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 re: solution (More confusing!) by Gamer)

Now that we know a + 10 + c = 25, and a + 1 + c = 16, we can use the second equation to figure out what a and c are.

a(10+c)=z^2, and since you know that a + c = 15 (subracting constants from the above equations gives you this), you can substitute (15-c) for a.

This gives (15-c)(c+10)=z^2, and converting this into quadratic form gives -c^2 + 5c + (150- z^2) = 0

Plugging -1 in for a, 5 for b and (150- z^2) for c in the quadratic formula, gives:

1/2(5�} �� (625-4(z^2)))=c

The only way c will be an integer is if the discriminant (d = (625-4(z^2))) is a perfect square. Since z has to be an integer, the only pairs that will work are (z=0, d=25) (z=10, d=15) and (z=12, d=7). Using a = 15-c once c is found gives (a,b,c) as (0,1,15) (5,1,10) and (9,1,6) because any any negative c would be disallowed.

Since the first two sets aren't applicable for base 10 (although they do "work" when tested), the only one possible given our restrictions is (9,1,6) or the number 916.

Posted by Gamer on 2003-05-15 13:01:28