Find the sum of the 10 given numbers. Then write an expression for the mean of all 11 numbers in terms of the unknown number. What condition must the numerator satisfy so that dividing by 11 gives an integer?

Once you know what type of numbers the unknown must be (for example, based on a remainder when divided by 11), list all such numbers between 20 and 80. Then remove any that are already in the data set or that would change the median away from 49.

In Desmos, type 441 to confirm the sum of the 10 given numbers (or type 22+24+37+43+45+49+53+54+56+58 and see the result). Then type (441 + x)/11 so Desmos shows the mean as a function of the unknown number x.

Click on x in (441 + x)/11 to create a slider. Edit the slider so that x goes from 20 to 80 and has step 1. Move the slider and watch the numerical value of (441 + x)/11; note the x values between 20 and 80 where the output is an integer.

From the x values that give an integer mean, ignore any that are already in the data set or are less than or equal to 49 (they would change the median). Among the remaining valid x values, the smallest one will be the least possible value of the largest integer in the data set; read that value of x from Desmos.

There are 11 integers in data set E. When the numbers are listed in increasing order, the median of 11 numbers is the 6th number.

We already have these 10 numbers, in order:

22, 24, 37, 43, 45, 49, 53, 54, 56, 58

Here, 49 is already the 6th number. For the median of the full 11-number data set to still be 49:

• The 6th number must remain 49.
• That means we must still have exactly 5 numbers less than 49.

If the new number were less than 49, then there would be 6 numbers less than 49, so the median would no longer be 49. Therefore, the unknown integer must be greater than 49.

Let $x$ be the unknown integer. The sum of the 10 known integers is:

So the total sum of all 11 integers is $441 + x$.

The mean (average) is

We are told this mean is an integer, so $441 + x$ must be a multiple of 11.

Since $40 \times 11 = 440$, we know $441$ is 1 more than a multiple of 11. To make $441 + x$ a multiple of 11, $x$ must be 10 more than some multiple of 11 (so that $1 + x$ is a multiple of 11). In other words, $x$ has remainder 10 when divided by 11.

Numbers between 20 and 80 that leave a remainder of 10 when divided by 11 are:

• $21$ (since $11 \times 1 + 10$)
• $32$ (since $11 \times 2 + 10$)
• $43$ (since $11 \times 3 + 10$)
• $54$ (since $11 \times 4 + 10$)
• $65$ (since $11 \times 5 + 10$)
• $76$ (since $11 \times 6 + 10$)

So, based only on the mean being an integer and the range 20 to 80, $x$ could be $21, 32, 43, 54, 65$, or $76$.

Now use all the conditions:

1. Distinct integers: $x$ cannot equal any of the existing numbers.

   • The existing numbers are 22, 24, 37, 43, 45, 49, 53, 54, 56, 58.
   • So $x$ cannot be $43$ or $54$ from our candidate list.

2. Median is 49: From Step 1, $x$ must be greater than 49.

   • That eliminates $21$ and $32$.

After removing $21, 32, 43$, and $54$, the only remaining possibilities for $x$ are 65 and 76.

Since $x$ will be larger than 58, $x$ itself will be the largest integer in the data set. We are asked for the least possible value of the largest integer, so we choose the smaller of the two possibilities:

$\boxed{65}$.