Weekly Challenge 35: Clickety Click and All The Sixes

7289 /www/nrich/html/content/id/7289/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/1398&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/2270&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/1333&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7293&part=solution">Last week's solution</a> </li> </ul> <div> What is the sum of: $$6 + 66 + 666 + 6666 + \cdots + 666666666\cdots6$$ where there are $n$ sixes in the last term? </div> <div class="framework"> Did you know ... ? Many functions, including the trigonometric and exponential functions that you meet in school, can be approximated by infinite power series and good approximations can be found using a finite number of terms. If the series is centred at zero then it can be written in the form $\Sigma_{n=0}^\infty a_nx^n$ where the coefficients depend on the derivative of the function at the origin. The infinite geometric series $1 + x + x^2 + \cdots $ which converges for $|x| < 1$ is the power series for the function $(1 - x )^{-1}$. </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This is the live weekly challenge. Our solution to this weekly challenge will appear sometime next Monday, when the next weekly challenge will appear. We don't publish user solutions for the weekly challenges in the same way as we do for the main problems on the site, but please do discuss the weekly challenges on the Ask NRICH -- follow the link from the main part of the challenge. Sometimes fascinating extensions and explorations will arise from the challenges. Why not share your thoughts, questions and findings on the Ask NRICH thread? We hope that you enjoy the challenges. If you have any comments please email Steve Hewson at post16.nrich@maths.org. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What can you say about the number $111111$? Can you write $666666$ as a series? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This caused a lot of thinking and Alex of Madras College gave the following proof. First consider $S_n = 1 + 11 + 111 + 1111 + \cdots$ to $n$ terms. Each individual term can be written and summed as a geometric series, for example $$1111 = 1 + 10 + 100 + 1000 = \frac{10^4-1}{10 - 1}$$ Hence $$S_n= \frac{10^1 - 1}{9} + \frac{10^2 - 1}{9} + \frac{10^3 - 1}{9} + \frac{10^4 - 1}{9} + ... + \frac{10^n - 1}{9}$$ $$= \frac{10 + 10^2 + 10^3 + 10^4 + ... +10^n }{9} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10}{81} - \frac{n}{9}$$ $$= \frac{10^{n+1}- 10 - 9n}{81}$$ So $6 + 66 + 666 + 6666 \cdots$ to $n$ terms is: $$6( 1 + 11 + 111 + 1111 + ... ) = \frac{2}{3}\Big[ \frac{10(10^n - 1)}{9}- n \Big]$$ Remarkably this result was submitted on the 1st of March by Chong Wenhao Edmund from Singapore, the earliest solution! No doubt the time difference helped. </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 35: Clickety Click and All The Sixes What is the sum of: 6 + 66 + 666 + 6666 ............+ 666666666...6 where there are n sixes in the last term? Sequences, Functions and Graphs Geometric sequence Numbers and the Number System Place value Advanced Algebra Summation of series Collections Weekly Challenge