Difference between revisions of "2004 AIME I Problems/Problem 2"

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== Problem ==
 
== Problem ==
 
[[Set]] <math> A </math> consists of <math> m </math> consecutive integers whose sum is <math> 2m,  </math>and set <math> B </math> consists of <math> 2m </math> consecutive integers whose sum is <math> m. </math> The absolute value of the difference between the greatest element of <math> A </math> and the greatest element of <math> B </math> is <math>99</math>. Find <math> m. </math>
 
[[Set]] <math> A </math> consists of <math> m </math> consecutive integers whose sum is <math> 2m,  </math>and set <math> B </math> consists of <math> 2m </math> consecutive integers whose sum is <math> m. </math> The absolute value of the difference between the greatest element of <math> A </math> and the greatest element of <math> B </math> is <math>99</math>. Find <math> m. </math>
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==Simple Solution==
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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, and the largest is 2 + (m-1)/2. That's because there are m-1 to go, and they are "evenly balanced" on either side.
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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}).
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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.
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From there, the largest element of set B is 1 + (m-1) = m.
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Solving, we get:
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1.5 + 0.5m +99 = m. There we go- m is equal to none other than 201.
  
 
== Solution ==
 
== Solution ==
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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.
 
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.
 
==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, and the largest is 2 + (m-1)/2. That's because there are m-1 to go, 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.
 
  
 
== See also ==
 
== See also ==

Revision as of 21:41, 28 February 2017

Problem

Set $A$ consists of $m$ consecutive integers whose sum is $2m,$and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$. Find $m.$

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, and the largest is 2 + (m-1)/2. That's because there are m-1 to go, 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: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$. So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$. Also, \[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$.

Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$. $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$.

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 $A$

\[\frac{m(f + l)}{2} = 2m\]

Where $f$ and $l$ represent the first and last terms of $A$. This comes from the sum of an arithmetic sequence.

Solving for $f+l$, we find the sum of the two terms is $4$.

Doing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$,

\[m(b+e) = m\]

and so $b+e = 1$.

Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$, and the difference between the greatest terms of the two two sequences is $99$ (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 $x$ is the last term of set A:

\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\]

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 $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).

Solving this equation we find $x = 102$. We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$. Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$, inclusive, and it is $201$.

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 (ProblemsAnswer KeyResources)
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

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