How many ordered triples (x, y, z) of odd positive integers satisfy x + y + z = 47?

Guest Oct 23, 2021

#2**+1 **

I will assume that they cannot be repeated.

there are (47-1)/2 = 23 odd positive numbers less than 47

43+4 1,3

41+6 1,5

39+8 1,7 3,5

37+10 1,9 3,7

35+12 1,11 3,9 5,7

33+14 1,13 3,11 5,9

31+16 1,15 3,13 5,11 7,9

29+18 1,17 3,15 5,13 7,11

27+20 1,19 3,17 5,15 7,13 9,11

25+22 1,21 9,13

------------------------------------------------------------------------------------

23+24 21, 3 19,5 17,7 15,9 13,11

21+26 19,7 17,9 15,11

19+28 17,11 15,13

1+1+2+2+3+3+4+4+5+5 5+3+2 = 1+1+2+2+3+3+4+4+5+5+5+3+2 = 40

**I get 40 odd triples with no repeats.**

Edit: I have changed this from 39 to 40 as there was an addition error at the very end.

Little numbers always get the best of me

Melody Oct 24, 2021

#3**+1 **

hmmm

let \(x=2a+1, y = 2b+1, z=2c+1\), so

\(2a+1+ 2b+1+2c+1=47\\ 2a+2b+2c=44\\ a+b+c=22\)

The problem reduces to:

"How many nonnegative ordered triples (a, b, c) satisfy \(a+b+c=22\)?"

which is a direct application of stars and bars, having an answer of \({22+3-1 \choose 3-1} = \frac{24!}{2!22!}=\boxed{276}\)

I'm getting the same answer as guest

textot Oct 24, 2021

#5**+1 **

Hi Textot,

Thanks for your contribution.

I really liked the way you started your answer.

You are of course correct,

a+b+c = 22

where a,b, and c are ordered integers all greater or equal to 0

will have exactly the same number of solutions.

Your use of stars and bars was also interesting but unfortunately, it was not appropriate for this problem as it does not take into account that the triplet must be ordered.

This means that your answer is somewhere in the order of 3! or 6 times too big.

I used your simplification of a+b+c=22

and made a probability contour map of it.

Instead of a,b, and c

I used

x, y-x and 22-y (if you add these up you get 22)

Now these restraints come into play

\(x\ge0\qquad (1)\\~\\ y-x>x\\ y\ge2x\qquad(2)\\~\\ 22-y\ge y-x\\ -2y\ge -x-22\\ 2y\le x+22\\ y\le \frac{x}{2}+11\qquad (3)\\~\\ 22-y<22\\ y\ge0\qquad (4) \)

I graphed all these here https://www.geogebra.org/classic/nzxqxyyv

this is the pic.

Every point in this region will give a triplet solution.

LaTex:

x\ge0\qquad (1)\\~\\

y-x>x\\

y\ge2x\qquad(2)\\~\\

22-y\ge y-x\\

-2y\ge -x-22\\

2y\le x+22\\

y\le \frac{x}{2}+11\qquad (3)\\~\\

22-y<22\\

y\ge0

Melody Oct 25, 2021

#6**+1 **

Thanks, Melody.

To fix my answer:

From the reduced problem, I noticed that the triples with repeated numbers are all in the form of (k, k, l), (k, l, k), and (l, k, k), where k is some number between 0 and 11 (inclusive) and l is a number that is forced by that particular value of k, meaning that there are 3*12=36 triples that are repeated.

We now have 276-36=240 triples that are not repeated. But now we can divide by 3! to account for the overcounted triples. That gives a final answer of 40, which matches your answer :)

textot
Oct 25, 2021