[lightw1thoutheat]

heres my work
let x=4444^4444
let A= s.o.d (x)
let B= s.o.d (A)
let C= s.o.d (B)
first, lets re-examine mod 9, because a number in decimal expansion is congruent to the sum of its digits mod 9. This is because each power of 10 is congruent to 1 mod 9.

so 4444=-2 mod 9
also see that (-2)^3=1 mod 9
(Note: = will mean congruent for a while)

so x=(-2)^4440 * (-2)^4 mod 9
and 3|4440, so
x=(-2)^3 * (-2)^4 mod 9
x=1 * (-2)^4 mod 9= -2 mod 9

so a= -2 mod 9

1000^4444< 4444^4444< 10000^4444
digits: 4444*3+1 < ? < 4444*4+1
so 4444^4444 has less than 18,000 digits
so A< 18000*9=16200 (6 digits)
so B< 9*6=54

the number <= 54 with the largest sum of digits is 49.

so s.o.d. (b) <4+9=13

we know b=-2 mod 9 and s.o.d (b)=-2 mod 9

so, -2= y mod 9 0=< y =< 13

so y=7

the answer is 7