Difference between revisions of "1986 AHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | {{ | + | Solution by 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 sock creates another pair. Thus the answer is <math>5+2\cdot(10-1) = \boxed{\textbf{(B) } 23}</math>. | ||
== See also == | == See also == | ||
Revision as of 11:55, 3 January 2017
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 
.
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. 
