Difference between revisions of "2018 AMC 10B Problems/Problem 14"
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To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have <math>223</math> numbers appear <math>9</math> times, <math>1</math> number appear once, and the mode appear <math>10</math> times, giving us a total of <math>223 + 1 + 1</math> = <math>\boxed{\textbf{(D) } 225}</math> | To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have <math>223</math> numbers appear <math>9</math> times, <math>1</math> number appear once, and the mode appear <math>10</math> times, giving us a total of <math>223 + 1 + 1</math> = <math>\boxed{\textbf{(D) } 225}</math> | ||
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+ | == Solution 2 (Setting up an Equation) == | ||
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+ | <p> As in the previous solution, we want to maximize the number of time each number appears to do so, we can set up an equation <math>10 + 9( x - 1 )</math> ≥ <math>2018</math> where <math>x</math> is the number of values. Notice how we can then rearrange the equation into <math>1 + 9 ( 1 )+9 ( x - 1 )</math> ≥ <math>2018</math> which becomes <math>9 x</math> ≥ <math>2017</math> then <math>x</math> ≥ <math>224</math> <math>1/9</math>. We cannot have a fraction of a value so we must round up to <math>225</math>. The solution is <math>\boxed{\textbf{(D) } 225}</math> <p> | ||
+ | Solution by Username_taken12 | ||
==See Also== | ==See Also== |
Revision as of 12:04, 1 August 2021
- The following problem is from both the 2018 AMC 12B #10 and 2018 AMC 10B #14, so both problems redirect to this page.
Problem
A list of positive integers has a unique mode, which occurs exactly times. What is the least number of distinct values that can occur in the list?
Solution
To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have numbers appear times, number appear once, and the mode appear times, giving us a total of =
Solution 2 (Setting up an Equation)
As in the previous solution, we want to maximize the number of time each number appears to do so, we can set up an equation ≥ where is the number of values. Notice how we can then rearrange the equation into ≥ which becomes ≥ then ≥ . We cannot have a fraction of a value so we must round up to . The solution is
Solution by Username_taken12
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.