+0

# Stuck

+1
141
6

Let n be the number of ordered quadruples \((x_1, x_2, x_3, x_4)\) of positive odd integers that satisfy the equation \(x_1+x_2+x_3+x_4 = 42\). Find n.

Oct 13, 2019

#1
+6

What do you have in mind to do first?

Oct 13, 2019
#2
+1

Here is a start of a list for you

1 3 5 33

1 3 7 31

1 3 9 29

1 3 11 27

Good luck! CalculatorUser  Oct 13, 2019
#3
+2

OK! Here is my attempt at this: I have used the first 17 odd numbers that sum up to 42. This is a partition problem summing up 4 odd numbers to equal 42. I have NOT considered the permutations of any one sum. For example: 1 + 3 + 5 + 33 =42 is considered as "one ordered quadruple", and NOT 4! = 24 permutations.

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33

1 - 33 + 5 + 3 + 1 = 42
2 - 31 + 7 + 3 + 1 = 42
3 - 29 + 9 + 3 + 1 = 42
4 - 29 + 7 + 5 + 1 = 42
5 - 27 + 11 + 3 + 1 = 42
6 - 27 + 9 + 5 + 1 = 42
7 - 27 + 7 + 5 + 3 = 42
8 - 25 + 13 + 3 + 1 = 42
9 - 25 + 11 + 5 + 1 = 42
10 - 25 + 9 + 7 + 1 = 42
11 - 25 + 9 + 5 + 3 = 42
12 - 23 + 15 + 3 + 1 = 42
13 - 23 + 13 + 5 + 1 = 42
14 - 23 + 11 + 7 + 1 = 42
15 - 23 + 11 + 5 + 3 = 42
16 - 23 + 9 + 7 + 3 = 42
17 - 21 + 17 + 3 + 1 = 42
18 - 21 + 15 + 5 + 1 = 42
19 - 21 + 13 + 7 + 1 = 42
20 - 21 + 13 + 5 + 3 = 42
21 - 21 + 11 + 9 + 1 = 42
22 - 21 + 11 + 7 + 3 = 42
23 - 21 + 9 + 7 + 5 = 42
24 - 19 + 17 + 5 + 1 = 42
25 - 19 + 15 + 7 + 1 = 42
26 - 19 + 13 + 9 + 1 = 42
27 - 19 + 13 + 7 + 3 = 42
28 - 19 + 11 + 9 + 3 = 42
29 - 19 + 11 + 7 + 5 = 42
30 - 17 + 17 + 5 + 3 = 42
31 - 17 + 15 + 9 + 1 = 42
32 - 17 + 15 + 7 + 3 = 42
33 - 17 + 13 + 11 + 1 = 42
34 - 17 + 13 + 9 + 3 = 42
35 - 17 + 13 + 7 + 5 = 42
36 - 17 + 11 + 9 + 5 = 42
37 - 15 + 13 + 11 + 3 = 42
38 - 15 + 13 + 9 + 5 = 42
39 - 15 + 11 + 9 + 7 = 42

Oct 13, 2019
#4
+1

Good job! Guest! However the problem is asking how many ordered pairs,     EDIT: THIS IS WRONG

so we have one last step

One ordered pair can be ordered in 4! = 24 ways

So using your 39 ordered pairs,

39 * 24 = 936

CalculatorUser  Oct 13, 2019
edited by CalculatorUser  Oct 14, 2019
#5
+1

No CU: it says "Let n be number of ordered "quadruples", or "sets of four", but not "pairs".

Oct 13, 2019
#6
+1

Aaaaaaaaah, I see!

CalculatorUser  Oct 14, 2019