Difference between revisions of "1991 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
− | Let <math> | + | === 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 | ||
<cmath> | <cmath> | ||
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</cmath> | </cmath> | ||
− | Solving the resulting quadratic equation <math> | + | 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> | + | 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> | + | In summary, the solution is that the maximum number of red socks is <math>r=\boxed{990}</math>. |
+ | === Solution 2 === | ||
+ | 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> | ||
+ | <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>. | ||
== 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 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 .
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 |