2022 AMC 10B Problem 10

Question

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?

\ 5

\ 7

\ 9

\ 11

\ 13

The unique mode must occupy the last two positions

Let the ordered integers be a,b,c,d,e

Let the integers in increasing order be

a,b,c,d,e.

The median is

c,

and the unique mode is

c+2.

Because the mode is larger than the median and is unique, the last two entries must both be

c+2,

so the list is

a,b,c,c+2,c+2.

The mean is

c-2,

so

\begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2. \end{aligned}

Hence

a+b+3c+4=5c-10,

so

a+b=2c-14.

To keep the mode unique,

a

and

b

must be distinct positive integers, both less than

c.

Since

a+b=2c-14

is even, the smallest such sum is

1+3=4,

so

2c-14\ge4,

giving

c\ge9.

The smallest possible mode is therefore

c+2=11,

and it is attainable with

1,3,9,11,11.

Thus, the answer is D .

Problem 10 in other years