Date: 01/07/98 at 01:15:46 From: Yvonne Tan Subject: Cycling digits I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth? I believed it was not telling the truth; however, I tried solving it by using two-digit numbers only. Could you please help me solve the problem? Thank you very much for your kind attention.
Date: 01/08/98 at 12:42:44 From: Doctor Mc Subject: Re: Cycling digits Hello Yvonne - nice problem. There is a solution, but it's rather large. Let n be the solution, and write n = 10a+b where a and b are integers and 0<=b<=9 (so b is the units digit). Now moving the units digits all the way to the left gives the number b (10^k)+a where k is the number of digits of a. So, we want 2n = 20 a + 2 b = a + (10^k) b. That is, 19a = ((10^k)-2) b. Now, since b cannot be divisible by 19, 19 must divide (10^k)-2. Here's where the numbers get large: the smallest k that works is 17. This gives a = 5263157894736842 b. Now, choosing b to be 1 doesn't work, since a has to have 17 digits. However, b = 2 does work. Thus the smallest solution is n = 105263157894736842. To practice the method, you might try to find a number that gives its triple when the units digit is moved over to the left. Of course, the numbers may get even larger! Good luck, and have fun with it. -Doctor Nick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 01/14/98 at 02:07:22 From: TAN KIAT PING YVONNE Subject: Re: Cycling digits Dearest Dr Math, Thank you very much for your assistance with the above-mentioned problem. I will continue to support your web site and even invite my friends to join in too. Thank you once again. Yours sincerely, Yvonne
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