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06-02-2012, 04:35 AM
If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?

orishorjo
06-02-2012, 12:08 PM
If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?
10+5+5=20
15+3+2=20
16+2+2=20
And many more

ahmed
06-02-2012, 12:39 PM
1+1+18=20

2+1+17=20

2+2+16=20

3+2+15=20

3+3+14=20

4+3+13=20

4+4+12=20

5+4+11=20

5+5+10=20

6+5+9=20

6+6+8=20

7+6+7=20

7+7+6=20

8+7+5=20

8+8+4=20

9+8+3=20

9+9+2=20

10+9+1=20

1 think there is 18 solutions:cool:

MAS1
06-02-2012, 03:45 PM
If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?

x = 1
---------
y = 1, z = 18
y = 2, z = 17
y = 3, z = 16
.
.
.
y = 18, z = 1

So there are 18 solutions when x = 1.

x = 2
---------
y = 1, z = 17
y = 2, z = 16
y = 3, z = 15
.
.
.
y = 17, z = 1

So there are 17 solutions when x = 2.

If we keep on going the same way for different values of x then there are

18 + 17 + 16 + ... + 2 + 1 = 171 solutions.