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

(Solution 2 (Algebra))
(9 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #10]] and [[2018 AMC 10B Problems|2018 AMC 10B #14]]}}
 +
 +
== Problem ==
 +
 
A list of <math>2018</math> positive integers has a unique mode, which occurs exactly <math>10</math> times. What is the least number of distinct values that can occur in the list?
 
A list of <math>2018</math> positive integers has a unique mode, which occurs exactly <math>10</math> times. What is the least number of distinct values that can occur in the list?
  
 
<math>\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234</math>
 
<math>\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234</math>
  
== Solution ==
+
== Solution 1 (Statistics) ==
 +
 
 +
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>
 +
 
 +
== 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 <math>10 + 9( x - 1 )\geq2018,</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 )\geq2018,</math> which becomes <math>9 x\geq2017,</math> or <math>x\geq224\frac19.</math> We cannot have a fraction of a value so we must round up to <math>\boxed{\textbf{(D)}\ 225}.</math>
 +
 
 +
~Username_taken12
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/M-z7SJwlsLY
  
To minimize the number of values, we want to maximize the number of times they appear. So, we could have 223 numbers appear 9 times, 1 number appear once, and the mode appear 10 times, giving us a total of <math>223 + 1 + 1</math> = <math>\boxed{(D) 225}</math>
+
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2018|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2018|ab=B|num-b=13|num-a=15}}
 +
{{AMC12 box|year=2018|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Combinatorics Problems]]

Revision as of 22:02, 27 May 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 they appear. 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png