Inequality x<y<z=<10 - How many positive integer solutions?

Question

Hi,
I really don't know how to solve the following:
Suppose we have this inequality:

x < y < z < =10

-How many triples can I find which satisfy it such that x + y = z?
(All of the solutions must be positive integers)
Could you please help me ?
The solution will be greatly appreciated .

I don't understand your answers.Can you be more specific?

Answer 1

ok, its straight forward.
we have x (1) x (2) y (3) z<=10
also we have:
(4) x+y=z

so,
using (2)&(4) we have:
y y => y-y< (x+y)-y (sub y from both sides)
=> 0 => x>0 (5)

again using (1)&(5) we have:
(6) 0
then using (3)&(4) we have:
(7) x+y<=10

also using (1)&(3)&(4) we have:
x x+x => 2x => 2x => 2x<10
=> x<5 (divide both sides by 2)

now we know that x<5 , thus using this and (6) & (7):
x=1 => y=2, 3, 4, 5, 6, 7, 8
x=2 => y=3, 4, 5, 6, 7, 8
x=3 => y=4, 5, 6, 7, 8
x=4 => y=5, 6, 7, 8

if we add each x to any of y's in front, we can have the corresponding z value.

Hence, 7+6+5+4=22 triples

Answer 2

Starting with 1, 2, ___, there are 8 answers since z could be 3, 4, 5, 6, 7, 8, 9, or 10
Starting with 1, 3, ___, there are 7
all the way to 1, 9, 10 with 1 answer. 8+7+..+1=36

Then 2, 3, __ = 6 ways; 2, 4, __ = 5 ways, etc for 6+..+1=21

Then 3,4, ___ has 4 ways 3,5,__ has 3, etc 4+..+1=10

Then 4,5 __ has 2 ways, 4,6,__ has 1 way so 2+1=3

No ways have x = 5 or higher. So add all of them up.

Answer 3

Good morning.

First, you state the problem wrong in the headline. It should be as it was in the body: x+y=z<=10.

Second, here's a hint: list each triple and count, e.g.
0+9=9, 1+8=9, 2+7=9...

Answer 4

Start with (x, y, z) = (1, 2, 3) then go to (1, 3, 4), (1, 4, 5) etc. Then go to (2, 3, 5), (2, 4, 6) and so on.
You'll probably see the pattern quickly but, if you don't, it's no big deal. Just count them up ☺

Doug

Answer 5

Nevermind I misread the question.