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There is a problem asking me to prove the product of 3 consecutive integers is always divisible by 6 by using induction and not using the fact that one of the 3 numbers must always be divisible by 3.

I know the standard solution to this problem where you can say at least one of the 3 numbers is divisible by 2 since it is even, and since there are 3 numbers, at least one of them is also divisible by 3, but I cannot really use these facts properly.

I just need a starting point or a hint if anyone can help.

Nikshith Konnuru
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I think you essentially need to use those facts, even in an induction argument.

Assume without loss of generality that we have $x=n(n+1)(n+2)$ where $n\geq 1$. For the base case, we have 6 divides $1(2)(3)=6$, which is true. For the inductive step, assume 6 divides $x=n(n+1)(n+2)$. Then we want to show that 6 divides $y=(n+1)(n+2)(n+3)$. Since 6 divides $x$, 2 divides $x$ and 3 also divides $x$. As 2 is prime, it must divide either $n$ or $n+1$ or $n+2$. If $n+1$ is even, 2 divides $n+1$, and if $n+1$ is odd, 2 divides $n+2$, so 2 divides $y$. Now 3 is prime, so it must also divide either $n$ or $n+1$ or $n+2$. If 3 divides $n+1$ or $n+2$, then 3 divides $y$ and so 6 divides $y$ as claimed. If 3 does not divide $n+1$ or $n+2$, then it has to divide $n+3$. Hence 3 divides $y$ and 6 divides $y$ as claimed.

Bill Dubuque
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