2015 AIME I Problems/Problem 12: Difference between revisions

Latest revision as of 16:38, 27 October 2025

Problem

Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ .

Hint

Use the Hockey Stick Identity in the form

$\binom{a}{a} + \binom{a+1}{a} + \binom{a+2}{a} + \dots + \binom{b}{a} = \binom{b+1}{a+1}.$

(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first $(b + 1)$ numbers with $(a + 1)$ elements whose least element is $i$ , for $1 \le i \le b - a + 1$ .)

Solution 1

Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, $\begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\ &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\ & + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\ & \dots \\ & + \binom{999}{999} \\ &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\ &= \binom{2016}{1001}. \end{align*}$ Using the definition of binomial coefficient and the identity $n! = n \cdot (n-1)!$ , we deduce that $M = \frac{2016}{1001} = \frac{288}{143}.$ The answer is $\boxed{431}.$

Potential Inspiration

For solution 1, the inspiration could be that once we select the set $\{a_1, a_2,...,a_{2015}\}$ , where $a_1<a_2<...<a_{2015}$ , we want to multiply each such set by $a_1$ and sum up through all such sets.

So, how do we scale each such set by $a_1$ ? We realize that if we were to choose one more element, specifically, less than $a_1$ , this could be done in $a_1-1$ ways.

Oh! So, now if we add $1$ to all the numbers in our set, they become $\{a_1+1, a_2+1,...,a_{2015}+1\}$ , so the number of ways to pick another number below the least number in this set is $a_1$ !

Solution 2

Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ . There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ ( $k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is equal to the sum. But choosing a set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is the same as choosing a 1001-element subset from $\left\{1,2,3,...,2016\right\}$ !

Thus, the average is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}$ . Our answer is $p+q=288+143=\boxed{431}$ .

Solution 3(NOT RIGOROUS)

Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$ :

For $p = 2$ and $q = 2$ , the answer is $1$ .

For $p = 3$ and $q = 2$ , the answer is $\frac43$ .

For $p = 4$ and $q = 2$ , the answer is $\frac53$ .

For $p = 3$ and $q = 3$ , the answer is $1$ .

For $p = 4$ and $q = 3$ , the answer is $\frac54$ .

For $p = 5$ and $q = 3$ , the answer is $\frac32$ .

At this point, we can see a pattern: our desired answer is always $\frac{p+1}{q+1}$ . Plugging in $p = 2015$ and $q = 1000$ , the answer is $\frac{2016}{1001}=\frac{288}{143}$ , so $288 + 143 = \boxed{431}$ .

Solution 4 (short)

In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is $\frac{2016}{1001}$ .

-tigershark22

VIEWER NOTE: This solution doesn't always work, for example, take $n^2$ on $[0,1]$ . The "average case" is $n=\frac{1}{2}\implies n^2=\frac{1}{4}$ but integrating $n^2$ from $0$ to $1$ we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.

Solution 5

When it says average, it really just means expected value. Let's do casework.

Our first case is when the smallest number in the subset is $1$ . There are $2014$ numbers to choose from, and $999$ numbers left to choose. So, this contributes $\dbinom{2014}{999}$ to our expected value.

Then, for the smallest number being $2$ , we have $2013$ numbers to choose from and $999$ numbers left to choose. So, this contributes $2\dbinom{2013}{999}$ .

Similarly, the smallest number being $3$ contributes $3\dbinom{2012}{999}$ to our expected value. We keep this up all the way to $1016\dbinom{999}{999}$ .

