Difference between revisions of "1991 AIME Problems/Problem 13"
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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>. | 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>. | ||
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+ | == 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>. | ||
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+ | 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>. | ||
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+ | 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>. | ||
== See also == | == See also == |
Revision as of 20:55, 23 April 2017
Problem
A drawer contains a mixture of red socks and blue socks, at most in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly 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 and denote the number of red and blue socks, respectively. Also, let . The probability that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
Solving the resulting quadratic equation , for in terms of , one obtains that
Now, since and are positive integers, it must be the case that , with . Hence, would correspond to the general solution. For the present case , and so one easily finds that is the largest possible integer satisfying the problem conditions.
In summary, the solution is that the maximum number of red socks is .
Solution 2
Let and denote the number of red and blue socks such that . Then by complementary counting, the number of ways to get a red and a blue sock must be equal to , so must be a perfect square . Clearly, , so the larger , the larger : is the largest perfect square below , and our answer is .
Solution 3
Let and denote the number of red and blue socks, respectively. In addition, let , the total number of socks in the drawer.
From the problem, it is clear that
Expanding, we get
Substituting for and cross multiplying, we get
Combining terms, we get
To make this expression factorable, we add to both sides, resulting in
From this equation, we can test values for the expression , which is the multiplication of two consecutive integers, until we find the highest value of or such that .
By testing and , we get that and . Testing values one integer higher, we get that and . Since is greater than , we conclude that is our answer.
Since it doesn't matter whether the number of blue or red socks is , we take the higher value for , thus the maximum number of red socks is .
Solution 4
As above, let , , and denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that , so .
Seeing that we can rewrite as , and remembering that , we have , so , which equals .
We now have , so and . Adding this to , we have . To maximize , we must use the positive square root and maximize . The largest possible value of is the largest perfect square less than 1991, which is . Therefore, .
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.