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.

Call ABC...XY=x, and Z=z. It follows that both are not negative and z<10. Also, x>0 if leading zeros are not allowed.

Then, (10x+z)*(k+1)=(100x+z) where k+1 is an integer,
which means z(k+1)-z=100x-10(k+1)x or zk=(9-k)10x

It then follows that k<=9, and thus z or k must be 5 if the right side is nonzero.

If z=5, then k must be even, and k>9-k, thus k=8 or 6, implying x=1 or 4, implying that 15 and 45 are solutions.

If k=5, then 5z=40x which implies z=8, x=1, giving the solution 18

However, it is not prohibited for z=0, only that x=0, so if 9-k=0 and z=0, then x can equal anything positive.

Posted by Gamer on 2007-01-03 13:55:26