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A. How many POSITIVE INTEGER solutions does the following

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A. How many POSITIVE INTEGER solutions does the following  [#permalink] New post 06 Dec 2003, 16:22
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A. How many POSITIVE INTEGER solutions does the following
equation have: x+y =20 .

19 or 37?

its not saying "unique solutions"?

similar question

B. How many POSITIVE INTEGER solutions does the following
equation have: x+y +z =100 .


again, no mention of unique solutions.
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 [#permalink] New post 06 Dec 2003, 19:35
I think what "they're saying" is that all variables must be positive integers.

If not, infinity would be the answer to each question.
So I say 19 for the first one,

And for the second:

Ways to make three positive integers sum up to-
0 0
1 0
2 0
3 1
4 3-----------1,1,2 three permutations
5 6-----------1,2,2 three permutations 3,1,1 three permutations
6 10-----------1,1,4 (3 perm) 1,2,3 (6 perm) 2,2,2 (1 perm)
7 15----------1,1,5 (3p) 1,2,4 (6p) 1,3,3 (3p) 2,2,3 (3p)
8 21----------1,1,6 (3) 1,2,5 (6) 1,3,4 (6) 2,2,4 (3) 2,3,3(3)

A pattern emerges!
But I can't think of the formula!!!


There must be a smarter way!

Each variable must be between 1 and 98, inclusive.
if X is 1, then Y+Z must equal 99, giving us 98 possible answers for Y, and one answer for Z under each possible Y.
if X is 2, then Y+Z must equal 98, giving us 97 possible answers for Y, and one answer for Z under each possible Y.
if X is 3, then Y+Z must equal 97, giving us 96 possible answers for Y, and one answer for Z under each possible Y.
...
if X is 97, then Y+Z must equal 3, giving us 2 possible answers for Y, and one answer for Z under each possible Y.
if X is 98, then Y+Z must equal 2, giving us 1 possible answer for Y, and one answer for Z .

So the answer is 98+97+96...+2+1, or 49.5*98, or 4851.
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 [#permalink] New post 06 Dec 2003, 19:47
A-ha!!

I figured out the formula!

It's ((n-2)(n-1))/2!!! for all N>=3
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 [#permalink] New post 06 Dec 2003, 19:48
stoolfi wrote:
I think what "they're saying" is that all variables must be positive integers.

If not, infinity would be the answer to each question.
So I say 19 for the first one,

And for the second:

Ways to make three positive integers sum up to-
0 0
1 0
2 0
3 1
4 3-----------1,1,2 three permutations
5 6-----------1,2,2 three permutations 3,1,1 three permutations
6 10-----------1,1,4 (3 perm) 1,2,3 (6 perm) 2,2,2 (1 perm)
7 15----------1,1,5 (3p) 1,2,4 (6p) 1,3,3 (3p) 2,2,3 (3p)
8 21----------1,1,6 (3) 1,2,5 (6) 1,3,4 (6) 2,2,4 (3) 2,3,3(3)

A pattern emerges!
But I can't think of the formula!!!


There must be a smarter way!

Each variable must be between 1 and 98, inclusive.
if X is 1, then Y+Z must equal 99, giving us 98 possible answers for Y, and one answer for Z under each possible Y.
if X is 2, then Y+Z must equal 98, giving us 97 possible answers for Y, and one answer for Z under each possible Y.
if X is 3, then Y+Z must equal 97, giving us 96 possible answers for Y, and one answer for Z under each possible Y.
...
if X is 97, then Y+Z must equal 3, giving us 2 possible answers for Y, and one answer for Z under each possible Y.
if X is 98, then Y+Z must equal 2, giving us 1 possible answer for Y, and one answer for Z .

So the answer is 98+97+96...+2+1, or 49.5*98, or 4851.


you got both of them Right.

EDIT : I got 37 by some wierd way.. stoolfi corrected it..that part has been deleted

I will have to get back to you with the "formula" , coz i dont see one too.

thanks for the great explanation though
keep posting

praetorian

Last edited by Praetorian on 06 Dec 2003, 20:24, edited 1 time in total.
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 [#permalink] New post 06 Dec 2003, 19:59
EVEN BETTER!

Combinations of positive integer formulas:

How many different ways to add two positive integers to reach a number?
(n-1)

How many different ways to add three positive integers to reach a number?
((n-2)(n-1))/2

How many different ways to add four positive integers to reach a number?
((n-3)(n-2)(n-1))/6

I could keep going. Math is fun! All my friends are getting wasted tonight and I'm enjoying math!!!
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 [#permalink] New post 06 Dec 2003, 20:03
Quote:
k...what i was trying to say is

x = 1, y = 19 ....x =19 , y =1 ( lets call this swap...an interchange)

Both these sum to 20... if i "interchange" every solution , i get 37


Nope. You don't.
1 19
2 18
3 17
4 16
5 15
6 14
7 13
8 12
9 11
10 10
11 9
12 8
13 7
14 6
15 5
16 4
17 3
18 2
19 1

By "interchanging" you go from 10 to 19, NOT 19 to 37!
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 [#permalink] New post 06 Dec 2003, 20:07
For five integers:

((n-1)(n-2)(n-3)(n-4))/24

for 6 integers:

((n-1)(n-2)(n-3)(n-4)(n-5))/120
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 [#permalink] New post 07 Dec 2003, 06:31
stoolfi wrote:
For five integers:

((n-1)(n-2)(n-3)(n-4))/24

for 6 integers:

((n-1)(n-2)(n-3)(n-4)(n-5))/120




the denominator could be written as (n-1)!
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 [#permalink] New post 07 Dec 2003, 06:37
praetorian123 wrote:
stoolfi wrote:
For five integers:

((n-1)(n-2)(n-3)(n-4))/24

for 6 integers:

((n-1)(n-2)(n-3)(n-4)(n-5))/120




the denominator could be written as (n-1)!


No. You are using the same symbol to represent 2 different things.

n is the sum. you need to use another variable, say k, to represent the number of integers you are adding together to reach that sum. That would be k-1.

Doesnt this simply look like (n-1)C(k-1)?
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MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 07 Dec 2003, 06:37
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A. How many POSITIVE INTEGER solutions does the following

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