4. (3 points) There are four groups of numbers with their averages defined as follows:
（1）The average of all multiples of 11 from 1 to 100810.
（2）The average of all multiples of 13 from 1 to 100810.
（3）The average of all multiples of 17 from 1 to 100810.
（4）The average of all multiples of 19 from 1 to 100810.
Among these four averages, the largest average value is . $\qquad$
【Analysis】Since any set of numbers arranged in a row forms an arithmetic sequence, and the average of an arithmetic sequence equals the average of the first and last terms, finding the average of the first and last terms will yield the maximum average value.

【Solution】Solution: Since any set of numbers arranged in a row forms an arithmetic sequence, and the average of an arithmetic sequence equals the average of the first and last terms,

the first and last terms of the four sequences are 11 and 100804, 13 and 100802, 17 and 100810, 19 and 100795, respectively. It is clear that the largest average is: $(17+100810) \div 2=50413.5$. Therefore, the answer is: (3).