2004 AIME I Problems/Problem 2
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 A has the properties of m integers that sum to 2m, it's obvious that the middle integer is just 2 (the average that "balances out" everything), and the largest is 2 + (m-1)/2. That's because there are m-1 to go after taking 2, and they are "evenly balanced" on either side.
Now, since 2m consecutive integers only sum to m (set B) and just m integers already got to 2m (set A), it's clear that max({A}) is 99 larger than max({B}).
From there, we see that set B's average is 0.5. How can that happen? Only if the middle TWO values are 0 and 1.
From there, the largest element of set B is 1 + (m-1) = m.
Solving, we get:
1.5 + 0.5m + 99 = m. There we go- m is equal to none other than 201.
Solution
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.
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.