Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added solution)
m (→‎Solution 4 (Sketchy solution to use when you don't have enough time))
 
(24 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
== 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>
  
−
== Solution ==
+
==Solution 1==
 +
 
 +
Note that since set <math>A</math> has <math>m</math> consecutive integers that sum to <math>2m</math>, the middle integer (i.e., the median) must be <math>2</math>.  Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>.
 +
 
 +
Further, we see that the median of set <math>B</math> is <math>0.5</math>, which means that the "middle two" integers of set <math>B</math> are <math>0</math> and <math>1</math>.  Therefore, the largest element in <math>B</math> is <math>1 + \frac{2m-2}{2} = m</math>.  <math>2 + \frac{m-1}{2} > m</math> if <math>m < 3</math>, which is clearly not possible, thus <math>2 + \frac{m-1}{2} < m</math>.
 +
 
 +
Solving, we get <cmath>\begin{align*}
 +
m - 2 - \frac{m-1}{2} &= 99\\ m-\frac{m}{2}+\frac{1}{2}&=101\\ \frac{m}{2}&=100\frac{1}{2}.\\ m &= \boxed{201}\end{align*}</cmath>
 +
 
 +
== Solution 2 ==
 
Let us give the [[element]]s of our sets names:
 
Let us give the [[element]]s of our sets names:
 
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>.  So we are given that
 
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>.  So we are given that
 
<cmath>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,</cmath>  
 
<cmath>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,</cmath>  
−
so <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math>. Also,
+
so <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math> (this is because <math>x = 2 - \frac{m-1}{2}</math> so plugging this into <math>x+(m-1)</math> yields <math>\frac{m+3}{2}</math>). Also,
 
<cmath>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,</cmath>
 
<cmath>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,</cmath>
 
so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.
 
so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.
Line 12: Line 21:
 
Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|</math>.  <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = \boxed{201}</math>.
 
Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|</math>.  <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = \boxed{201}</math>.
  
−
== Solution 2 ==  
+
== Solution 3 ==  
  
 
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.
 
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.
Line 36: Line 45:
 
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 <math>4</math> and <math>1</math> respectively (add <math>x</math> and <math>(-x+4)</math> to see what i mean).
 
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 <math>4</math> and <math>1</math> respectively (add <math>x</math> and <math>(-x+4)</math> to see what i mean).
  
−
Solving this equation we find <math>x = 102</math>.  We know the first and last terms have to sum to <math>4</math> so we find the first term of the sequence is <math>-98</math>.  Now, the solution is in clear sight, we just find the number of integers between <math>-98</math> and <math>102</math>, inclusive, and it is <math>201</math>.
+
Solving this equation we find <math>x = 102</math>.  We know the first and last terms have to sum to <math>4</math> so we find the first term of the sequence is <math>-98</math>.  Now, the solution is in clear sight, we just find the number of integers between <math>-98</math> and <math>102</math>, inclusive, and it is <math>m = \boxed{201}</math>.
  
 
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==
+
== Solution 4 (Sketchy Solution for Speedrunners) ==
  
−
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.  
+
First, calculate the average of set <math>A</math> and set <math>B</math>. It's obvious that they are <math>2</math> and <math>1/2</math> respectively.  
 +
Let's look at both sets. Obviously, there is an odd number of integers in the set with <math>2</math> being in the middle, which means that <math>m</math> is an odd number and that the number of consecutive integers on each side of <math>2</math> are equal. In set <math>B</math>, it is clear that it contains an even number of integers, but since the number in the middle is <math>1/2</math>, we know that the range of the consecutive numbers on both sides will be <math>(x</math> to <math>0)</math> and <math>(1</math> to <math>-y)</math>.  
  
−
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}).
+
Nothing seems useful right now, but let's try plugging an odd number, <math>3</math>, for <math>m</math> in set <math>B</math>. We see that there are <math>6</math> consecutive integers and <math>3</math> on both sides of <math>1/2</math>. After plugging this into set <math>A</math>, we find that the set equals <cmath>{1,2,3}</cmath>. From there, we find the absolute value of the difference of both of the greatest values, and get 0.  
  
−
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.  
+
Let's try plugging in another odd number, <math>55</math>. We see that the resulting set of numbers is <math>(-54</math> to <math>0)</math>, and <math>(1</math> to  <math>55)</math>. We then plug this into set <math>A</math>, and find that the set of numbers is <math>(-25</math> to <math>-29)</math> which indeed results in the average being <math>2</math>. We then find the difference of the greatest values to be 26.  
  
−
From there, the largest element of set B is 1 + (m-1) = m.
+
From here, we see a pattern that can be proven by more trial and error. When we make <math>m</math> equal to <math>3</math>, then the difference is <math>0</math> whearas when we make it <math>55</math>, then the difference is <math>26</math>. <math>55-3</math> equals to <math>52</math> and <math>26-0</math> is just <math>0</math>. We then see that <math>m</math> increases twice as fast as the difference. So when the difference is <math>99</math>, it increased <math>99</math> from when it was <math>0</math>, which means that <math>m</math> increased by <math>99*2</math> which is <math>198</math>. We then add this to our initial <math>m</math> of <math>3</math>, and get <math>\boxed{201}</math> as our answer.
  
