1986 AHSME Problems/Problem 17

Problem

A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)

$\textbf{(A)}\ 21\qquad \textbf{(B)}\ 23\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 50$

Solution 1

~ e_power_pi_times_i

Suppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is $5+2\cdot(10-1) = \boxed{\textbf{(B) } 23}$ .

Solution 2

~ Levieee (formatted shalomkeshet)

We have to choose $10$ $pairs$ of socks, i.e $20$ socks

We choose $20$ socks in the worst possible scenario there will be $8$ $pairs$ and $4$ socks that are not paired, so we choose $4$ more socks to this, but $24$ can be the sum of $4$ $odd$ $numbers$ (like $7,7,7,3$ )

This means there might be $11$ $pairs$ and a few unpaired. If we choose $23$ , it cannot be formed by summing $4$ $odd$ $numbers$ . Therefore, $23$ must have $10$ $pairs$ of socks in it. Note that $21$ cannot happen because in the worst case scenario, we have $8$ $pairs$ , and if $4$ $unpaired$ , then pulling out one more will leave us with $9$ $pairs$ . However, the question demands $10 pairs$ . $22$ cannot happen either because if we pull out another sock it can be the same colour as the one we pulled out before, meaning it cannot be paired, but if we pull out another one and in the worst case scenario if the colour is same it still forms a pair with the last sock that we pulled out, therefore, the answer is $\boxed{\textbf{(B) } 23}$ .

Solution 3

Since there are $4$ colors of socks, any selected set of socks could contain at most $4$ unpaired socks. So a set of socks which contains $9$ or fewer matched pairs could have at most $4 + 2 \cdot 9 = 22$ socks. This could happen if $19$ red socks and $1$ sock of each other color are selected. So the minimum number of socks needed to guarantee $10$ matched pairs is $22 + 1 = \boxed{\textbf{(B) } 23}$ .

-j314andrews

Solution 4

Let $w$ , $x$ , $y$ , and $z$ be the number of red, green, blue, and black socks selected, respectively. Then the number of matched pairs of socks is $\left\lfloor \frac{w}{2} \right\rfloor + \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{y}{2} \right\rfloor + \left\lfloor \frac{z}{2} \right\rfloor$ . Note that for any integer $n$ , $\left\lfloor \frac{n}{2} \right\rfloor = \frac{n}{2}$ if $n$ is even and $\left\lfloor \frac{n}{2} \right\rfloor = \frac{n-1}{2}$ if $n$ is odd. So $\left\lfloor \frac{w}{2} \right\rfloor + \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{y}{2} \right\rfloor + \left\lfloor \frac{z}{2} \right\rfloor \geq \frac{w-1}{2} + \frac{x-1}{2} + \frac{y-1}{2} + \frac{z-1}{2} = \frac{w+x+y+z}{2} - 2$ . Thus if $\left\lfloor \frac{w}{2} \right\rfloor + \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{y}{2} \right\rfloor + \left\lfloor \frac{z}{2} \right\rfloor \leq 9$ , then $\frac{w+x+y+z}{2} - 2 \leq 9$ and $w+x+y+z \leq 22$ . That is, a set of socks containing $9$ or fewer matched pairs contains at most $22$ socks. Several equality cases exist, such as $(w, x, y, z) = (19, 1, 1, 1)$ . So the minimum number of socks needed to guarantee $10$ matched pairs is $22 + 1 = \boxed{\textbf{(B) } 23}$ .

-j314andrews

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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