\[4n^3+6n^2+14n\]

is divisible by 12 for every integer \[n\]

. We can do this by looking at the remainders of \[, \]

on division by 6 - the factor 2 means that \[n(2n^2+3n+7)\]

needs to be divisible by 6.\[n\] |
Remainder of \[n\] on Division By 6 |
Remainder of \[2n^2+3n+7\] After Division By 6 |
Remainder of \[n(2n^2+3n+7)\] After Division By 6 |

\[6k+1\] |
1 | 12 | 0 |

\[6k+2\] |
2 | 21/6=3 remainder 3 | 0 |

\[6k+3\] |
3 | 34/6=5 remainder 4 | 0 |

\[6k+4\] |
4 | 51/6=8 remainder 3 | 0 |

\[6k+5\] |
5 | 72/6=12 remainder 0 | 0 |

\[6k+6\] |
0 | 97 | 0 |