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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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 Puzzle Solution With Explanation Comment 13 of 13 |

Let the original number be (10x+y) consisting of precisely r digits. Then by the problem:

10^(r-1)*y + x = m(10x+y); where m> = 2
Or, x/y = [10^(r-1) - m]/[10m - 1]

m is clearly less than 10. Otherwise, 10^(r-1)*y + x is a r+1 digit number , which is a contradiction.
Now, (10m-1) divides 10^(r-1) - m
or, (10m-1) divides 10m-1 + 10^r -10m = 10^r -1........(*)

We set up the following table in conformity with (*).The third column for Minimum r-1, displays the smallest value of r-1 such that (10m-1) divides 10^(r-1)- m.

m.....(10m -1).........Minimum r-1

2.........19...................18
3.........29..................28
4.........39....................5
5.........49..................42
6.........59..................58
7.........69..................22
8.........79..................78
9.........89..................88

From the above table, the minimum value of r-1 = 5 corresponding to m =4. So the  original number has precisely 6 digits. Accordingly:

x/y = 999996/ 39 = 2564 = 2564t /t

The minimum value of t such that 10x+y has precisely 6 digits is feasible whenever t=4, giving: (x, y) = (10256, 4) so that the required original number is 102564.

A direct check reveals that :
410256/ 10256 = 4, which is indeed in conformity with all the provisions of the given problem.

 Posted by K Sengupta on 2007-06-07 05:34:24

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