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 Number = Sum (Digit)*Product (Digit) (Posted on 2010-06-21)
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.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 Solution & thoughts | Comment 2 of 10 |
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

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