4. (3 points) Rearrange the numbers $39, 41, 44, 45, 47, 52, 55$ into a sequence such that the sum of any three consecutive numbers is a multiple of 3. Among all such sequences, the maximum value of the fourth number is ( )
A. 44
B. 45
C. 47
D. 52
【Analysis】Divide each given number by 3, the remainders are: $0, 2, 2, 0, 2, 1, 1$; Arrange the remainders, the sum of every three adjacent numbers is 3, then rearrange them as: $2, 1, 0, 2, 1, 0, 2$; Since the fourth number is 2, the largest number with a remainder of 2 is 47.

【Solution】Solution: Since $39 \div 3=13, 41 \div 3=13 \cdots 2, 44 \div 3=14 \cdots 2, 45 \div 3=15, 47 \div 3=15 \cdots$
$2, 52 \div 3=17 \cdots 1, 55 \div 3=18 \cdots 1$,
the remainders are: $0, 2, 2, 0, 2, 1, 1$;
Arrange the remainders so that the sum of every three adjacent numbers is 3, rearrange them as: $2, 1, 0, 2, 1, 0, 2$; Since the fourth number is 2, the largest number with a remainder of 2 is 47.

Therefore, the answer is: $C$.