Difference between revisions of "1991 AIME Problems/Problem 13"

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== Solution ==
 
== Solution ==
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
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=== Solution 1 ===
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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
  
 
<cmath>
 
<cmath>
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</cmath>
 
</cmath>
  
Solving the resulting quadratic equation <math>r_{}^{2}-rt+t(t-1)/4=0</math>, for <math>r_{}^{} </math> in terms of <math>t_{}^{}</math>, one obtains that  
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Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math>, for <math>r</math> in terms of <math>t</math>, one obtains that  
  
 
<cmath>
 
<cmath>
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</cmath>
 
</cmath>
  
Now, since <math>r_{}^{}</math> and <math>t_{}^{}</math> are positive integers, it must be the case that <math>t_{}^{}=n^{2}</math>, with <math>n\in\mathbb{N}</math>. Hence, <math>r=n(n\pm 1)/2</math> would correspond to the general solution. For the present case <math>t\leq 1991</math>, and so one easily finds that <math>n_{}^{}=44</math> is the largest possible integer satisfying the problem conditions.  
+
Now, since <math>r</math> and <math>t</math> are positive integers, it must be the case that <math>t=n^{2}</math>, with <math>n\in\mathbb{N}</math>. Hence, <math>r=n(n\pm 1)/2</math> would correspond to the general solution. For the present case <math>t\leq 1991</math>, and so one easily finds that <math>n=44</math> is the largest possible integer satisfying the problem conditions.  
  
In summary, the solution is that the maximum number of red socks is <math>r_{}^{}=\boxed{990}</math>.
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In summary, the solution is that the maximum number of red socks is <math>r=\boxed{990}</math>.
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=== Solution 2 ===
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Let <math>r</math> and <math>b</math> denote the number of red and blue socks such that <math>r+b\le1991</math>. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to <math>1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)</math>
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<math>=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2</math>
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<math>=(r-b)^2=r+b</math>, so <math>r+b</math> must be a perfect square <math>k^2</math>. Clearly, <math>r=\frac{k^2+k}2</math>, so the larger <math>k</math>, the larger <math>r</math>: <math>k^2=44^2</math> is the largest perfect square below <math>1991</math>, and our answer is <math>\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:38, 24 March 2009

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?

Solution

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 $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$.

See also

1991 AIME (ProblemsAnswer KeyResources)
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
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