Difference between revisions of "1998 AIME Problems/Problem 7"
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&= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ | &= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ | ||
&= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ | &= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ | ||
− | &= 28812 - \frac{47(47 + 1)(2(47) + 1) | + | &= 28812 - \frac{\frac{47(47 + 1)(2(47) + 1)}{6} - \frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\ |
&= 19600 | &= 19600 | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 11:41, 13 August 2023
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.