A Logic Puzzle

Let's suppose each digit is a value, as the hint says. From $f(1111)=4$ , we see that $\boxed{f(1)=1}$ .

From $f(1000)=4$ , since $f(1)=1$ , we must have $\boxed{f(0)=1}$ .

From $f(8008)=6$ , since $f(0)=1$ , we have $\boxed{f(8)=2}$ .

From $f(1357)=4$ , since $f(1)=1$ , we have $f(357)=3$

From $f(3817)=6$ , since $f(8)=2$ and $f(1)=1$ we have $f(37)=3$

This means $\boxed{f(5)=0}$

From $f(6518)=4$ , since $f(5)=0$ , $f(1)=1$ , and $f(8)=2$ , we have $\boxed{f(6)=1}$ .

From $f(4567)=2$ , since $f(5)=0$ and $f(6)=1$ , we have $f(47)=1$

From $f(1234)=3$ , since $f(1)=1$ we have $f(234)=2$ . This also means that $f(3)\le 2$ .

Looking at $f(47)=1$ , we see that $f(7)\le 1$ .

Looking that $f(37)=3$ , we see that we must have $\boxed{f(3)=2}$ and $\boxed{f(7)=1}$ for the two inequalities to be true.

Looking at $f(4567)=2$ , since $f(5)=0$ , $f(6)=f(7)=1$ , we have $\boxed{f(4)=0}$ .

Looking at $f(1234)=3$ , since we have $f(1)=1$ , $f(3)=2$ , and $f(4)=0$ , we have $\boxed{f(2)=0}$

We conclude that $f(2)=0$ , $f(0)=1$ , $f(1)=1$ , and $f(4)=0$ .

Thus, $f(2014)=0+1+1+0=\boxed{2}$ .

The 'logic' behind this puzzle was beyond me. I miraculously guessed it right in one try.