Question 171·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

The median is the sixth value when $11$ values are in order. Five listed values are below $49$, and four are above it. So the missing integer $x$ must be greater than $49$. A value below $49$ would move $49$ out of sixth place, and another $49$ would break the distinctness rule.

Type the ten values as $a=[22,24,37,43,45,49,53,54,56,58]$, then type $\operatorname{total}(a)$. Desmos shows $441$, the sum before the missing value is added.

The mean is the total divided by the count. Since $x$ is an integer greater than $49$, it is at least $50$. Type $\frac{441+50}{11}$; Desmos shows $44.636\ldots$. So if the mean is an integer, it must be at least $45$.

Call the integer mean $m$. The total is the known $441$ plus $x$. An integer mean makes the total a multiple of the number of values. So $441+x=11m$. Start with the smallest possible $m$, which is $45$.

For $m=45$, subtract the known total from the required total. Type $11(45)-441$; Desmos shows $54$. But $54$ is already listed, so this value would not be distinct.

Try the next integer mean, $46$. Type $11(46)-441$; Desmos shows $65$. This missing value is distinct, lies between $20$ and $80$, and stays above $49$, so the median remains $49$. It also exceeds the listed maximum, $58$, making $65$ the largest value. The only smaller possible integer mean failed. Grid in 65.