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# A. How many POSITIVE INTEGER solutions does the following

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CEO
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A. How many POSITIVE INTEGER solutions does the following  [#permalink]  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.
Director
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[#permalink]  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.
Director
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[#permalink]  06 Dec 2003, 19:47
A-ha!!

I figured out the formula!

It's ((n-2)(n-1))/2!!! for all N>=3
CEO
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[#permalink]  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.
Director
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[#permalink]  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!!!
Director
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[#permalink]  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!
Director
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[#permalink]  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
CEO
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Kudos [?]: 717 [0], given: 781

[#permalink]  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)!
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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[#permalink]  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)?
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
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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