answer:1. Let us recall the given polynomial function: [ f(x) = a_{0} x^{n} + a_{1} x^{n-1} + cdots + a_{n-1} x + a_{n} ] We are given that (f(x)) has integer coefficients (a_{0}, a_{1}, ldots, a_{n}) and both (f(2)) and (f(3)) are divisible by 6. 2. Starting with the evaluations: [ f(2) = a_{0} 2^{n} + a_{1} 2^{n-1} + cdots + a_{n-1} 2 + a_{n} ] and [ f(3) = a_{0} 3^{n} + a_{1} 3^{n-1} + cdots + a_{n-1} 3 + a_{n} ] 3. Since (f(2)) is divisible by 6, it must be divisible by both 2 and 3 individually. - From (f(2)) being divisible by 2, every term except the constant term (a_{n}) contains at least one factor of 2 (since they are powers of 2). This implies that (a_{n}) must be divisible by 2. - Similarly, from (f(3)) being divisible by 3, every term except the constant term (a_{n}) contains at least one factor of 3 (since they are powers of 3). This implies that (a_{n}) must be divisible by 3. 4. Since (a_{n}) is divisible by both 2 and 3, it must be divisible by their least common multiple, which is 6. 5. We now analyze the divisibility of (f(5)): [ f(5) = a_{0} (2+3)^{n} + a_{1} (2+3)^{n-1} + cdots + a_{n-1} (2+3) + a_{n} ] 6. Using the binomial theorem to expand ((2+3)^{k}): [ f(5) = a_{0} sum_{k=0}^{n} binom{n}{k} 2^{k} 3^{n-k} + a_{1} sum_{k=0}^{n-1} binom{n-1}{k} 2^{k} 3^{n-1-k} + cdots + a_{n-1} sum_{k=0}^{1} binom{1}{k} 2^{k} 3^{1-k} + a_{n} ] 7. Combining the given information: [ f(5) = f(2) + f(3) - a_{n} + a_{0} left(C_{n}^{1} 2^{n-1} cdot 3 + C_{n}^{2} 2^{n-2} cdot 3^{2} + cdots + C_{n}^{n-1} 2 cdot 3^{n-1} right) ] [ + a_{1} left(C_{n-1}^{1} 2^{n-2} cdot 3 + C_{n-1}^{2} 2^{n-3} cdot 3^{2} + cdots + C_{n-1}^{n-2} 2 cdot 3^{n-2} right) ] [ + cdots + a_{n-2} cdot C_{2}^{1} cdot 2 cdot 3 ] 8. Reviewing each term, we observe that every term contains a multiples of 6 and thus (f(5)) must be divisible by 6 as each term remains divisible by 6. 9. Conclusion: [ f(5) text{ is divisible by } 6. ] [ boxed{6} ]

answer:Since the graph of the power function f(x)=x^{a} passes through the point (3, frac{1}{9}), we have 3^{a}= frac{1}{9}, solving for a gives a=-2. Hence, f(x)=x^{-2}, and f(2)=(2)^{-2}= frac{1}{4}. Therefore, the answer is: boxed{frac{1}{4}}. We deduce f(x)=x^{-2} from the given conditions, from which we can find the value of f(2). This problem tests the method of finding function values and is a basic question. When solving, carefully read the question and pay attention to the reasonable application of function properties.

answer:Let (a), (b), and (c) be the dimensions of the prism, such that (ab=60), (bc=70), and (ca=84). We are tasked with finding the volume (abc). Start by multiplying the equations for the areas: [ (ab)(bc)(ca) = 60 cdot 70 cdot 84. ] This simplifies to: [ a^2b^2c^2 = 60 cdot 70 cdot 84. ] Taking the square root of both sides gives: [ abc = sqrt{60 cdot 70 cdot 84}. ] Breaking down the numbers into prime factors, we get: [ 60 = 2^2 cdot 3 cdot 5, quad 70 = 2 cdot 5 cdot 7, quad 84 = 2^2 cdot 3 cdot 7. ] Thus, [ abc = sqrt{(2^2 cdot 3 cdot 5) cdot (2 cdot 5 cdot 7) cdot (2^2 cdot 3 cdot 7)} = sqrt{2^5 cdot 3^2 cdot 5^2 cdot 7^2}. ] Simplifying further, [ abc = 2^{5/2} cdot 3 cdot 5 cdot 7 = 4 cdot sqrt{2} cdot 3 cdot 5 cdot 7 = 420 sqrt{2}. ] Rounding to the nearest whole number gives: [ abc approx 420 cdot 1.414 = 593.88 approx boxed{594} text{ cubic units}. ]

answer:1. **Define the Expressions**: Let C = 7 + sqrt{50} and D = 7 - sqrt{50}. 2. **Application of Conjugate Properties**: As C and D are conjugates, their expansions help simplify, especially using the binomial theorem: [ C^{21} + D^{21} = 2left[binom{21}{0}7^{21} + binom{21}{2}7^{19}50 + cdotsright] ] 3. **Focusing on Units Digit**: Consider units digit calculations using modular 10 arithmetic: - Since 7^{21} modulo 10 needs to be calculated, remember (7^2 equiv 49 equiv -1 pmod{10}). Thus, 7^4 equiv 1pmod{10} and hence 7^{21} equiv 7 times (7^4)^5 equiv 7 times 1^5 equiv 7 pmod{10}. 4. **Other Higher Power Terms**: Consider the higher powers terms such as binom{21}{2}7^{19}50 pmod{10}, but recall all even powers of sqrt{50} become 50 with some power, reducing modulo 10 still results in 0, simplifying contribution calculations. 5. **Summarize All Contributions**: [ (C^{21} + D^{21}) pmod{10} = 2 times 7 equiv 14 equiv 4 pmod{10} ] Thus, the units digit of C^{21} + D^{21} is 4. The final answer is boxed{textbf{(B)} 4}