Question 615·Hard·Math

The mean is the total divided by the number of values, so multiply the given $9$ by $12$. Type $9\cdot12$ in Desmos; it shows $108$. That is the total the nine values must share, not a value that must appear in the list.

Write the values in order as $x_1\le x_2\le\cdots\le x_9$. The median is the middle value, so call the fifth value $x_5=m$. The five values in positions $5$ through $9$ must each be at least $m$.

The range is largest minus smallest, so $x_9-x_1=16$. Isolate the smallest value:

Since $x_9\ge m$, use that lower limit:

So raising the median raises the smallest possible value, too.

The first four values are each at least $x_1$, so each is at least $m-16$. The last five are each at least $m$. Since their total must be $108$, even their lower bound cannot exceed $108$:

Type $4(m-16)+5m\sim108$. The $\sim$ asks Desmos to find where the lower bound reaches the fixed total. Under PARAMETERS, it shows $m\approx19.111$. The bound grows as $m$ grows, and $m$ must be an integer, so test the greatest integer below that limit.

To test the greatest integer below the regression limit, make the largest value equal to that candidate median. The range then makes the smallest value $16$ less. Start with four copies of the smallest and five copies of the median. If their total is short, raise one of the four smaller values without changing either endpoint or the median.

The greatest integer below $19.111$ is $19$. Make the largest value $19$ too; the range then makes the smallest $19-16=3$. Type $4(19-16)+5(19)$; Desmos shows $107$, one short of $108$. Raise one $3$ to $4$, leaving the endpoints and fifth value unchanged. Type the list as $A=[3,3,3,4,19,19,19,19,19]$. Then type $\operatorname{mean}(A)$, $\max(A)-\min(A)$, and $\operatorname{median}(A)$. Desmos shows $12$, $16$, and $19$. The list meets every condition, while the bound rules out a larger median. The greatest possible median is $19$. Grid in 19.