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Difference between revisions of "2022 AMC 10B Problems/Problem 10"

(Created page with "==Problem== Camila writes down five positive integers. The unique mode of these integers is <math>2</math> greater than their median, and the median is <math>2</math> greater...")
 
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Let <math>a</math> and <math>b</math> be the two smallest integers such that <math>a<b.</math> The sorted list is <cmath>a,b,M,M+2,M+2.</cmath>
 
Let <math>a</math> and <math>b</math> be the two smallest integers such that <math>a<b.</math> The sorted list is <cmath>a,b,M,M+2,M+2.</cmath>
 +
Since the median is <math>2</math> greater than their arithmetic mean, we have <math>\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,</math> or <cmath>a+b+14=2M.</cmath>
 +
Note that <math>a+b</math> must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let <math>a=1</math> and <math>b=3,</math> from which <math>M=9</math> and <math>M+2=\boxed{\textbf{(D)}\ 11}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:51, 17 November 2022

Problem

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution

Let $M$ be the median. It follows that the two largest integers are $M+2.$

Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{\textbf{(D)}\ 11}.$

~MRENTHUSIASM

See Also

2022 AMC 10B (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 10 Problems and Solutions
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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