### Purr-fection

What is the smallest perfect square that ends with the four digits 9009?

### Old Nuts

In turn 4 people throw away three nuts from a pile and hide a quarter of the remainder finally leaving a multiple of 4 nuts. How many nuts were at the start?

### Mod 7

Find the remainder when 3^{2001} is divided by 7.

# Weekly Challenge 8: Sixinit

##### Stage: 5 Short Challenge Level:

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.