Let a, b, c be the digits of the three digit number, e.g. a=1, b=2, c=3 for 123. None of the digits can be zero, since then the product would be zero.Originally Posted by Chuck_3000
The two digit number made from a and c is 10a + c, so the condition you are interested in is
Solve for a:
Now c = 1,2,3,...,9, so for b = 1, a would have to be negative. Thus b=1 is impossible.
The idea then is to look at all larger values of b and try values of c such that this fraction is an integer between 1 and 9.
For b>1, we must have
which leads to
So for b = 4, e.g., we only need to look at c = 1, 2,3.
Using this information and trying all cases, you can find the solutions is reasonable time. I found four solutions.