Difference between revisions of "1986 AHSME Problems/Problem 17"
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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>. | 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 by Levieee | ||
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| + | 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}$. | ||
== See also == | == See also == | ||
Revision as of 06:13, 25 July 2024
Problem
A drawer in a darkened room contains 
red socks, 
green socks, 
blue socks and 
black socks.
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn.
What is the smallest number of socks that must be selected to guarantee that the selection contains at least 
pairs?
(A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)
Solution
Solution by e_power_pi_times_i
Suppose that you wish to draw one pair of socks from the drawer. Then you would pick 
socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get pairs. This is because drawing the same sock results in a pair every 
of that sock, whereas drawing another sock creates another pair. Thus the answer is 
.
Solution by Levieee
we have to choose 10 pairs of socks i.e 20 socks so we choose 20 socks in the worst possible scenario there will be 8 pairs and 4 socks which arent paired, so we choose 4 more socks to this, but 24 can be the sum of 4 odd numbers eg (7,7,7,3), so it means there might be 11 pairs, and a few unpaired so if we choose 23, 23 cant be formed by summing 4 odd numbers therefore 23 must have 10 pairs of socks in it also 21 cant happen because in the worst case scenario we have 8 pairs and if 4 unpaired so pulling out one more will leave us with 9 pairs but the question demands 10, also 22 cant happen because pulling out another sock it can be the same colour as the one we pulled out before therefore it cant pair but if we pull out another one and in the worst case scenario if the colour is same it still forms a pair with the last sock that we pulled out therefore the answer is \boxed{\textbf{(B) } 23}$.
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. 
