六个正整数的平均数为 15，中位数为 15，众数为 16。
这些数字的极差是 7。求这六个数字。 答 ... ... ... ... ... ...

1. Use the mean to find the total, and use the median to restrict the middle two numbers.

Arrange the six positive integers in ascending order:

$a \leq b \leq c \leq d \leq e \leq f$.

Mean $=15$, so

$a+b+c+d+e+f=6\times 15=90.$

For six numbers, the median is the average of the 3rd and 4th numbers:

$\frac{c+d}{2}=15$

$c+d=30.$

2. Use the mode and range to determine the six integers.

The mode is $16$, so $16$ must occur more often than any other number.

Since $c+d=30$ and the data are ordered, the middle pair must allow $16$ to appear as the mode. The possible middle pair is

$c=14,\quad d=16.$

So the numbers are

$a,\ b,\ 14,\ 16,\ e,\ f.$

The range is $7$, so

$f-a=7.$

Also,

$a+b+14+16+e+f=90$

$a+b+e+f=60.$

Trying values consistent with the order and range gives

$a=12,\quad f=19.$

Then

$12+b+e+19=60$

$b+e=29.$

To make $16$ the mode, take $e=16$, so

$b=13.$

Hence the six numbers are

$12,\ 13,\ 14,\ 16,\ 16,\ 19.$