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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.
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.
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