1. If b is a positive integer than 200, then how many integer pairs (a,b) satisfy the equation a/b= 2/9
I assume you meant:
If b is a positive integer < 200
a/b = 2/9
9a = 2b
a = 2b/9
So b must be a multiple of 9 to have a as an integer
the smallest value of b is 9
and the largest value of b is 198
so we have:
9 18 27 .... 180 189 198
to find out how many terms are in this arithmetic sequence ...
term(n) = a + (n-1)d
198 = 9 + (n-1)(9)
189 = 9n - 9
198 = 9n
n = 22
there are 22 such ordered pairs
check:
the pairs are (2,9) , (4,18) , (6,27) , ... (42, 189) , (44, 198)
in each case a/b = 2/9
e.g.1 :
27/6 = 2/9
189/42 = 2/9
....
etc