Weekly Challenge 8: Sixinit

7292 /www/nrich/html/content/id/7292/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/2036&part=">Try this next</a> </li> <li> <a href="http://plus.maths.org/issue12/features/codes/">Read all about it</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/4350">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7060&part=solution">Last week's solution</a> </li> </ul> <div style=""> Choose any whole number $n$ and calculate $n^3 + 11n$. Is the answer a multiple of 6 for all choices of $n$? The challenge is to explain why or why not. </div> <div class="framework"> 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.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 8: Sixinit Choose any whole number n, cube it and add 11n. Is the answer always divisible by 6? If so why? Numbers and the Number System Modulus arithmetic Admin Short problems Collections Weekly Challenge