618 /www/nrich/html/content/99/01/six1/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Can you find what the last two digits of the number 4¹⁹⁹⁹ are?<br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <span class="editorial">The last two digits of the number 41999 are 44. The fact that this problem focuses on the last two digits suggests we should use 'clock' or modulus arithmetic where we are only interested in the remainders after division by 100.</span> <p class="editorial">Well done all of you who explained carefully how you found the pattern in the last two digits of powers of 4 and decided that the answer must be 44. Congratulations to the following people for their solutions: Claire and Rhona of Madras College, St Andrew's; Angela, Geoffrey, Rachel, David and James of Hethersett High School, Norwich; Bithian and Guobin of The Chinese High School, Singapore; and The Key Stage 3 Maths Club at Strabane Grammar School.</p> <p>Proving that the pattern really does go on repeating itself indefinitely amounts to looking at multiples of 100 plus the last two digits, in other words, using arithmetic modulo 100.</p> <p>We write 4 ^<em>a</em> = 100 <em>b</em> + <em>c</em> where <em>a</em> , <em>b</em> and <em>c</em> are whole numbers and 0 <mdo:image alt="leq" src="lte.gif"></mdo:image> <em>c</em> &lt; 99.</p> <p>The first few terms in the cyclic pattern are:</p> <table border="0"> <tbody> <tr> <th>Power of 4 ( <em>a</em> )</th> <th>Result</th> <th><em>b</em></th> <th><em>c</em></th> </tr> <tr> <td align="right">1</td> <td align="right">4</td> <td align="right">0</td> <td align="right">4</td> </tr> <tr> <td align="right">2</td> <td align="right">16</td> <td align="right">0</td> <td align="right">16</td> </tr> <tr> <td align="right">3</td> <td align="right">64</td> <td align="right">0</td> <td align="right">64</td> </tr> <tr> <td align="right">4</td> <td align="right">256</td> <td align="right">2</td> <td align="right">56</td> </tr> <tr> <td align="right">5</td> <td align="right">1024</td> <td align="right">10</td> <td align="right">24</td> </tr> <tr> <td align="right">6</td> <td align="right">4096</td> <td align="right">40</td> <td align="right">96</td> </tr> <tr> <td align="right">7</td> <td align="right">16384</td> <td align="right">163</td> <td align="right">84</td> </tr> <tr> <td align="right">8</td> <td align="right">65536</td> <td align="right">655</td> <td align="right">36</td> </tr> <tr> <td align="right">9</td> <td align="right">262144</td> <td align="right">2621</td> <td align="right">44</td> </tr> <tr> <td align="right">10</td> <td align="right">1048576</td> <td align="right">10485</td> <td align="right">76</td> </tr> <tr> <td align="right">11</td> <td align="right">4194304</td> <td align="right">41943</td> <td align="right">4</td> </tr> <tr> <td align="right">12</td> <td align="right">16777216</td> <td align="right">167772</td> <td align="right">16</td> </tr> <tr> <td align="right">13</td> <td align="right">67108864</td> <td align="right">671088</td> <td align="right">64</td> </tr> <tr> <td align="right">14</td> <td align="right">268435456</td> <td align="right">2684354</td> <td align="right">56</td> </tr> <tr> <td align="right">...</td> <td align="right"><br></br></td> <td align="right"><br></br></td> <td align="right">...</td> </tr> <tr> <td align="right">10k</td> <td align="right"><br></br></td> <td align="right"><br></br></td> <td align="right">76</td> </tr> <tr> <td align="right">10k+1</td> <td align="right"><br></br></td> <td align="right"><br></br></td> <td align="right">4</td> </tr> </tbody> </table> <p>The steps in the argument, given in words and in the language of modulus arithmetic, are:</p> <ul> <li>4 ¹⁹⁹¹ is some multiple of 100 (say b) plus 4, i.e. 4 ¹⁹⁹¹ is congruent to 4 modulo 100, which is written 4 ¹⁹⁹¹ <mdo:image src="congruent.gif" alt=""></mdo:image> 4 (mod 100).</li> <li>4 ⁸ is some multiple of 100 (say B) plus 36, i.e. 4 ⁸ is congruent to 36 modulo 100, which is written 4 ⁸ <mdo:image src="congruent.gif" alt=""></mdo:image> 36 (mod 100).</li> <li>4 ¹⁹⁹⁹ is 4 ¹⁹⁹¹ multiplied by 4 ⁸ , or (100b+4)(100B+36).</li> <li>Hence 4 ¹⁹⁹⁹ is some multiple of 100, plus 4 times 36, giving 44 as the last two digits.</li> <li>4 ¹⁹⁹⁹ <mdo:image src="congruent.gif" alt=""></mdo:image> 4 ¹⁹⁹¹ x 4 ⁸ <mdo:image src="congruent.gif" alt=""></mdo:image> 4 x 4 ⁸ <mdo:image src="congruent.gif" alt=""></mdo:image> 4 ⁹ <mdo:image src="congruent.gif" alt=""></mdo:image> 44 (mod 100).</li> </ul> <span class="editorial">To prove this rigorously needs a proof that 4</span> <sup class="editorial">10k+1</sup> <mdo:image src="congruent.gif" alt=""></mdo:image> <span class="editorial">4 (mod 100) which can be done using the axiom of mathematical induction or by other methods. It also requires the use of some simple algebra to show that 4</span> <sup class="editorial">1999</sup> <mdo:image src="congruent.gif" alt=""></mdo:image> <span class="editorial">4</span> <sup class="editorial">9</sup> <span class="editorial">(mod 100) as outlined in the bullet points above. Some students may like to write a full account of this argument for publication in the NRICH 'Inspirations' section.</span> <br></br></mdoxml> 2 5 0 0 1 0 0 Can you find what the last two digits of the number 4^1999 are? Numbers and the Number System Factors and multiples Calculations and Numerical Methods Multiplication & division Numbers and the Number System Powers & roots Admin Workshop