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 Rare Rotative Relationship! (Posted on 2004-09-23)
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?

 See The Solution Submitted by Old Original Oskar! Rating: 4.7500 (4 votes)

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 The algebraic way | Comment 10 of 13 |

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

-Jer

 Posted by Jer on 2004-09-27 13:11:37

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