know...
look u solve it with smth tht in english may b called *****''mathematical induction''*****
do u know it?
No. Actually, I am in grade 10. So far we have not been taught. Actually, it is related to Euclid's Division Lemma, our teacher gave, I believe.
Euclid's????? Idk..maybe....I'm sorry that's how we solve them in here...
Ok. no problem. are u in high school or college? have u taken in maths olympiad? can u suggest some books and preparation tips? u r from which country?
I'm in hs,12th grade....I'm from Albania...and...i'm not very good at maths....so no olympiad
No problem sister. I m from India. Plz tell me why after my name (good answer) tab is not coming. I wanted to get medals!!!!!
take nC3 this is a integer. but nC3= n(n-1)(n-2)/3!=n(n-1)(n-2)/6 but we have nC3 is integer so n(n-1)(n-2)/6=k where k is integer so n(n-1)(n-2) is divisible by 6.
u can also prove by induction
n(n+1)(n+2) = n(n^2 +3n +2) =n^3 +3n^2 + 2n try factoring out a 6 \[6(\frac{n^{3}}{6} + \frac{n^{2}}{2} + \frac{n}{3})\] combine fractions, LCD of 6 \[6(\frac{n^{3} + 3n^{2} + 2n}{6})\] 6's cancel leaving the original expression therefore a 6 can be pulled out of expression evenly product is divisible by 6
let f(n) =n(n+1)(n+2)=6k n^3 +3n^2+2n=6k putting m=n+1 f(n+1)=(n+1)^3+3(n+1)^2+2(n+1)= n^3 + 6n^2 +11n+6 =n^3+3n^2+2n+3(n^2+3n+2) =6k+3(n+1)(n+2) now any one of n+1 or n+2 must be even so f(n+1) =6k+6l so f(n+1) is divided by 6. so by induction we can say that n(n+1)(n+2) is divided by six

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