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 "1991 AIME Problems/Problem 13"

(→‎Solution 4)
 
(9 intermediate revisions by 5 users not shown)
Line 2: Line 2:
 
A drawer contains a mixture of red socks and blue socks, at most <math>1991</math> in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly <math>\frac{1}{2}</math> that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?  
 
A drawer contains a mixture of red socks and blue socks, at most <math>1991</math> in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly <math>\frac{1}{2}</math> that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?  
  
−
== Solution ==
+
== Solutions ==
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. Also, let <math>t=r+b</math>. The probability <math>P</math> that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
 
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. Also, let <math>t=r+b</math>. The probability <math>P</math> that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
Line 29: Line 29:
 
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. In addition, let <math>t = r + b</math>, the total number of socks in the drawer.
 
Let <math>r</math> and <math>b</math> denote the number of red and blue socks, respectively. In addition, let <math>t = r + b</math>, the total number of socks in the drawer.
  
−
From the problem, it is clear that <math>\frac{r(r-1)}{t(t-1)} + \frac{b(b-a)}{t(t-1)} = \frac{1}{2}</math>
+
From the problem, it is clear that <math>\frac{r(r-1)}{t(t-1)} + \frac{b(b-1)}{t(t-1)} = \frac{1}{2}</math>
  
 
Expanding, we get <math>\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}</math>
 
Expanding, we get <math>\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}</math>
Line 43: Line 43:
 
By testing <math>(b - r - 1) = 43</math> and <math>(b - r) = 44</math>, we get that <math>r = 43(22) = 946</math> and <math>b = 990</math>. Testing values one integer higher, we get that <math>r = 990</math> and <math>b = 1035</math>. Since <math>990 + 1035 = 2025</math> is greater than <math>1991</math>, we conclude that <math>(946, 990)</math> is our answer.  
 
By testing <math>(b - r - 1) = 43</math> and <math>(b - r) = 44</math>, we get that <math>r = 43(22) = 946</math> and <math>b = 990</math>. Testing values one integer higher, we get that <math>r = 990</math> and <math>b = 1035</math>. Since <math>990 + 1035 = 2025</math> is greater than <math>1991</math>, we conclude that <math>(946, 990)</math> is our answer.  
  
−
Since it doesn't matter whether the number of blue or red socks is <math>990</math>, we take the higher value for <math>r</math>, thus the maximum number of red socks is <math>r=\boxed{990}</math>.
+
Since it doesn't matter whether the number of blue or red socks is <math>990</math>, we take the lower value for <math>r</math>, thus the maximum number of red socks is <math>r=\boxed{990}</math>.
 +
 
 +
=== Solution 4 ===
 +
As above, let <math>r</math>, <math>b</math>, and <math>t</math> denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that <math>\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}</math>, so <math>r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}</math>.
 +
 
 +
Seeing that we can rewrite <math>r^2+b^2</math> as <math>(r+b)^2-2rb</math>, and remembering that <math>r+b=t</math>, we have <math>\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t</math>, so <math>2rb=\frac{t^2}{2}-\frac{t}{2}</math>, which equals <math>r^2+b^2-t</math>.
 +
 
 +
We now have <math>r^2+b^2-t=2rb</math>, so <math>r^2-2rb+b^2=t</math> and <math>r-b=\pm\sqrt{t}</math>. Adding this to <math>r+b=t</math>, we have <math>2r=t\pm\sqrt{t}</math>. To maximize <math>r</math>, we must use the positive square root and maximize <math>t</math>. The largest possible value of <math>t</math> is the largest perfect square less than 1991, which is <math>1936=44^2</math>. Therefore, <math>r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}</math>.
 +
 
 +
Solution by Zeroman
 +
 
 +
===Solution 5===
 +
Let <math>r</math> be the number of socks that are red, and <math>t</math> be the total number of socks. We get:
 +
 
 +
<math>2(r(r-1)+(t-r)(t-r-1))=t(t-1)</math>
 +
Expanding the left hand side and the right hand side, we get:
 +
<math>4r^2-4rt+2t^2-2t = t^2-t</math>
 +
 
