2022 AMC 10B Problems/Problem 10

The following problem is from both the 2022 AMC 10B #10 and 2022 AMC 12B #7, so both problems redirect to this page.

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

Solution 2

We can also easily test all the answer choices.

For answer choice A, the mode is 5, the median is 3, and the arithmetic mean is 1. However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of 1 while having a mode of 5,

Trying answer choice B, the mode is 7, the median is 5, and the arithmetic mean is 3. From the arithmetic mean, we know that all the numbers have to sum to 15. We know three of the numbers: _,_,5,7,7. This exceeds the sum of 15,

Now we try answer choice C. The mode is 9, the median is 7, and the arithmetic mean is 5. From the arithmetic mean, we know that the list sums to 25. Three of the numbers are _,_,7,9,9, which is exactly 25. However, our list needs positive integers, so this won't work,

Since we were really close on choice C, we can intuitively feel that the answer is probably going to be D. We can confirm this by creating a list that satisfies the problem and choice D: 1,3,9,11,11.

So, our answer is $\boxed{\textbf{(D)}\ 11}.$

Video Solution 1

https://youtu.be/2tx9GEbIRxU

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2022 AMC 10B Problems/Problem 10

Contents

Problem

Solution

Solution 2

Video Solution 1

See Also