Difference between revisions of "2004 AIME I Problems/Problem 2"
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Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>, | Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>, | ||
| − | <cmath> | + | <cmath>m(b+e) = m</cmath>. |
and so <math>b+e = 1</math> | and so <math>b+e = 1</math> | ||
Revision as of 23:43, 28 February 2011
Contents
Problem
Set 
consists of 
consecutive integers whose sum is 
and set 
consists of 
consecutive integers whose sum is 
The absolute value of the difference between the greatest element of and the greatest element of is 
. Find
Solution
Let us give the elements of our sets names:
and 
. So we are given that
![\[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\]](http://latex.artofproblemsolving.com/1/b/8/1b8dfbff7e5ceb7325804f9a79567c01abf0f143.png)
so 
and 
. Also,
![\[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\]](http://latex.artofproblemsolving.com/5/2/b/52b45e04e3ac1ea0f24bc439776292b71a495059.png)
so 
so 
and 
.
Then by the given, 
. is a positive integer so we must have 
and so 
.
Solution 2
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set
![\[\frac{m(f + l)}{2} = 2m\]](http://latex.artofproblemsolving.com/3/f/5/3f5919e59735f991327f66721373db1cb345cc37.png)
,
Where 
and 
represent the first and last terms of . This comes from the sum of an arithmetic sequence.
Solving for 
, we find the sum of the two terms is 
Doing the same for set B, and setting up the equation with 
and 
being the first and last terms of set ,
![\[m(b+e) = m\]](http://latex.artofproblemsolving.com/b/5/b/b5b64f2bc9d1fdc62d70b3ec8a1dee76ddd8b414.png)
.
and so 
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set has half the number of elements as set , and the difference between the greatest terms of the two two sequences is (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where 
is the last term of set A:
![\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\]](http://latex.artofproblemsolving.com/1/0/9/109c612d7a4208832b8ca5ddd7dff6523b1dafc9.png)
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to and 
respectively (add and 
to see what i mean).
Solving this equation we find 
. We know the first and last terms have to sum to so we find the first term of the sequence is 
. Now, the solution is in clear sight, we just find the number of integers between and 
, inclusive, and it is 
.
Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.
See also
| 2004 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
