N is a positive base ten integer having at least 2-digits but at most 4-digits, 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 2-digit 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 4-digit solutions.
Why stop at 4-digits? 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 2010-06-21 14:09:07