2002 AMC 10A Problem 21

The sum is $8\cdot 8=64.$ The collection $6,$ $6,$ $6,$ $8,$ $8,$ $8,$ $8,$ $14$ has mean, median, unique mode, and range all equal to $8,$ so $14$ is attainable. If the largest were at least $16,$ the range condition would make the smallest at least $8.$ A mean of $8$ would then force all eight integers to equal $8,$ contradicting the range. If the largest were $15,$ the range $8$ forces the smallest to be $7,$ so all eight integers are at least $7.$ The other seven then sum to $64-15=49=7\cdot 7,$ forcing every one of them to equal $7.$ But then the median and mode would be $7,$ not $8,$ a contradiction. Thus, the correct answer is D.