 +
And, moving terms, we will get that:
 +
<math>4r^2-4rt+t^2 = t</math>
 +
 
 +
We notice that the left side is a perfect square.
 +
<math>(2r-t)^2 = t</math>
 +
 
 +
Thus <math>t</math> is a perfect square. And, the higher <math>t</math> is, the higher <math>r</math> will be. So, we should set <math>t = 44^2 = 1936</math>
 +
 
 +
And, we see, <math>2r-1936 = \pm44</math> We will use the positive root, to get that <math>2r-1936 = 44</math>, <math>2r = 1980</math>, and <math>r = \boxed{990}</math>.
 +
 
 +
- AlexLikeMath
 +
 
 +
==Solution 6==
 +
Let <math>r</math> and <math>b</math> denote the red socks and blue socks, respectively. Thus the equation in question is:
 +
 
 +
<math>\frac{r(r-1)+b(b-1)}{(r+b)(r+b-1)}=\frac{1}{2}</math>
 +
 
 +
<math>\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b</math>
 +
 
 +
<math>\Rightarrow r^2+b^2-r-b-2rb=0</math>
 +
 
 +
<math>\Rightarrow (r-b)^2=r+b\le 1991</math>
 +
 
 +
Because we wish to maximize <math>r</math>, we have <math>r\ge b</math> and thus <math>r-b\le 44</math> as both <math>r</math> and <math>b</math> must be integers and <math>{44}^2=1936</math> is the largest square less than or equal to <math>1991</math>. We now know that the maximum difference between the number of socks is <math>44</math>. Now we return to an earlier equation:
 +
 
 +
<math>r^2+b^2-r-b-2rb=0</math>
 +
 
 +
<math>\Rightarrow r^2-(2b+1)r+(b^2-b)=0</math>
 +
 
 +
Solving by the [[Quadratic formula]], we have:
 +
 
 +
<math>r=\frac{2b+1\pm\sqrt{4b^2+4b+1-4b^2+4b}}{2}</math>
 +
 
 +
<math>\Rightarrow r=\frac{2b+1\pm\sqrt{8b+1}}{2}</math>
 +
 
 +
<math>\Rightarrow r-b=\frac{1\pm\sqrt{8b+1}}{2}</math>
 +
 
 +
<math>\Rightarrow 44\ge r-b=\frac{1\pm\sqrt{8b+1}}{2}</math>
 +
 
 +
<math>\Rightarrow 44\ge\frac{1\pm\sqrt{8b+1}}{2}</math>
 +
 
 +
<math>\Rightarrow \pm\sqrt{8b+1}\le 87</math>
 +
 
 +
<math>\Rightarrow b\le 946</math> (Here we use the positive sign to maximize <math>r</math>.)
 +
 
 +
Thus the optimal case would be <math>b=946</math> as <math>r</math> increases only when <math>b</math> increases. In addition, <math>\sqrt{8b+1}=87</math> as we are using the equality case. We then plug back in:
 +
 
 +
<math>r=\frac{2b+1\pm\sqrt{8b+1}}{2}</math>
 +
 
 +
<math>\Rightarrow r=\frac{1893+87}{2}</math> (using the established <math>\sqrt{8b+1}=87</math>)
 +
 
 +
<math>\Rightarrow r=\frac{1980}{2}</math>
 +
 
 +
<math>\Rightarrow r=\boxed{990}</math>
 +
 
 +
~eevee9406
  
 
== See also ==
 
== See also ==

Latest revision as of 19:24, 5 February 2024

Problem

A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Solutions

Solution 1

Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by

\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\]

Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$, for $r$ in terms of $t$, one obtains that

\[r=\frac{t\pm\sqrt{t}}{2}\, .\]

Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$, with $n\in\mathbb{N}$. Hence, $r=n(n\pm 1)/2$ would correspond to the general solution. For the present case $t\leq 1991$, and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions.

In summary, the solution is that the maximum number of red socks is $r=\boxed{990}$.

Solution 2

Let $r$ and $b$ denote the number of red and blue socks such that $r+b\le1991$. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$ $=(r-b)^2=r+b$, so $r+b$ must be a perfect square $k^2$. Clearly, $r=\frac{k^2+k}2$, so the larger $k$, the larger $r$: $k^2=44^2$ is the largest perfect square below $1991$, and our answer is $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$.


