Cycling Digits

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