What is the smallest number so that if you move its last digit to the beginning (for example, turning 1234 into 4123) you get a new number that is an integer multiple of the original number?
This is not the smallest solution, but it will show a method of finding solutions.
I cheated a little by looking at the solutions thatt Charlie gave to guide me.
Call the original number 10a + b where b is the last digit and a is all of the other digits. The transformation yields 10^n*b + a for some integer b. Thus given that the second is m times the first:
m(10a+b) = 10^n*b + a
which can be rewritten as
a(10m - 1) = b(10^n - 5)
This has too many variables, so I cheated and saw that there is a solution when m = 5. This means
a(49) = b(10^n - 5)
b is a 1 digit factor of either a or 49. We don't know what a is but b could be 7. Dividing both sides by 7 yields
7a = 10^n - 5
The first two whole number solutions are n=5 a=14285 and n=11 a=14285714285
and the original numbers would be 142857 and 142857142857
The only thing I am unsure of of how to proceed with the assumption that b is a factor of a instead of 10^m - 1
Posted by Jer
on 2004-09-27 13:11:37