Determine all possible positive integers, two or more digits long, like ABC...XYZ, such that ABC...XY0Z is a multiple of it.

Note: digits may be repeated.

Solutions: 15, 18 and 45

Rewrite the numbers as xy and x0y

The numbers share a relationship of:

(10x + y) * z = 100x + y     therefore

x and y are positive forcing z to be positive.  The maximum value for z is 9. Testing the values of z (9,8,7 . .  .1)

For 9:

 9(10x + y) = 100x + y

   90x + 9y = 100x + y

             8y = 10x

               y = 5/4 x    (x,y) = (4,5)  and 45 * 9 = 405

For 8:

8(10x + y) = 100x + y

             7y = 20x     (x,y) = no solution

For 7:  

7(....)

             6y = 30x     (x,y) = (1,5)  and 15 * 7 = 105

For 6:    5y = 40x     (x,y) = (1,8)  and 18 * 6 = 108

For 5:    4y = 50x     This forces y to be greater than 10 and will have no valid solution

For 4, 3, 2, 1 see "5"

Edited on January 3, 2007, 12:56 pm

Posted by Leming on 2007-01-03 11:54:55