Euler is Everywhere! #1

For this problem, we use the fact that $a^{\phi (N)} \equiv 1 \pmod{N}$ iff $\gcd(a, N) = 1,$ where $\phi (N)$ is Euler's totient function. (This is Euler's generalization of Fermat's little theorem.)

For the first part of the problem, $\phi (2015) = 1440,$ and since $1441$ and $2015$ are relatively prime, $1441^{1441} \equiv 1441 \pmod{2015} \equiv -574 \pmod{2015}.$ Thus, $A = 574.$

For the second part of the problem, we use the fact that $n^{\phi (2014)} \not\equiv 1 \pmod{2014}$ if $n$ and $2014$ are not relatively prime. Since $\phi (2014) = 936,$ and the totient function is the number of positive integers less than or equal to $2014$ that are relatively prime to $2014,$ the number of values of $n$ that do not satisfy the equation is just $2014 - 936 = 1078 = B.$

For the third part of the problem, since $123456789$ and $2013$ share a common factor of $3,$ even though $\phi (2013) = 1200$ and $a^{6000} \equiv a^{1200} \pmod{2013},$ the congruence is not satisfied, so $C = 2.$

Finally, $\left | AC - B \right | = \left | 574(2) - 1078 \right | = \left | 1148 - 1078 \right | = \boxed{70}.$