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