There are only 13 multiples of both 3 and 5, since
[tex]200 = 3\cdot5\cdot13 + 5[/tex]
From these integers, we eliminate any that are also divisible by 4 or 7.
[tex]3\cdot5\cdot4 = 60 \implies \{60,120,180\}[/tex]
[tex]3\cdot5\cdot7 = 105 \implies \{105\}[/tex]
so there are 13 - 4 = 9 such integers.