First we find LCM of 6 and 8, which is 24. Then we find the factors of 24 in the specified range. 96 Is close but not in, so then we get 120, 144, 168, and 192, which is 4 solutions so far. These all have remainder 0 when divided by either number. But what about other remainders? We notice that every number after the multiples of 24 is a solution if it is less than the multiple plus 6, because the remainder is anywhere from 0 - 5. So this is 6 in total for each multiple, so 6 * 4 is 24. But there's one more. Back to 96. The first one above 100 and last one before 102, is a solution. So the answer is $\boxed{25}$

include <iostream.h>

 #include   <conio.h>
 main()

{ clrscr(); int i,j=0; for(i=101;i<200;i++) if(i%6==i%8) j++; cout<<j; getch(); return 0; }

Remainder when $101$ is divided by $8$ is equal to when $101$ is divided by $6$ . The next time the remainder is same is after $18$ numbers. If we check the remainders for $100,99,98, \dots$ , they are $4,3,2, \dots 0$ respectively. We notice that the remainders are the same after a period of $18$ numbers.

So, $18$ numbers after $101$ , as we observed, the remainders are synchronized, thus, at number $102 + 18 = 120$ . Now, we have $6$ numbers of common remainders, $0,1,2,3,4,5$ . Thus, we have come up to $120 + 6 = 126$ . Now, the remainders will synchronize again in $18$ steps, thus, at $126 + 18 = 144$ . Now, there will be $6$ numbers of the same remainder. In short, every $24$ numbers, we have $6$ numbers with common remainders. Thus, above $101$ , there are a total of $4$ multiples of $24$ that fit our criteria and the first number $101$ . Thus, we have a total of $1 + 6 + 6 + 6 + 6 = \boxed{25}$ numbers.

LCM of 6 and 8 is 24.........choose the multiple of 24 between 100 and 200 where 100 and 200 are not included..... the numbers are 120,144,168,192.........for each cases there are up to (number+5) have same reminder....so total number of cases are......... .multiplication.(4 ,6)=24..........here we r forgetting a number 101 i.e............(multiplication(24*4)+5)...............so total is (24+1)=25...........