Find a three-digit number containing three different digits where the following are all perfect squares:

(A) The sum of the first digit and the number formed by the second and third digits;

(B) The first digit multiplied by the number formed by the second and third digits and

(C) The sum of the three digits.

(In reply to solution by Charlie)

Ok Charlie... You beat me to it, but I have a non-brute force solution.

If no leading zeroes are allowed (just like the condition in the post below), then this will work:

First, think of the 3 digits as a,b,c, and the 3 square roots as x,y,z. All of these must be integers to be digits.

The three things we know are a+b+c=y^2, a+10b+c=x^2, a*(10b+c)=z^2.

Taking the first two, using linear combination, it can be shown that 9b = y^2 - x^2. If y = w+x (since 9b is not negative positive, y must be greater than or equal to x, which means w is not negative), 9b = (w+x)^2 - x^2, which simplifies to 9b = 2wx + w^2.

Given the values 1 to 9 as possible values for w (10 to 18 would repeat):

(x2 stands for x squared, y2 stands for y squared)

W-X-y2--x2-B
1-4-025-16-1
2-8-100-64-4
3-3-036-09-2
4-7-121-49-8
5-2-049-04-3
6-6-036-36-12
7-1-064-01-4
8-5-169-25-16
9-0-081-00-5

You can figure out x from w. Since 9b=(2x+w)w, (2x+w)w must be divisible by 9. Except for a number divisible by 3 (see below) there is only one x value for each w. (x can't be zero, or a,b,c = 0, or another disallowed value)

(w+x)^2=y^2 is more than 27, no a,b,c all less than 10 will work. This means all but w=1 and w=3 (the only special case whose square isn't greater than 27) can be eliminated immediatley

w=3 doesn't work because x=0 is disallowed, and if x=3, (3+3)^2 is greater than 27.

So w = 1, and x must equal 4. This means that y^2 = (1+4)^2, or 25, and x^2 = 16, and b = ((2(4)+1)1)/9, or 1.

Posted by Gamer on 2003-05-15 12:44:25