This makes our sum $\dbinom{2014}{999} + 2\dbinom{2013}{999} + 3\dbinom{2012}{999} + \cdots + 1016\dbinom{999}{999}.$

To evaluate this, we can split each binomial into groups of $1$ , like this: $$ (Error compiling LaTeX. Unknown error_msg) \binom{2014}{999} + \binom{2013}{999} + \binom{2013}{999} + \binom{2012}{999} + \binom{2012}{999} + \binom{2012}{999} + \cdots + \underbrace{\binom{999}{999} + \binom{999}{999} + \cdots + \binom{999}{999}}_{\text{1016 $\binom{999}{999}$ 's}}. $Now, we can turn this into:$ \left( \binom{2014}{999} + \binom{2013}{999} + \cdots + \binom{999}{999} \right) + \left( \binom{2013}{999} + \binom{2012}{999} + \cdots + \binom{999}{999} \right) + \cdots + \left( \binom{1000}{999} + \binom{999}{999} \right) + \binom{999}{999}. $Now, by the Hockey Stick Identity, we can further simplify this into$ \dbinom{2015}{1000} + \dbinom{2014}{1000} + \cdots + \dbinom{1001}{1000} + \dbinom{1000}{1000}. $Further simplifying this, we get$ \dbinom{2016}{1001}. $$ (Error compiling LaTeX. Unknown error_msg)

If this is our sum and there are $\dbinom{2015}{1000}$ possible subsets, our average is simply $\dfrac{\dbinom{2016}{1001}}{\dbinom{2015}{1000}} = \dfrac{2016!}{1001!\,1015!} \times \dfrac{1000!\,1015!}{2015!} = \dfrac{2016}{1001} = \dfrac{288}{143} \Longrightarrow \boxed{431}.$

