A (rewritten!) Prime Product

The first step in solving this multi-part problem is to generalize Suraj's ratio. The square's area is ${ s }^{ 2 }$ , and the rectangle's area is ${ s }^{ 2 }-{ a }^{ 2 }$ . The ratio, then, is $\frac { { s }^{ 2 } }{ { s }^{ 2 }-{ a }^{ 2 } }$ . The friend mentions the product of this ratio with different values, so we replace $a$ with $1$ and $s$ with ${ P }_{ n }$ (the $n$ th prime). Now we set up the product as

$\prod _{ n=2 }^{ \infty }{ \frac { { { P }_{ n } }^{ 2 } }{ { { P }_{ n } }^{ 2 }-1 } }$

Notice that $n=2$ because we start with the second prime number in the problem. This product is well known as a Euler Product , which (when $n$ is initialized as $1$ ) is equivalent to

$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 }$

(This equivalence can be found without prior knowledge of the Euler Product, but it would be more difficult)

However, since $n$ is initialized as the second prime, we must take $\frac { { \pi }^{ 2 } }{ 6 }$ and divide it by the term not included in our product (where $n=1$ ), $\frac { 4 }{ 3 }$ . Our product, then, is $\frac { { \pi }^{ 2 } }{ 8 }$ . Divide by $\pi$ , and it is clear to see that the radius of our "circle" is $\sqrt { \frac { 1 }{ 8 } }$ . Multiplying this by $1000$ then rounding up, we get $\boxed{354}$ , our solution.