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## 'Weekly Challenge 8: Sixinit' printed from http://nrich.maths.org/

Yatir Halevi, Maccabim and Reut High-School, Israel proved that
$n^3+11n$ is always divisible by $6$ using modular arithmetic
modulus $6$.

For each $n$, we can have a remainder of either: $0$, $1$, $2$,
$3$, $4$ or $5$ when divided by $6$ and this is called the residue
modulo $6$.

For $n^3$ we get the following residues: $0$, $1$, $2$, $3$, $4$,
$5$ respectively (to $n$). For $11n$ we get the following residues:
$0$, $5$, $4$, $3$, $2$, $1$ respectively (to $n$). Combining $n^3$
and $11n$ (respectively) we get a residue $0$ because: $0+0=0$ (mod
$6$), $1+5=6=0$ (mod $6$), $2+4=6=0$ (mod $6$), $3+3=6=0$ (mod
$6$), $4+2=6=0$ (mod $6$), $5+1=6=0$ (mod $6$). This means that we
get a zero residue when dividing by $6$, or in other words,
$(n^3+11n)$ is a multiple of $6$ or $6$ divides $n^3+11n$.

Kookhyun Lee gave another proof that all the terms are divisible by
$6$. Consider: $$n^3 + 11n = n^3 + 12n - n = n(n^2-1) + 12n.$$ This
must be a multiple of 6 because $n(n^2-1)$ can be written as
$(n-1)\times n \times (n+1)$. Any multiple of three consecutive
integers is a multiple of $6$ because it contains a multiple of two
(an even number) and a multiple of three.