2006-11-08 14:51:22 · 7 answers · asked by chriskingwong 1 in Science & Mathematics ➔ Mathematics
2, 8 and 16
2006-11-08 15:03:56 · answer #1 · answered by martina_ie 3 · 0⤊ 0⤋
Here's one possibility:
16^2 + 8^2 + 2^2 = 256 + 64 + 4 = 324.
That was by inspection (fancy way to say
trial and error.) I assume there are other solutions.
2006-11-08 23:07:28 · answer #2 · answered by banjuja58 4 · 0⤊ 0⤋
the integers are8,8 and 14
=8^2+8^2+14^2
=64+64+196
=128+196
=324
2006-11-08 23:05:10 · answer #3 · answered by raj 7 · 0⤊ 1⤋
2, 8. 16
2006-11-08 23:10:00 · answer #4 · answered by Dr. J. 6 · 0⤊ 0⤋
(x + y + z)^2 = 324
x + y + z = 18
combination of x + y + z =18
1 1 16
1 2 15 ... many more
2006-11-08 22:57:16 · answer #5 · answered by The Clueless Philospher 2 · 0⤊ 1⤋
first integer is x^2.
second integer is (x+1)^2
third integer is (x+2)^2
We now have:
x^2 + (x+1)^2 + (x+2)^2 = 324.
Now, you solve for x.
Guido
2006-11-08 23:01:17 · answer #6 · answered by Anonymous · 0⤊ 1⤋
I am assuming you mean consecutive integers, because otherwise there are many solutions.
2006-11-08 22:56:41 · answer #7 · answered by man of questions 3 · 0⤊ 0⤋