Prove that for any integer the number $2n^3+3n^2+7n$ is divisible by $6$.
Problem
Source: Kosovo MO 2014 Grade 9, Problem 1
Tags: number theory
07.06.2021 21:47
$$2n^3+3n^2+7n \pmod{6} = 2n^3+3n^2+n = n(2n+1)(n+1)$$ We proceed with casework. $$n \equiv 0\pmod{6}$$where $n$ is divisible by $6$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 1\pmod{6}$$$2n+1 \equiv 2\pmod{6}$ and $n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 2\pmod{6}$$$n \equiv 2\pmod{6}$ and $n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 3\pmod{6}$$$n \equiv 3\pmod{6}$ and $n+1 \equiv 4\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 4\pmod{6}$$$n \equiv 4\pmod{6}$ and $2n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 5\pmod{6}$$where $n+1$ is divisible by $6$ thus $2n^3+3n^2+n$ is also divisible by $6$ We have exhausted all cases and we have proven that it works for all of them thus we are done.
07.06.2021 21:56
you can shorten the solution by noting that the expression is always divisible by $2$ and it is not difficult to see that the expression is always divisible by $3$
08.06.2021 00:03
Apply induction
29.05.2022 14:03
Let $S = 2n^3+3n^2+7n$. $S \equiv n^2+n = n(n+1) \equiv 0 \pmod 2$ $S \equiv -n^3+n = -n(n-1)(n+1) \equiv 0 \pmod 3$ Done.
29.05.2022 15:23
Note that $2n^3+3n^2+7n\equiv 2n^3+3n^2+n\pmod 6$. We will prove that $6\mid 2n^3+3n^2+n$ for each $n$, which solves the problem. Claim: $2n^3+3n^2+n\equiv 0\pmod 2$. Note that $2n^3+3n^2+n\equiv 3n^2+n=n(3n+1)\pmod 2$. If $n$ is odd, then $3n+1$ is even, so $n(3n+1)$ is even, and if $n$ is even, we also have $n(3n+1)$ is even. $\blacksquare$ Claim: $2n^3+3n^2+n\equiv 0\pmod 3$. Note that $2n^3+3n^2+n\equiv 2n^3+n=n(2n^2+1)\pmod 3$. If $3\mid n$, then clearly $3\mid n(2n^2+1)$. If $3\nmid n$, then $n^2\equiv 1\pmod 3$, so $3\mid 2n^2+1\mid n(2n^2+1)$. $\blacksquare$ So $2n^3+3n^2+n\equiv 0\pmod 6$, as desired.
29.05.2022 17:11
Solution: $n(2{{n}^{2}}+3n+7)=n\left( 3(n(n+1)+2)+1-{{n}^{2}} \right)$ it is enought that $6\left| (n-1)n(n+1) \right.$ and this is true, because product of 2 consecutive integer number it is divisible by 2, and from 3 too.
31.05.2022 07:31
Notice $$2n^3+3n^2+7n\equiv 2n^3+3n^2+n\equiv n(n+1)(2n+1)\pmod{6}.$$Since $$\frac{n(n+1)(2n+1)}{6}=1^2+2^2+\dots+n^2\in\mathbb{N}$$by induction, $6\mid n(n+1)(2n+1).$ $\square$
06.05.2023 13:33
$n \equiv 0 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 1 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 2 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 3 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 4 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 5 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ Therefore, $6|2n^3+3n^2+7n$ $\blacksquare$
22.12.2024 07:59
Use CRT and win.