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?

(In reply to

Answer by K Sengupta)

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.