Weekly challenge 31: Common Divisor

7042 /www/nrich/html/content/id/7042/ 1 2011-03-15T10:57:24 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/704&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/6413&part=">Try this next</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/4352&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7091&part=solution">Last week's solution</a> </li> </ul> <div> Find the largest integer which divides every member of the following sequence: $$ 1^5-1,\ 2^5-2,\ 3^5-3,\cdots\ n^5-n.$$ </div> <div class="framework"> Did you know ... ? Although number theory - the study of the natural numbers - does not typically feature in school curricula it plays a leading role in university at first year and beyond. Having a good grasp of the fundamentals of number theory is useful across all disciplines of mathematics. Moreover, problems in number theory are a great leisure past time as many require only minimal knowledge of mathematical 'content'.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Given the sequence $1^5-1,\ 2^5-2,\ 3^5-3,\cdots \ n^5-n$ we see that \[n^5 - n = n(n^4 - 1) = n(n - 1)(n + 1)(n^2 + 1)\] and it is quite easy to see that $n(n-1)(n+1)(n^2+1)$ is divisible by $2$, $3$ and $5$ for all values of $n$. As $n$, $(n-1)$ and $(n+1)$ are three consecutive integers their product must be divisible by $2$ and by $3$. If none of these numbers is divisible by $5$ then $n$ is either of the form $5k+2$ or $5k+3$ for some integer $k$ and in both of these cases we can check that $n^2 + 1$ is divisible by $5$. Since $2$, $3$ and $5$ are coprime therefore $n^5 - n$ is divisible by $2 \times 3 \times 5$ i.e by $30$. Since the second term of the sequence is $2^5-2 = 30$ therefore the divisor cannot be greater than $30$. Therefore $30$ is the largest number that d ivides each member of the sequence. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Given the sequence $1^5-1,\ 2^5-2,\ 3^5-3,\cdots \ n^5-n$ we see that \[n^5 - n = n(n^4 - 1) = n(n - 1)(n + 1)(n^2 + 1)\] and it is quite easy to see that $n(n-1)(n+1)(n^2+1)$ is divisible by $2$, $3$ and $5$ for all values of $n$. As $n$, $(n-1)$ and $(n+1)$ are three consecutive integers their product must be divisible by $2$ and by $3$. If none of these numbers is divisible by $5$ then $n$ is either of the form $5k+2$ or $5k+3$ for some integer $k$ and in both of these cases we can check that $n^2 + 1$ is divisible by $5$. Since $2$, $3$ and $5$ are coprime therefore $n^5 - n$ is divisible by $2 \times 3 \times 5$ i.e by $30$. Since the second term of the sequence is $2^5-2 = 30$ therefore the divisor cannot be greater than $30$. Therefore $30$ is the largest number that d ivides each member of the sequence. </mdoxml> 2 3 0 0 0 0 1 Weekly challenge 31: Common Divisor A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Stage 5 Core Mapping Document Polynomials AS Admin Stage 5 - Core Mapping Secondary Mapping Document DisplayCabinet