Weekly Problem 22

5765 /www/nrich/html/content/id/5765/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">The following sequence continues indefinitely: 27 = 3 x 3 x 3 207 = 3 x 3 x 23 2,007 = 3 x 3 x 223 20,007 = 3 x 3 x 2223....... Which of the following integers is a multiple of 81? A: 200,007 B: 20,000,007 C: 2,000,000,007 D: 200,000,000,007 E: 20,000,000,000,007 If you liked this problem, <a href="http://nrich.maths.org/public/viewer.php?obj_id=480">here is an NRICH task</a> which challenges you to use similar mathematical ideas. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Because each of the five options given is a member of the sequence, each is also a multiple of 9. So we require the third factor of the number, that is the factor consisting of several 2s followed by a single 3, to be a multiple of 9 also. This will be true if and only if the sum of its digits is a multiple of 9. For each number in the sequence, the number of 2s in its third factor equals the number of 0s in that number. The options given have 4, 6,8, 10 or 12 zeros, corresponding with their third factors having digit sums of 11, 15, 19,23 and 27 respectively. Of these, only 27 is a multiple of 9 so the correct answer is 20,000,000,000,007.</mdoxml> 2 3 0 0 1 1 0 Weekly Problem 22 - 2008 The following sequence continues indefinitely... Which of these integers is a multiple of 81? Numbers and the Number System Factors and multiples Sequences, Functions and Graphs Sequences Secondary Mapping Document Factors, multiples and primes Secondary Mapping Document DisplayCabinet