Solution 3

Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$, the total number of socks in the drawer.

From the problem, it is clear that $\frac{r(r-1)}{t(t-1)} + \frac{b(b-1)}{t(t-1)} = \frac{1}{2}$

Expanding, we get $\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}$

Substituting $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$

Combining terms, we get $b^2 - 2br + r^2 - b - r = 0$

To make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$

From this equation, we can test values for the expression $(b - r - 1)(b - r)$, which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \leq 1991$.

By testing $(b - r - 1) = 43$ and $(b - r) = 44$, we get that $r = 43(22) = 946$ and $b = 990$. Testing values one integer higher, we get that $r = 990$ and $b = 1035$. Since $990 + 1035 = 2025$ is greater than $1991$, we conclude that $(946, 990)$ is our answer.

Since it doesn't matter whether the number of blue or red socks is $990$, we take the lower value for $r$, thus the maximum number of red socks is $r=\boxed{990}$.

Solution 4

As above, let $r$, $b$, and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}$, so $r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}$.

Seeing that we can rewrite $r^2+b^2$ as $(r+b)^2-2rb$, and remembering that $r+b=t$, we have $\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t$, so $2rb=\frac{t^2}{2}-\frac{t}{2}$, which equals $r^2+b^2-t$.

We now have $r^2+b^2-t=2rb$, so $r^2-2rb+b^2=t$ and $r-b=\pm\sqrt{t}$. Adding this to $r+b=t$, we have $2r=t\pm\sqrt{t}$. To maximize $r$, we must use the positive square root and maximize $t$. The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$. Therefore, $r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}$.

Solution by Zeroman

Solution 5

Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get:

$2(r(r-1)+(t-r)(t-r-1))=t(t-1)$ Expanding the left hand side and the right hand side, we get: $4r^2-4rt+2t^2-2t = t^2-t$

And, moving terms, we will get that: $4r^2-4rt+t^2 = t$

We notice that the left side is a perfect square. $(2r-t)^2 = t$

Thus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$

And, we see, $2r-1936 = \pm44$ We will use the positive root, to get that $2r-1936 = 44$, $2r = 1980$, and $r = \boxed{990}$.

- AlexLikeMath

Solution 6

Let $r$ and $b$ denote the red socks and blue socks, respectively. Thus the equation in question is:

$\frac{r(r-1)+b(b-1)}{(r+b)(r+b-1)}=\frac{1}{2}$

$\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b$

$\Rightarrow r^2+b^2-r-b-2rb=0$

$\Rightarrow (r-b)^2=r+b\le 1991$

Because we wish to maximize $r$, we have $r\ge b$ and thus $r-b\le 44$ as both $r$ and $b$ must be integers and ${44}^2=1936$ is the largest square less than or equal to $1991$. We now know that the maximum difference between the number of socks is $44$. Now we return to an earlier equation:

$r^2+b^2-r-b-2rb=0$

$\Rightarrow r^2-(2b+1)r+(b^2-b)=0$

Solving by the Quadratic formula, we have:

$r=\frac{2b+1\pm\sqrt{4b^2+4b+1-4b^2+4b}}{2}$

$\Rightarrow r=\frac{2b+1\pm\sqrt{8b+1}}{2}$

$\Rightarrow r-b=\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow 44\ge r-b=\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow 44\ge\frac{1\pm\sqrt{8b+1}}{2}$

$\Rightarrow \pm\sqrt{8b+1}\le 87$

$\Rightarrow b\le 946$ (Here we use the positive sign to maximize $r$.)

Thus the optimal case would be $b=946$ as $r$ increases only when $b$ increases. In addition, $\sqrt{8b+1}=87$ as we are using the equality case. We then plug back in:

$r=\frac{2b+1\pm\sqrt{8b+1}}{2}$

$\Rightarrow r=\frac{1893+87}{2}$ (using the established $\sqrt{8b+1}=87$)

$\Rightarrow r=\frac{1980}{2}$

$\Rightarrow r=\boxed{990}$

~eevee9406

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_13&oldid=214628"