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Difference between revisions of "1986 AHSME Problems/Problem 17"

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==Solution==
 
==Solution==
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Solution by e_power_pi_times_i
 
  
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===Solution 1===
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~ e_power_pi_times_i
  
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Suppose that you wish to draw one pair of socks from the drawer. Then you would pick <math>5</math> 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 <math>10</math> pairs. This is because drawing the same sock results in a pair every <math>2</math> of that sock, whereas drawing another *ock creates another pair. Thus the answer is <math>5+2\cdot(10-1) = \boxed{\textbf{(B) } 23}</math>.
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Suppose that you wish to draw one pair of socks from the drawer. Then you would pick <math>5</math> 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 <math>10</math> pairs. This is because drawing the same sock results in a pair every <math>2</math> of that sock, whereas drawing another sock creates another pair. Thus the answer is <math>5+2\cdot(10-1) = \boxed{\textbf{(B) } 23}</math>.
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===Solution 2===
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~ Levieee (formatted shalomkeshet)
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We have to choose <math>10</math> <math>pairs</math> of socks, i.e <math>20</math> socks
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We choose <math>20</math> socks in the worst possible scenario there will be <math>8</math> <math>pairs</math> and <math>4</math> socks that are not paired, so we choose <math>4</math> more socks to this, but <math>24</math> can be the sum of <math>4</math> <math>odd</math> <math>numbers</math> (like <math>7,7,7,3</math>)
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This means there might be <math>11</math> <math>pairs</math> and a few unpaired. If we choose <math>23</math>, it cannot be formed by summing <math>4</math> <math>odd</math> <math>numbers</math>. Therefore, <math>23</math> must have <math>10</math> <math>pairs</math> of socks in it. Note that <math>21</math> cannot happen because in the worst case scenario, we have <math>8</math> <math>pairs</math>, and if <math>4</math> <math>unpaired</math>, then pulling out one more will leave us with <math>9</math> <math>pairs</math>. However, the question demands <math>10 pairs</math>. <math>22</math> 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 <math>\boxed{\textbf{(B) } 23}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:21, 31 October 2024

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 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 $\boxed{\textbf{(B) } 23}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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