Difference between revisions of "2004 AIME I Problems/Problem 2"
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[[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 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 == | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | == Solution == | ||
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>. | ||
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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 | + | == 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. | ||
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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. | ||
+ | |||
+ | == Solution 4 (Sketchy solution to use when you don't have enough time) == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 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. | ||
+ | == 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> | ||
+ | |||
+ | -Heavytoothpaste | ||
== See also == | == See also == |
Latest revision as of 21:53, 19 May 2023
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 1
Note that since set has
consecutive integers that sum to
, the middle integer (i.e., the median) must be
. Therefore, the largest element in
is
.
Further, we see that the median of set is
, which means that the "middle two" integers of set
are
and
. Therefore, the largest element in
is
.
if
, which is clearly not possible, thus
.
Solving, we get
Solution 2
Let us give the elements of our sets names:
and
. So we are given that
so
and
(this is because
so plugging this into
yields
). Also,
so
so
and
.
Then by the given, .
is a positive integer so we must have
and so
.
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
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 4 (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.
Solution 5
Let the first term of be
and the first term of
be
. There are
elements in
so
is
. Adding these up, we get
. Set
contains the numbers
. Summing these up, we get
. The problem gives us that the absolute value of the difference of the largest terms in
and
is
. The largest term in
is
and the largest term in
is
so
. From the first two equations we get, we can get that
. Now, we make a guess and assume that
(if we get a negative value for
, we can try
). From here we get that
. Solving for
, we get that the answer is
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.