*Originally posted by Thomaster*

**How did you find it?**

For the 5 digits - label them abcde:

4*5*5=100, so even if two of the numbers are 1s, then the others must be lower than 4,5,5. Obviously reducing one of those and increasing the 1s is only making things worse as x^2 is convex. If that is so, then the sum of numbers must be lower than 1+1+4+5+5=16 => a=1.

If a=1, then c+d+e=9 for all possible b!

Moreover, b*c*d*e = 10*d+e <=> b*c = 10/e+1/d. b*c is an integer, so we need to find pairs of (e,d) that result in integers. These are {(1,2),(1,5),(2,4),(3,6)}. Let Sn be the set of solutions for that equation when b=n

S1 = {(6,1,2),(3,1,5),(3,2,4),(2,3,6)}. But c+d+e=9, so we can eliminate (2,3,6). The other 3 are the solutions I've presented.

For S2 to S9 is easy, we only need to divide the c in S1 by b, check if they are integers and if not, discard them. That leaves us with:

S2 = {(3,1,2),(1,3,6)}. None sum up to 9. Eliminate.

S3 = {(2,1,2),(1,2,4),(1,1,5)}. None sum up to 9. Eliminate.

S4 = {}

S5 = {}

S6 = {(1,1,2)} Eliminate.

Rest are empty sets.

Three solutions then:

11612

11315

11324

Edit - Mmm... I assumed digits couldn't be zero. If they can, then the product is zero and d=e=0. We then have a+b+c=10*a+b <=> c=9*a as the only condition => a=1, c=9 and b can be anything. So we also have new possibilities:

(1,b,9,0,0) for b ={0,1,...,9}

(0,b,0,0,0) for b ={0,1,...,9}

Edit - The statement about sum<16 is not correct. However, there are only a few combinations for which sum>=20 and product<100 and it's easy to check they don't work (e.g. 11199), so a=1.