2004 AIME I Problems/Problem 2
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
Simple Solution
Look at the problem... consecutive integers. Now, since set has the properties of integers that sum to , it's obvious that the middle integer is just (the average that "balances out" everything), and the largest is . That's because there are to go after taking , and they are "evenly balanced" on either side.
From there, we see that set 's average is . How can that happen? Only if the middle TWO values are and . Since set has integers, its maximum must be larger than the maximum of set unless , which is impossible.
From there, the largest element of set is .
Solving, we get There we go- is equal to none other than .
Solution 1
Let us give the elements of our sets names: and . So we are given that so and . Also, 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
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 ,
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:
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.
Solution 3 (Sketchy solution to use when you don't have enough time)
First, calculate the average of set and set . It's obvious that they are and respectively. Let's look at both sets. Obviously, there is an odd number of integers in the set with being in the middle, which means that is an odd number and that the number of consecutive integers on each side of are equal. In set , it is clear that it contains an even number of integers, but since the number in the middle is , we know that the range of the consecutive numbers on both sides will be to and to .
Nothing seems useful right now, but let's try plugging an odd number, , for in set . We see that there are consecutive integers and on both sides of . After plugging this into set , we find that the set equals . From there, we find the absolute value of the difference of both of the greatest values, and get 0.
Let's try plugging in another odd number, . We see that the resulting set of numbers is to , and to . We then plug this into set , and find that the set of numbers is to which indeed results in the average being . We then find the difference of the greatest values to be 26.
From here, we see a pattern that can be proven by more trial and error. When we make equal to , then the difference is whearas when we make it , then the difference is . equals to and is just . We then see that increases twice as fast as the difference. So when the difference is , it increased from when it was , which means that increased by which is . We then add this to our initial of , and get as our answer.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.