−
Solving, we get:
+
== Solution 5 ==
 +
Let the first term of <math>A</math> be <math>a</math> and the first term of <math>B</math> be <math>b</math>. There are <math>m</math> elements in <math>A</math> so <math>A</math> is <math>a, a+1, a+2,...,a+m-1</math>. Adding these up, we get <math>\frac{2a+m-1}{2}\cdot m = 2m \implies 2a+m=5</math>. Set <math>B</math> contains the numbers <math>b, b+1, b+2,...,b+2m-1</math>. Summing these up, we get <math>\frac{2b+2m-1}{2}\cdot 2m =m \implies 2b+2m=2</math>. The problem gives us that the absolute value of the difference of the largest terms in <math>A</math> and <math>B</math> is <math>99</math>. The largest term in <math>A</math> is <math>a+m-1</math> and the largest term in <math>B</math> is <math>b+2m-1</math> so <math>|b-a+m|=99</math>. From the first two equations we get, we can get that <math>2(b-a)+m=-3</math>. Now, we make a guess and assume that <math>b-a+m=99</math> (if we get a negative value for <math>m</math>, we can try <math>b-a+m=-99</math>). From here we get that <math>b-a=-102</math>. Solving for <math>m</math>, we get that the answer is <math>\boxed{201}</math>
  
−
1.5 + 0.5m + 99 = m. There we go- m is equal to none other than 201.
+
-Heavytoothpaste
  
 
== See also ==
 
== See also ==

Latest revision as of 17:50, 29 December 2024

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 $m.$

Solution 1

Note that since set $A$ has $m$ consecutive integers that sum to $2m$, the middle integer (i.e., the median) must be $2$. Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$.

Further, we see that the median of set $B$ is $0.5$, which means that the "middle two" integers of set $B$ are $0$ and $1$. Therefore, the largest element in $B$ is $1 + \frac{2m-2}{2} = m$. $2 + \frac{m-1}{2} > m$ if $m < 3$, which is clearly not possible, thus $2 + \frac{m-1}{2} < m$.

Solving, we get \begin{align*} m - 2 - \frac{m-1}{2} &= 99\\ m-\frac{m}{2}+\frac{1}{2}&=101\\ \frac{m}{2}&=100\frac{1}{2}.\\ m &= \boxed{201}\end{align*}

Solution 2

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 $x + (m - 1) = \frac{m + 3}2$ (this is because $x = 2 - \frac{m-1}{2}$ so plugging this into $x+(m-1)$ yields $\frac{m+3}{2}$). 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 3

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 $m = \boxed{201}$.

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.

Solution 4 (Sketchy Solution for Speedrunners)

First, calculate the average of set $A$ and set $B$. It's obvious that they are $2$ and $1/2$ respectively. Let's look at both sets. Obviously, there is an odd number of integers in the set with $2$ being in the middle, which means that $m$ is an odd number and that the number of consecutive integers on each side of $2$ are equal. In set $B$, it is clear that it contains an even number of integers, but since the number in the middle is $1/2$, we know that the range of the consecutive numbers on both sides will be $(x$ to $0)$ and $(1$ to $-y)$.

Nothing seems useful right now, but let's try plugging an odd number, $3$, for $m$ in set $B$. We see that there are $6$ consecutive integers and $3$ on both sides of $1/2$. After plugging this into set $A$, we find that the set equals \[{1,2,3}\]. 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, $55$. We see that the resulting set of numbers is $(-54$ to $0)$, and $(1$ to $55)$. We then plug this into set $A$, and find that the set of numbers is $(-25$ to $-29)$ which indeed results in the average being $2$. 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 $m$ equal to $3$, then the difference is $0$ whearas when we make it $55$, then the difference is $26$. $55-3$ equals to $52$ and $26-0$ is just $0$. We then see that $m$ increases twice as fast as the difference. So when the difference is $99$, it increased $99$ from when it was $0$, which means that $m$ increased by $99*2$ which is $198$. We then add this to our initial $m$ of $3$, and get $\boxed{201}$ as our answer.

Solution 5

Let the first term of $A$ be $a$ and the first term of $B$ be $b$. There are $m$ elements in $A$ so $A$ is $a, a+1, a+2,...,a+m-1$. Adding these up, we get $\frac{2a+m-1}{2}\cdot m = 2m \implies 2a+m=5$. Set $B$ contains the numbers $b, b+1, b+2,...,b+2m-1$. Summing these up, we get $\frac{2b+2m-1}{2}\cdot 2m =m \implies 2b+2m=2$. The problem gives us that the absolute value of the difference of the largest terms in $A$ and $B$ is $99$. The largest term in $A$ is $a+m-1$ and the largest term in $B$ is $b+2m-1$ so $|b-a+m|=99$. From the first two equations we get, we can get that $2(b-a)+m=-3$. Now, we make a guess and assume that $b-a+m=99$ (if we get a negative value for $m$, we can try $b-a+m=-99$). From here we get that $b-a=-102$. Solving for $m$, we get that the answer is $\boxed{201}$

-Heavytoothpaste

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_2&oldid=238312"