Find a three digit number that fits the following criteria:

1)The digits are all different.
2)When each digit is squared and added together the number is the same as the number formed by the 2nd and 3rd digits in the number.
3)When the number formed by the last two digits is squared then added to (the first digit squared and added to itself) it is the same as the original number.

How many different three digit numbers are there that fit the criteria?

the unique solution is 420:

let the number be 'xyz'
the following system of equations gives the solution:
x²+y²+z²=10y+z
(10y+z)²+x²+x=100x+10y+z
y,z = 0,1,2,...9, x=1,2,3,...9

working on the equations we obtain:
4x²+4(y-5)²+(2z-1)²=101
99(y²-x)+20yz=0

the second equation means that one of following is true:
- (y²-x)<0 and the product yz is multiple of 11 (imposible)
- y²=x and yz=0
if y=0, x=0 and the solution is not a 3 digit number

Thus:
z=0
y ²=x
x²+(y-5)²=25

the last equation has possible solutions:
x=3 and y-5=4 or
x=4 and y-5=3 or
x=3 and y-5=-4 or
x=4 and y-5=-3
but only one verify the condition of y²=x

so, the final solution (and unique) is
x=4, y=2, z=0, the number beeing 420

Posted by luminita on 2004-01-04 14:35:41