6

I came across the following problem:

A list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?

From the information, I got the following information:

  • 11 integers with a mean of 10 means a total must be 110.
  • A unique mode means that there must be at least two 8's.
  • if there are two 8's, and the median is 9, there must be at least two numbers greater than 9 also.

However, I can't figure out exactly what to do from here. Any help would be appreciated!

jj172
  • 359

4 Answers4

10

Taking the numbers to be arranged smallest to largest, we have:

$x_1+x_2+x_3+x_4+x_5+9+x_7+x_8+x_9+x_{10}+x_{11}=110$

We want to make $x_{11}$ as large as possible, while keeping everything else positive, and making sure $8$ is the unique mode. This means making the sum of $x_1$ through $x_{10}$ as small as possible, while respecting the same constraints.

As a first case, we try using exactly two $8$'s, so every other number can only occur once. This gives us:

$1+2+3+8+8+9+10+11+12+13+x_{11}=110$,

or $x_{11}=33$.

As a second case, let's try three $8$'s, which allows us to use each other number twice:

$1+1+8+8+8+9+9+10+10+11+x_{11}=110$,

or $x_{11}=35$.

That seemed to help, so let's try four $8$'s:

$1+8+8+8+8+9+9+9+10+10+x_{11}=110$,

or $x_{11}=30$.

Finally, with five $8$'s, we have:

$8+8+8+8+8+9+9+9+9+10+x_{11}=110$,

or $x_{11}=24$

This exhausts the cases.

Barry Cipra
  • 79,832
G Tony Jacobs
  • 31,218
1

HINT

If you let frequency of mode = $3$, you can cookup below dataset :

$$1,1,8,8,8,\boxed 9, 9,10,10,11,x$$

Since $8$ occurred 3 times, any other number can occur atmost two times.
You need to find the value of $x$ such that mean = $10$

Also, try letting frequency of mode=$2$ or $3$ and see what happens to the maximum possible value

AgentS
  • 12,195
1

Your bullet points look correct, though your third requires a little thought to justify.

A simpler point is that you know from the median that there are no more than five numbers strictly below $9$ and no more than five numbers strictly above $9$. It would be helpful if the five smallest could be as small as possible within the constraints so as to make the biggest in the whole set as large as possible. You also want four of the five largest to be as small as possible. So, subject to the constraints, consider

  • if five of the five smallest are $8$s then you could have $8,8,8,8,8,9,9,9,9,10,24$
  • if four of the five smallest are $8$s then you could have $1,8,8,8,8,9,9,9,10,10,30$
  • if three of the five smallest are $8$s then you could have $1,1,8,8,8,9,9,10,10,11,35$
  • if two of the five smallest are $8$s then you could have $1,2,3,8,8,9,10,11,12,13,33$

and choose the pattern with the largest maximum.

As an additional comment, I think it would be neater if there were no missing integers between those which do appear, in a sense making the distribution more unimodal (arguably the examples above have local modes at their maximum values), which can be met by a result like $8,8,8,8,9,9,10,11,12,13,14$, but this is probably not what your question means by a unique mode.

Barry Cipra
  • 79,832
Henry
  • 157,058
1

You're right that the numbers sum to $110$, and you're right that at least two of them must be $8$s. The third key is that for the median of $11$ numbers to be $9$, then one of them must be a $9$, five of them must be less than (or equal to) $9$, and five must be greater than (or equal to) $9$.

You'd really like to use the numbers $1,1,1,8,8,9,9,9,9,9,N$ where $N=110-$ the sum of the small stuff, but that would violate the condition that $8$ be a unique mode. If you only have two $8$'s, the best you can do, keeping small stuff small, is

$$1,2,3,8,8,9,10,11,12,13,N$$

with $N=110-(1+2+3+8+8+9+10+11+12+13)=110-77=33$. But you can do better by turning the $3$ into an $8$, in which case you can turn the $2$ into a $1$, the $13$ into a $9$, and the $12$ into a $10$:

$$1,1,8,8,8,9,9,10,10,11,N$$

with $N=110-(1+1+8+8+8+9+9+10+10+11)=110-75=35$. It's easy to see you can't do better than this, because you would have to turn a $1$ into an $8$ but could at best turn the $11$ into a $9$.

Barry Cipra
  • 79,832