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.
Puzzle Solution With Explanation
