Name: GG Who is asking: Student Level of the question: Middle Question: The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number. GG, I have three responss to your question. From Penny Look at the problem this way. You have a five digit number that you make into a six digit number in two different ways, by appending a 2 at each end of the number. Suppose that F is the five digit number. To place a 2 in the units position first multiply F by 10 to put a zero in the units place, and then add 2. Thus one of the two numbers is 10F + 2 To put a 2 at the start of the number you can add 200000 to F. Thus the second of the two numbers is 200000 + F Can you complete the problem now? Penny From Chris You can easily guess the answer by knowing what the decimal expansion of 1/7 looks like. But there is a really neat way to solve this kind of puzzle. Think of your number as being the repeating block of a repeating decimal: x = .abcde2abcde2... Your problem says that if we define y to be the repeating decimal y = .2abcde2abcde2..., then these two rational numbers satisfy the equation, x = 10y - 2 = 3y. now solve the right-hand equation for y and use that solution to find your desired number x. Chris and from Paul Let the original number be ABCDE2, where A, B, C, D and E are digits. And the given information means: 2ABCDE 3 ------------- ABCDE2 Consider 3 E This must produce a number whose one's digit is 2, which only works when E = 4 (3 4 = 12). We now have: 2ABCD4 3 ------------- ABCD42 Consider 3 D 3 E = 12 so we must carry the 1, so 3 D + 1 must produce a number whose one's digit is 4, which only works when D = 1 (3 1 + 1 = 4). Note there is no carry to the next column. We now have: 2ABC14 3 ------------- ABC142 3 x C must produce a number whose one's digit is 1, which only works when C = 7 (3 x 7 = 21). Note that we must carry the 2. We now have: 2AB714 3 ------------- AB7142 3 x B + 2 must produce a number whose one's digit is 7, which only works when B = 5 (3 x 5 + 2 = 17). Note that we must carry the 1. We now have: 2A5714 3 ------------- A57142 3 A + 1 must produce a number whose one's digit is 5, which only works when A = 8 (3 8 + 1 = 25). Note that we must carry the 2. We now have: 285714 3 ------------- 857142 Note that this calculation works. Therefore, the original number is 857142. Paul