Easy as $3.14$ question

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!

Answer 1

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.

Answer 2

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.

Answer 3

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 $2403 = \mathit{anything\;at\;all}$.

Answer 4

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.