Difference between revisions of "2017 UNCO Math Contest II Problems/Problem 7"
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| − | (a) If we let <math>n</math> be the maximum number of balls we can choose such that there is no set of <math>5</math> balls, then the answer is <math>n+1</math>. To calculate <math>n</math>, notice that there can be at most <math>10</math> balls from each color that satisfies the requirements, so for each of the <math>4</math> colors, we can choose at most <math>10</math> balls. Thus, the answer is <math>4\times10+1=\boxed{41</math> | + | (a) If we let <math>n</math> be the maximum number of balls we can choose such that there is no set of <math>5</math> balls, then the answer is <math>n+1</math>. To calculate <math>n</math>, notice that there can be at most <math>10</math> balls from each color that satisfies the requirements, so for each of the <math>4</math> colors, we can choose at most <math>10</math> balls. Thus, the answer is <math>4\times10+1=\boxed{41}</math> |
(b) Proceed as above | (b) Proceed as above | ||
Latest revision as of 00:09, 17 January 2023
Problem
A box of 48 balls contains balls numbered 1, 2, 3, . . ., 12 in each of four different colors. Without ever looking at any of the balls, you choose balls at random from the box and put them in a bag.
(a) If you must be sure that when you finish, the bag contains at least one set of five balls whose numbers are consecutive, then what is the smallest number of balls you can put in the bag? (For example, a set of balls, in any combination of colors, with numbers 3, 4, 5, 6, and 7 is a set of five whose numbers are consecutive.)
(b) If instead you must be sure that the bag contains at least one set of five balls all in the same color and with consecutive numbers, then what is the smallest number of balls you can put in the bag? Remember to justify answers for maximum credit.
Solution
(a) If we let 
be the maximum number of balls we can choose such that there is no set of 
balls, then the answer is 
. To calculate , notice that there can be at most 
balls from each color that satisfies the requirements, so for each of the 
colors, we can choose at most balls. Thus, the answer is 
(b) Proceed as above
See also
| 2017 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
