Since eleven is odd, there must be a centra element, which in this case is the sixth. So, we can write 11 consecutive integers as

Since the arithmetic mean is the sum of the values, divided by how many they are, we have

So, the mean of 11 consecutive integers is the 6th integer.
Try this solution:
1. Note, that 100 is divisible by 4, and 999 is not divisible by it, only 996. This is an arithmetic sequence.
2. a1;a2;a3;a4;...a(n) the sequence, where a1=100; a2=104; a3=108; a4=112; ... etc., and a(n)=996. n=?
3. using a formula for n-term of the sequence: a(n)=a1+d(n-1), where a(n)=996; a1=100 and d=4 (according to the condition ' is divisible by 4'). Then 100+4(n-1)=996; ⇒ 4n=900; ⇒ n=225 (including 100).
answer: 225
1. x = 2
2. x = 9
3. a = 3
4. x = 13
5. c = 3
6. s = 3
7. x = 7
8. c = 0
9. b = 1
10. c = 4
11. x = 4
12. x = 8
13. x = 12
14. y = 10
15. x = 11
16. x = 10
17. x = 6
18. x = 11