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Difference between revisions of "2018 AMC 10B Problems/Problem 14"

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== Solution 1 (Statistics) ==
 
== Solution 1 (Statistics) ==
  
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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 = \boxed{\textbf{(D)}\ 225}.</math>
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To minimize the number of distinct values, we want to maximize the number of times a number appears. 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 = \boxed{\textbf{(D)}\ 225}.</math>
  
 
== Solution 2 (Algebra) ==
 
== Solution 2 (Algebra) ==
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~Username_taken12
 
~Username_taken12
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/M-z7SJwlsLY
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~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Latest revision as of 14:07, 14 June 2023

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 $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$

Solution 1 (Statistics)

To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \boxed{\textbf{(D)}\ 225}.$

Solution 2 (Algebra)

As in Solution 1, we want to maximize the number of time each number appears to do so. We can set up an equation $10 + 9( x - 1 )\geq2018,$ where $x$ is the number of values. Notice how we can then rearrange the equation into $1 + 9 ( 1 )+9 ( x - 1 )\geq2018,$ which becomes $9 x\geq2017,$ or $x\geq224\frac19.$ We cannot have a fraction of a value so we must round up to $\boxed{\textbf{(D)}\ 225}.$

~Username_taken12

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/M-z7SJwlsLY

~Education, the Study of Everything

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

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