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)

I made mistakes here... First x=3, w=3 also has x=3, w=6, and the B value for x=3, w=3 is 3. (Also, the other multiples of 3 and 9 for 3, 6, 9 were discarded because the y^2 value was too high)

I can eliminate X=6,7,8,9 since X^2 is over 27 and x=5 because y^2 is over 108, and can also eliminate x=0 or 9 because 0 and 81 aren't allowed. Also, x=1 can be eliminated because that would mean at least one leading zero.

Then the list would read:

x-w-y2-x2-B
2-5-49-04-3
3-3-36-09-3 OR
3-6-81-09-8
4-1-25-16-1

Note that with a + b + c = 4 (x=2), none of the combinations work. (220,310,130,211,121,112)

With x=3 (both ones) a+b = 6, and a+b=1, and like before none of the combinations work... so w must = 1

Posted by Gamer on 2003-05-19 13:36:12