Easy as $3.14$ question I've got the following question with the hint "Easy as $3.14$".
$$2003 = 2\\1337 = 4\\7877 = 3\\5732 = 4\\8315 = 3\\4466 = 0\\0354 = 2\\8923 = 2\\8888 = 0\\0000 = 0\\4059 = 1\\2403 = \ ?$$
I need to solve the question mark. 
Does anyone have an idea how I can solve it? Thanks!
 A: This is a spin on an old problem - count the circles in a number f.e. $8888 = 8$, or $6655 = 2$. This explains why the question is titled "Easy as 3.14", it is referring to a closed loop. There is no mathematical formula for calculating it, you have to look typographically:
$1111 = 4,$ all 4 of the numbers don't have a loop in them.
$0000 = 0,$  all 4 of the numbers have a loop in them.
So, the answer by this logic would be:
$2403 = 2$, as $\{ 4,0\}$ have loops.
A: This is a very suspect question. One that someone may find many answers to. In this case I believe the desired solution is that for four digits $abcd$. Then $abcd =$ the number of primes out of $\{a,b,c,d\}$. Hence $2403 = 2$ is probably the intended answer.
Edit : 1 is also included as a prime here. Although in general it is not considered to be one.
A: I notice a pattern which ist$ 4-n$
Where $ n $ is how many digits are written by overlapping. 
Eg 1, 2, 3, 5 and 7 don't overlap
4, 6, 8, 9 and 0 overlap.
Assuming this the answer is 2.
My reasoning works for all the numbers you gave but might not be the desired solution. 
A: The first two equations imply that $2001=0$ and $1333=0$, but $2001$ and $1333$ are coprime, so we must be working modulo $1$. (And the rest of the equations are indeed true modulo $1$). Therefore the correct answer is that $2403 = \mathit{anything\;at\;all}$.
