Sixinit
Choose any whole number n, cube it, add 11n, and divide by 6. What do you notice?
Age
16 to 18
Challenge level
Problem
Choose any whole number $n$, calculate $n^3 + 11n$, and divide by 6.
For what values of $n$ is your answer a whole number?
Can you explain why?
Did you know ... ?
Modular arithmetic is widely used to verify automatically whether a number has been correctly entered to a system. Each identification number (e.g. for passports, bank accounts, credit cards, ISBN book numbers and so on) obeys a rule which makes it easy to check (most of the time) whether or not the number has been copied correctly. For this reason such numbers are also called check codes.
Modular arithmetic is widely used to verify automatically whether a number has been correctly entered to a system. Each identification number (e.g. for passports, bank accounts, credit cards, ISBN book numbers and so on) obeys a rule which makes it easy to check (most of the time) whether or not the number has been copied correctly. For this reason such numbers are also called check codes.
Student Solutions
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.