Question 129097: Find three consecutive odd numbers such that the sum of their squares is a four-digit number with all four digits the same?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! Let x, x+2, x+4 be the numbers.
x^2+(x+2)^2+(x+4)^2
x^2+x^2+4x+4+x^2+8x+16
3x^2+12x+20=5555 I tried all combinations from 1111 t0 5555 on TI-89 calculator until I got an integer answer for x.
3x^2+12x-5535=0
3(x^2+4x-1845)=0
x^2+4x =1845
x^2+4x+4=1845+4 Complete the square.
(x+2)^2=1849
x+2=+-43
x=-2+-43
x=-45, x=41 (-45 is extraneous.)
x=41
The numbers are 41,43,45
.
Check:
41^2+43^2+45^2
=1681+1849+2025
=5555
.
Ed
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