The troubling four 4s...

You probably should have started off the question by mentioning the fact that it is an IMO problem.

Let N $= 4444^{4444}$ . The maximum number of digits in N is less than $4444\times4 = 17776$ . Hence, the maximum possible value of A is $17776\times9 = 159984$ .

Similarly, the maximum possible value of B is $45$ & the sum of the digits of B $\leq 12$ .

Observe, that N $≡$ A $≡$ B $(mod 9)$ & $4444$ $≡$ $7$ $(mod 9)$ .

Since $7^{3}$ $≡$ $1$ $(mod 9)$ , we get, $4444^{4444}≡7^{4444}≡7$ $(mod 9)$ as $4444≡1 (mod3)$ .

Hence, the sum of the digits of B is $\boxed{7}$ .

Note that 4444^{4444}<10000^{4444}=\left(10^4\right)^{4444}=10^{17776}

Therefore 4444^{4444} has less than 17776 digits. This shows that A<9\cdot 17775=159975. The sum of the digits of A is then maximized when A=99999, so B\leq 45. Note that out of all of the positive integers less than or equal to 45, the maximal sum of the digits is 12.

It's not hard to prove that any base-10 number is equivalent to the sum of its digits modulo 9. Therefore 4444^{4444}\equiv A\equiv B(\bmod{9}). This motivates us to compute X, where 1\leq X \leq 12, such that 4444^{4444}\equiv X(\bmod{9}). The easiest way to do this is by searching for a pattern. Note that

4444^1\equiv 7(\bmod 9)\4444^2\equiv 4(\bmod 9)\4444^3\equiv 1(\bmod 9)

and since 4444=3\times 1481+1,

4444^{4444}\equiv 4444^{3\times1481+1}\equiv \left(4444^3\right)^{1481}\times 4444\equiv 1\times 4444\equiv 7(\bmod{9})

Thus, X=7, which means that the sum of the digits of B is \boxed{7}.