2002 AMC 10A Problem 21

Let the largest value be $M$. The constraints: total $= 8 \cdot 8 = 64$; the smallest value is $M - 8$; the 4th and 5th values average to $8$; and $8$ appears strictly more often than any other value.

Try $M = 15$. Then every value is at least $7$. With mean $8$, the seven other values sum to $49$, so they average exactly $7$ and all equal $7$: then $7$ is the mode, contradiction. Larger $M$ fails the same way (the minimum would be $8$ or more, making the total exceed $64$).

Try $M = 14$. Minimum $7$... but we may use $6$ as the minimum only if $M - 8 = 6$, i.e. $M = 14$ gives minimum $6$. Take $6, 6, 6, 8, 8, 8, 8, 14$: sum $64$, median $\frac{8+8}{2} = 8$, mode $8$ (four times versus three), range $14 - 6 = 8$. All conditions hold.

The answer is $\boxed{\textbf{(D)}\ 14}$.

Optimization with statistics constraints is a balancing act: the mean fixes the sum, so raising one value forces others down; the range ties the minimum to the maximum; and the unique mode caps how many low values you may repeat. Test the largest candidate, show it breaks a constraint, then construct an explicit example for the next one.

• Ignoring that the mode must be unique, or forgetting the mean fixes the total at 64.
• Stopping at a valid example (like a maximum of $12$) without checking whether a larger maximum is also achievable.