How many solutions does the equation x + y + z = 13 have, where x,y,z are nonnegative integers less than 6?

The simplest way to solve this problem is to just list the possibilities. There aren't very many.

3 + 5 + 5 = 13

4 + 4 + 5 = 13

4 + 5 + 4 = 13

5 + 3 + 5 = 13

5 + 4 + 4 = 13

5 + 5 + 3 = 13

Thus, there are six solutions.

To solve this problem more systematically, first we'd find the smallest possible value any of the variables could take. Since the problem statement gives the largest possible value as 5, we have:

⌈(13 - x) / 2⌉ = 5

It turns out that the smallest possible value is 3, requiring that both the other variables equal 5. There is one solution with x = 3. If we set x = 4, then we can subtract one from either other variable, implying there are two solutions with x = 4. If we set x = 5, then we can subtract two from either other variable, or one from each, implying that there are three solutions with x = 5. Thus, there are a total of 1 + 2 + 3 = 6 solutions.

Alternately, once we found that the variables must be between 3 and 5 inclusive, we could simply have calculated the permutations of 3 numbers taken 3 at a time, which is 6.

Well, the nonnegative integers less than 6 are 0, 1, 2, 3, 4, and 5. If x, y, and z all have to be different numbers, then this problem has no solution since at leat one number must be used twice. However, if at least one number is used twice, these are your solutions:

x=3, y=5, z=5

x=4, y=4, z=5

x=4, y=5, z=4

x=5, y=3, z=5

x=5, y=4, z=4

x=5, y=5, z=3

So there would be six total.

You can do this semi systematically by realising that 13/3 > 4 so there has to be at least 2 fours or 2 fives in every answer.

3 + 5 + 5

4 + 4 + 5

4 + 5 + 4

5 + 3 + 5

5 + 4 + 4

5 + 5 + 3

so 6 in total

It is clear that x, y, z must be at least 3 and at most 5, so 13= 5+5+3=4+4+5 and the permutations of the 3 variables. So there are 6 different solutions

Only 2:

(5,5,3), (5,4,4)

(3,5,5)

(4,4,5)

(4,5,4)

(5,3,5)

(5,4,4)

(5,5,3)

6 triples