Power mad!

6401 /www/nrich/html/content/id/6401/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Powers of numbers behave in surprising ways. Can you find convincing arguments that explain why all the statements below are true? <div style="float: left;"><mdo:image alt="This power pylon is unlikely to help..." class="insetimage" height="365" src="Pylon.jpg" width="180"></mdo:image></div> a) $2^{1}, 2^{2}, 2^{3},......, 2^{99}$ are never multiples of $10$. <div>b) $2^{1} + 3^{1}$, $2^{3} + 3^{3}$, $2^{5} + 3^{5}$, ......, $2^{99} + 3^{99}$ are all multiples of $5$.</div> c) $1^{99} + 2^{99} + 3^{99}$ is even d) $1^{99} + 2^{99} + 3^{99} + 4^{99}$ is a multiple of $5$ e) $1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}$ is a multiple of $5$ f) $2^{99} + 3^{99} + 4^{99} + 5^{99} + 6^{99}$ is a multiple of $5$. g) $3^{99} + 4^{99} + 5^{99} + 6^{99} + 7^{99}$ is a multiple of $5$. h) $1^{x} + 2^{x} + 3^{x} + 4^{x} + 5^{x}$ is a multiple of $5$ when x is odd. What other surprising results can you find? Can you explain why they are true? <a href="http://nrich.maths.org/public/viewer.php?obj_id=6824&part=">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> An excellent solution to this problem was sent in by Edward from Carre's Grammar School. Edward took a sophisticated approach and used modular arithmetic - if you are not familiar with modular arithmetic <a style="font-weight: bold;" href="http://nrich.maths.org/public/viewer.php?obj_id=4350">this article</a> will help you follow his reasoning. Here is an outline of what he wrote. (a) $2^1,\ldots,2^{99}$ can never be multiples on ten since any multiple of ten is divisible by $5$, and of course no power of $2$ is divisible by $5$. (b) Edward provided two solutions to this part. Method 1 We use induction. Clearly $2^1+3^1$ is divisible by $5$. Now for the induction hypothesis, suppose $2^k+3^k$ is divisible by $5$ for $k$ odd, i.e. $2^k+3^k=5m$ for some integer $m$. Then $2^{k+2}+3^{k+2}=4\times 2^k+9\times 3^k=4(2^k+3^k)+5(3^k)=4(5m)+5(3^k)=5(4m+3^k)$ and so is divisible by $5$. This completes the induction step, and we have shown that $2^k+3^k$ is divisible by $5$ for any odd $k$. Method 2 This method uses modular arithmetic. Edward observed the following rules: 1. If $a\equiv b$ (mod $m$) then $a^n\equiv b^n$ (mod $m$) 2. If $a\equiv b$ (mod $m$) and $c\equiv d$ (mod $m$) then $a+c\equiv b+d$ (mod $m$) Now $2\equiv 2$ (mod $5$) and $3\equiv -2$ (mod $5$), so for any odd $k$, $3^k\equiv (-2)^k\equiv -2^k$ (mod $5$) using rule (1), and therefore $2^k+3^k\equiv 2^k-2^k\equiv 0$ (mod $5$) by rule (2). In other words, $5$ divides $2^k+3^k$ for any odd $k$. Edward solved the remaining parts using modular arithmetic. (c) $2^{99}\equiv 0$ (mod $2$), and $1^{99}+3^{99}\equiv 1+1\equiv 0$ (mod $2$) so $1^{99}+2^{99}+3^{99}$ is divisible by $2$. (d) $2^{99}+3^{99}\equiv 0$ (mod $5$) by part (b). Similarly, since $4\equiv -1$ (mod $5$), $1^{99}+4^{99}\equiv 1^{99}+(-1)^{99}\equiv 0$ (mod $5$). Therefore $1^{99}+2^{99}+3^{99}+4^{99}$ is divisible by $5$. We will leave parts (e)-(h) for the reader. See if you can use modular arithmetic to prove them. We also received a nice solution from Yike, who concentrated on the last digits of the numbers and found various patterns. Can you prove that these patterns hold for all natural numbers? Why not try using modular arithmetic? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> We like to offer students plenty of opportunities to work mathematically. This problem requires students to analyse patterns, make some generalisations, pose their own questions and explain their findings - all key aspects of good mathematical thinking. <h3>Possible approach</h3> The first few questions are intended to highlight the impact of the units digits on the product of two or more numbers. On the board, write and ask for solutions to 12 x 23 42 x 73 652 x 9883 17852 x 35703 "What do all the solutions have in common?" "What can you tell me about the product of 543789542 and 56398643?" Offer the first couple of problems, give students some time to think about them and then share their convincing arguments. Ensure that the significance of the units digit is well understood. Ask them to work on the next five problems and emphasise the need to provide convincing arguments. This is ideal for a paired activity. "We have discovered that powers of numbers do behave in surprising ways - can you find any other unexpected results?" Ask the students to contribute to a whole class display of their results and their proofs. <h3>Key questions</h3> <div>What patterns can you find in the units digit of ascending powers of 2, 3, 4...?</div> <div>How can you be sure the patterns will continue?</div> <h3>Possible extension</h3> This is an open ended activity which already offers plenty of opportunities for extension work. <h3>Possible support</h3> You might suggest that students draw up 'power tables' so that the cyclical nature of the units digits becomes apparent. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> What is the impact of the units digits on the product of two or more numbers? e.g. what do you notice about the units digits in all the answers to the following: <div style="margin-left: 40px;">2 x 3</div> <div style="margin-left: 40px;">42 x 53</div> <div style="margin-left: 40px;">652 x 43</div> What can you say about the patterns of the units digits of powers of 2, 3, 4, 5...? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><a href="/content/id/6401/icon.jpg">fg</a>h </mdoxml> 2 4 0 0 1 0 0 Power mad! 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