-jb2015007

Video Solution

https://www.youtube.com/watch?v=t-yHTkGkMK4 ~anellipticcurveoverq

Generalization

https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2

@@ Line 1: / Line 1: @@
 ==Problem==
-Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
+Consider all 1000-element subsets of the set <math>\{1, 2, 3, ... , 2015\}</math>.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
 ==Hint==
@@ Line 8: / Line 8: @@
 <cmath>\binom{a}{a} + \binom{a+1}{a} + \binom{a+2}{a} + \dots + \binom{b}{a} = \binom{b+1}{a+1}.</cmath>
-(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first <math>(b + 1)</math> numbers with <math>(a + 1)</math> elements whose least element is <math>i</math>, for <math>1 \le i \le b - a</math>.)
+(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first <math>(b + 1)</math> numbers with <math>(a + 1)</math> elements whose least element is <math>i</math>, for <math>1 \le i \le b - a + 1</math>.)
-==Solution==
 ===Solution 1===
 Let <math>M</math> be the desired mean. Then because <math>\dbinom{2015}{1000}</math> subsets have 1000 elements and <math>\dbinom{2015 - i}{999}</math> have <math>i</math> as their least element,
@@ Line 27: / Line 26: @@
 <cmath>M = \frac{2016}{1001} = \frac{288}{143}.</cmath>
 The answer is <math>\boxed{431}.</math>
+==Potential Inspiration==
+For solution 1, the inspiration could be that once we select the set <math>\{a_1, a_2,...,a_{2015}\}</math>, where <math>a_1<a_2<...<a_{2015}</math>, we want to multiply each such set by <math>a_1</math> and sum up through all such sets.
+So, how do we scale each such set by <math>a_1</math>? We realize that if we were to choose one more element, specifically, less than <math>a_1</math>, this could be done in <math>a_1-1</math> ways.
+Oh! So, now if we add <math>1</math> to all the numbers in our set, they become <math>\{a_1+1, a_2+1,...,a_{2015}+1\}</math>, so the number of ways to pick another number below the least number in this set is <math>a_1</math>!
 ===Solution 2===
@@ Line 32: / Line 38: @@
 Thus, the average is <math>\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}</math>. Our answer is <math>p+q=288+143=\boxed{431}</math>.
+===Solution 3(NOT RIGOROUS)===
+Let <math>p</math> be the size of the large set and <math>q</math> be the size of the subset (i.e. in this problem, <math>p = 2015</math> and <math>q = 1000</math>). We can easily find the answers for smaller values of <math>p</math> and <math>q</math>:
+For <math>p = 2</math> and <math>q = 2</math>, the answer is <math>1</math>.
+For <math>p = 3</math> and <math>q = 2</math>, the answer is <math>\frac43</math>.
+For <math>p = 4</math> and <math>q = 2</math>, the answer is <math>\frac53</math>.
+For <math>p = 3</math> and <math>q = 3</math>, the answer is <math>1</math>.
+For <math>p = 4</math> and <math>q = 3</math>, the answer is <math>\frac54</math>.
+For <math>p = 5</math> and <math>q = 3</math>, the answer is <math>\frac32</math>.
+At this point, we can see a pattern: our desired answer is always <math>\frac{p+1}{q+1}</math>. Plugging in <math>p = 2015</math> and <math>q = 1000</math>, the answer is <math>\frac{2016}{1001}=\frac{288}{143}</math>, so <math>288 + 143 = \boxed{431}</math>.
+===Solution 4 (short)===
+In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is <math>\frac{2016}{1001}</math>.
+-tigershark22
+VIEWER NOTE: This solution doesn't always work, for example, take <math>n^2</math> on <math>[0,1]</math>. The "average case" is <math>n=\frac{1}{2}\implies n^2=\frac{1}{4}</math> but integrating <math>n^2</math> from <math>0</math> to <math>1</math> we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.
+==Solution 5==
+When it says average, it really just means expected value. Let's do casework.
+Our first case is when the smallest number in the subset is <math>1</math>. There are <math>2014</math> numbers to choose from, and <math>999</math> numbers left to choose. So, this contributes <math>\dbinom{2014}{999}</math> to our expected value.
+Then, for the smallest number being <math>2</math>, we have <math>2013</math> numbers to choose from and <math>999</math> numbers left to choose. So, this contributes <math>2\dbinom{2013}{999}</math>.
+Similarly, the smallest number being <math>3</math> contributes <math>3\dbinom{2012}{999}</math> to our expected value. We keep this up all the way to <math>1016\dbinom{999}{999}</math>.
+This makes our sum
+<cmath>
+\dbinom{2014}{999} + 2\dbinom{2013}{999} + 3\dbinom{2012}{999} + \cdots + 1016\dbinom{999}{999}.
+</cmath>
+To evaluate this, we can split each binomial into groups of <math>1</math>, like this:
+<math></math>
+\binom{2014}{999} + \binom{2013}{999} + \binom{2013}{999} + \binom{2012}{999} + \binom{2012}{999} + \binom{2012}{999} + \cdots +
+\underbrace{\binom{999}{999} + \binom{999}{999} + \cdots + \binom{999}{999}}_{\text{1016 <math>\binom{999}{999}</math>'s}}.
+<cmath>
+Now, we can turn this into:
+</cmath>
+\left( \binom{2014}{999} + \binom{2013}{999} + \cdots + \binom{999}{999} \right) +
+\left( \binom{2013}{999} + \binom{2012}{999} + \cdots + \binom{999}{999} \right) + \cdots +
+\left( \binom{1000}{999} + \binom{999}{999} \right) + \binom{999}{999}.
+<cmath>
+Now, by the Hockey Stick Identity, we can further simplify this into
+</cmath>
+\dbinom{2015}{1000} + \dbinom{2014}{1000} + \cdots + \dbinom{1001}{1000} + \dbinom{1000}{1000}.
+<cmath>
+Further simplifying this, we get
+</cmath>
+\dbinom{2016}{1001}.
+<math></math>
+If this is our sum and there are <math>\dbinom{2015}{1000}</math> possible subsets, our average is simply
+<cmath>
+\dfrac{\dbinom{2016}{1001}}{\dbinom{2015}{1000}} =
+\dfrac{2016!}{1001!\,1015!} \times \dfrac{1000!\,1015!}{2015!} =
+\dfrac{2016}{1001} = \dfrac{288}{143} \Longrightarrow \boxed{431}.
+</cmath>
+-jb2015007
+==Video Solution==
+https://www.youtube.com/watch?v=t-yHTkGkMK4
+~anellipticcurveoverq
+== Generalization ==
+https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2
 == See also ==

2015 AIME I (Problems • Answer Key • Resources)
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