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

Solution by 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 by Levieee

we have to choose 10 pairs of socks i.e 20 socks so we choose 20 socks in the worst possible scenario there will be 8 pairs and 4 socks which arent paired, so we choose 4 more socks to this, but 24 can be the sum of 4 odd numbers eg (7,7,7,3), so it means there might be 11 pairs, and a few unpaired so if we choose 23, 23 cant be formed by summing 4 odd numbers therefore 23 must have 10 pairs of socks in it also 21 cant happen because in the worst case scenario we have 8 pairs and if 4 unpaired so pulling out one more will leave us with 9 pairs but the question demands 10, also 22 cant happen because pulling out another sock it can be the same colour as the one we pulled out before therefore it cant pair 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 $\boxed{\textbf{(B) } 23}$ .

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1986 AHSME Problems/Problem 17

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