Find all integers n greater than nine that satisfy
the following:

The integer formed by inserting the zero digit
between the units and the tens digits of n is
a multiple of n.   

All multiples of 10 have this property.  n=10a becomes 100a which is 10n.

If n=10a+b with 0<b<10 and a>0, adding the zero creates
100a+b and we seek the ratio to be an integer:

(100a+b)/(10a+b)=m
solving for a gives
a=b(1-m)/(10(m-10))

Since m<10 this makes for an easy search.  For each m, try to find a b that makes a a whole number.

There are three solutions:
m=6, b=8, a=1
m=7, b=5, a=1
m=9, b=5, a=4

corresponding to
18*6=108
15*7=105
45*9=405

Posted by Jer on 2016-03-21 12:30:42