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

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 $5+2\cdot(10-1) = \boxed{23}$.
23
Level 5
Counting & Probability
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:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the smallest number of socks needed to guarantee at least 10 pairs, we first need to understand the concept of pairs. A pair consists of two socks of the same color. 

We have four colors of socks: red, green, blue, and black. The maximum number of pairs we can form from each color is half the number of socks of that color. Thus, we can form:

- Red: 100 socks → 50 pairs
- Green: 80 socks → 40 pairs
- Blue: 60 socks → 30 pairs
- Black: 40 socks → 20 pairs

The total number of pairs available is 50 + 40 + 30 + 20 = 140 pairs. However, we need to ensure that we have at least 10 pairs regardless of the colors drawn.

To find the worst-case scenario, we consider the maximum number of socks we can draw without forming 10 pairs. Each pair requires 2 socks, so to avoid forming 10 pairs, we can have at most 9 pairs, which means we can have 18 socks of the same color without forming the 10th pair.

If we take 18 socks of each color, we have:

- Red: 18 socks → 9 pairs
- Green: 18 socks → 9 pairs
- Blue: 18 socks → 9 pairs
- Black: 18 socks → 9 pairs

This gives us a total of 18 + 18 + 18 + 18 = 72 socks, which results in 36 pairs, but we are only interested in the maximum number of socks drawn without reaching 10 pairs.

To ensure we have at least 10 pairs, we need to draw one more sock after reaching the maximum of 9 pairs from any color. Therefore, we need to draw:

- 18 socks from three colors (54 socks total) 
- 9 socks from the fourth color (to avoid forming the 10th pair)

This results in 54 + 9 = 63 socks. However, this is not the optimal way to guarantee 10 pairs.

Instead, we can calculate the minimum number of socks needed to ensure at least 10 pairs. If we draw 22 socks, we can still avoid forming 10 pairs. However, drawing one more sock (23 socks total) guarantees that at least one color will have enough socks to form the 10th pair.

Thus, the smallest number of socks that must be selected to guarantee at least 10 pairs is 23.<|END_OF_LATENT|> The final answer is $23$.