N is a positive base ten integer having at least 2digits but at most 4digits, which is obtained by multiplying the sum of its digits with the product of its digits. It is known that N cannot contain any leading zero.
Determine all possible value(s) of N.
It is not too find 2 digit solutions analytically.
10a+b = ab * (a+b)
10a+b = a^2 + ab^2
0 = a^2 + (b^2  10)a  b
Which is quadratic in a, the discriminant is not a perfect square for any value of b from 1 to 9.
So there are no 2digit solutions.
For 3 or 4 digits I didnt see an easy way to do this analytically so I just wrote a quick program to check.
Assuming my programming skills are sufficient the only numbers that work are:
135 = 1*3*5*(1+3+5)
144 = 1*4*4*(1+4+4)
And there are no 4digit solutions.
Why stop at 4digits? Maybe someone using something faster than a graphing calculator can try more.
There is an upper bound (can you find one?) to the number of digits N can possibly have. The lowest upper bound I found is beyond what brute force can achieve. Hopefully someone has an analytic idea.

Posted by Jer
on 20100621 14:09:07 