Solve an IMO problem in 5 minutes (1975 Problem 4)

Let A be the sum of the digits of 4444^4444, B the sum of digits of A, and C the sum of digits of B.
Find C.

[basubk]

Nice, I was looking for a understand able solution and I got one.... thanks

[yurenchu]

X = 4444^4444

number of digits of X:
log10(X) = log10(4444^4444) = 4444 * log10(4444) < 4444 * log10(10000) = 4444 * 4 = (1111 * 16)
A = {sum of digits of X} ≤ 9 * ceil(log10(X)) ≤ 9 * (1111*16) = 9999 * 16 < 160, 000
B = {sum of digits of A} ≤ 5 * 9 = 45
C = {sum of digits of B} ≤ {sum of digits of 39} = 12

4444^4444 (mod 9) ≡ A (mod 9) ≡ B (mod 9) ≡ C (mod 9)

4444 = 7 (mod 9)
7^2 = 49 = 4 (mod 9)
7^3 = 28 = 1 (mod 9)

4444^4444 (mod 9) =
= 7^4444 (mod 9)
= 7^(1 + 3*1481) (mod 9)
= 7 * (7^3)^1481 (mod 9)
= 7 * (1^1481) (mod 9)
= 7 (mod 9)

==> C = 7 .