Difference between revisions of "1986 AHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | 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 <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>. | 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>. | ||
| + | ===Solution 2=== | ||
| + | ~ Levieee (formatted shalomkeshet) | ||
| + | We have to choose <math>10</math> <math>pairs</math> of socks, i.e <math>20</math> socks | ||
| − | + | 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>) | |
| − | + | 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 
red socks, 
green socks, 
blue socks and 
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 
pairs?
(A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)
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 
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 pairs. This is because drawing the same sock results in a pair every 
of that sock, whereas drawing another sock creates another pair. Thus the answer is 
.
Solution 2
~ Levieee (formatted shalomkeshet)
We have to choose 
of socks, i.e 
socks
We choose socks in the worst possible scenario there will be 
and 
socks that are not paired, so we choose more socks to this, but 
can be the sum of 
(like 
)
This means there might be 
and a few unpaired. If we choose 
, it cannot be formed by summing . Therefore, must have of socks in it. Note that 
cannot happen because in the worst case scenario, we have , and if 
, then pulling out one more will leave us with 
. However, the question demands 
. 
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
