1998 AIME Problems/Problem 7
Contents
Problem
Let be the number of ordered quadruples of positive odd integers that satisfy Find
Solution 1 (Clever Stars and Bars Manipulation)
We want . This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set , as for all integers , will be odd. Substituting we get Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us . Computing this and dividing by 100 gives us an answer of . ~smartguy888
Solution 2
Define . Then , so .
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is , and .
Solution 3
Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have stones left. Because we want an odd number in each box, we pair the stones, creating sets of . Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).
Our problem can now be restated: how many ways are there to partition a line of stones? We can easily solve this by using sticks to separate the stones into groups, and this is the same as arranging a line of sticks and stones. Our answer is therefore
Solution 4
Let and . Then , where are positive even integers ranging from .
-When , and . This accounts for solutions.
-When , and . This accounts for solutions.
We quickly see that the total number of acceptable ordered pairs is:
Therefore, .
(This solution uses the sum of squares identity to calculate and .)
-baker77
Solution 5
We write the generating functions for each of the terms, and obtain as the generating function for the sum of the numbers. We seek the coefficient, or the coefficient in Now we simplify this as and in general we want that the coefficient of is We see the coefficient so we let and so the coefficient is in which
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.