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.
Besides all numbers ending in zero which all become 10 times greater, I only found 3 numbers that fit: 18, 15 and 45.
18*6=108
15*7=105
45*9=405
For a multiple of a number to end in the same digit as the number, the number is either even (and the multiple ends in 6) or the number ends in five (and the multiple is odd).
First the case where then last digit is even.
A quick search finds it isn't possible to have a number ending in 2 or 6 that gives a 0 in the tens place. Besides the solution found,
34*6=204, 84*6=504, 68*6=408 are all candidates.
Trying to fit a number to the first:
(100A+34)*6=1000A+304
A=.25 so it doesn't work.
The other two candidates don't work similarly.
In the case where the last digit is 5 the only other candidate comes from:
35*3=105 which can be shown not to work as above.
None of the solutions found can be expanded to more digits so there are all of them.

Posted by Jer
on 20070103 11:32:30 