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(B) We need to find the greatest $k$ such that $10^k$ divides $9^{11} + 11^9$.First,express the terms using binomial expansion:$9^{11} = (10-1)^{11} =

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$2$

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(B) We need to find the greatest $k$ such that $10^k$ divides $9^{11} + 11^9$.First,express the terms using binomial expansion:$9^{11} = (10-1)^{11} = \binom{11}{0}10^{11} - \binom{11}{1}10^{10} + \dots - \binom{11}{10}10^1 + \binom{11}{11}(-1)^{11} = \dots - 110 + 10 - 1 = \dots + 100 - 10 - 1 = \dots + 89$.More precisely,$9^{11} = (10-1)^{11} = \sum_{r=0}^{11} \binom{11}{r} 10^{11-r} (-1)^r = \dots + \binom{11}{9}10^2 - \binom{11}{10}10^1 + \binom{11}{11} = \dots + 5500 - 110 + 1 = \dots + 5391$.Similarly,$11^9 = (10+1)^9 = \sum_{r=0}^{9} \binom{9}{r} 10^r = 1 + \binom{9}{1}10 + \binom{9}{2}10^2 + \dots = 1 + 90 + 3600 + \dots = 3691 + \dots$.Adding them: $9^{11} + 11^9 = (\dots + 5391) + (3691 + \dots) = \dots + 9082$.Let's check modulo $100$:$9^{11} = (10-1)^{11} \equiv -1 + 11(10) = 109 \equiv 9 \pmod{100}$.$11^9 = (1+10)^9 \equiv 1 + 9(10) + \binom{9}{2}100 \equiv 91 \pmod{100}$.$9^{11} + 11^9 \equiv 9 + 91 = 100 \equiv 0 \pmod{100}$.Thus,$100$ divides the sum,so $k \ge 2$.Checking modulo $1000$:$9^{11} = (10-1)^{11} = -1 + 11(10) - 55(100) = -1 + 110 - 5500 \equiv 109 \pmod{1000}$.$11^9 = (1+10)^9 = 1 + 9(10) + 36(100) + 84(1000) \equiv 1 + 90 + 3600 \equiv 3691 \equiv 691 \pmod{1000}$.$9^{11} + 11^9 \equiv 109 + 691 = 800 \pmod{1000}$.Since $800$ is not divisible by $1000$,the greatest value is $k = 2$.

If $k$ is a positive integer and $10^k$ is a divisor of the number $9^{11}+11^9$,then the greatest value of $k$ is

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