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

(Solution)
(Solution)
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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>
 
<math>=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2</math>
 
<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>.
 
<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>.
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===Solution 3===
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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.
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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>
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Expanding, we get <math>\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}</math>
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Substituting <math>t</math> for <math>r + b</math> and cross multiplying, we get <math>2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b</math>
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Combining terms, we get <math>b^2 - 2br + r^2 - b - r = 0</math>
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To make this expression factorable, we add <math>2r</math> to both sides, resulting in <math>(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r</math>
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From this equation, we can test values for the expression <math>(b - r - 1)(b - r)</math>, which is the multiplication of two consecutive integers, until we find the highest value of <math>b</math> or <math>r</math> such that <math>b + r \leq 1991</math>.
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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.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 17:53, 16 May 2013

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


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

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 higher value for $r$, thus the maximum number of red socks is $r=\boxed{990}$.

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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