answer:To find the value of k, we need to simplify the given equation: (1/2)^16 * (1/81)^k = 1/(18^16) First, let's simplify the right side of the equation by expressing 18 as a product of its prime factors: 18 = 2 * 9 18 = 2 * (3^2) 18^16 = (2 * 3^2)^16 18^16 = (2^16 * (3^2)^16) 18^16 = (2^16 * 3^(2*16)) 18^16 = (2^16 * 3^32) Now, let's rewrite the left side of the equation using the same base for the exponents: (1/2)^16 * (1/81)^k (1/2)^16 * (1/(3^4))^k (1/2)^16 * (1^k / (3^4)^k) (1/2)^16 * (1 / 3^(4k)) Now, we can equate the exponents of the prime factors on both sides of the equation: 2^16 * 3^(4k) = 2^16 * 3^32 Since the bases are the same, we can equate the exponents: 4k = 32 Now, we can solve for k: k = 32 / 4 k = 8 Therefore, the value of k is boxed{8} .

answer:To solve the problem, we will show step-by-step that the least value of ( k ) (the number of chosen numbers) to ensure at least one pair of numbers such that one is divisible by the other is 1004. Step-by-Step Solution: 1. **Representation of Each Number:** - Any number ( n ) from the first 2005 natural numbers can be written in the form ( 2^{a_i} b_i ), where ( a_i geq 0 ) and ( b_i ) is odd. Here: [ n = 2^{a_i} b_i ] where ( b_i ) is the odd part of ( n ). 2. **Total Number of Odd Numbers:** - Next, we observe that there are 1003 odd numbers in the first 2005 natural numbers, specifically: [ {1, 3, 5, 7, ldots, 2005} ] 3. **Pigeonhole Principle Application:** - Suppose we arbitrarily choose 1004 numbers from the first 2005 natural numbers and write each chosen number ( n_i ) in the form ( 2^{a_i} b_i ). Since we have 1004 numbers and only 1003 possible odd numbers (( b_i )), the Pigeonhole Principle guarantees that at least two of ( b_i ) values must be the same. 4. **Condition for Divisibility:** - Let's say two chosen numbers have the same odd part, i.e., ( b_{alpha} = b_{beta} ). For these numbers: [ n_alpha = 2^{a_alpha} b_{alpha} quad text{and} quad n_beta = 2^{a_beta} b_{beta} ] Since ( b_{alpha} = b_{beta} ), the condition for divisibility is: [ n_alpha mid n_beta text{ or } n_beta mid n_alpha ] As long as ( a_alpha neq a_beta ), one of these numbers will be divisible by the other because one will have a higher power of 2. 5. **Identify Counterexamples for ( k < 1004 ):** - Consider the highest consecutive set of 1003 numbers from ( 1003 ) to ( 2005 ). None of these numbers can be paired such that one divides the other because they are constructed such that their respective factors of 2 (powers) are not able to divide each other cleanly due to the range constraint. 6. **Conclusion:** - Therefore, any set of 1003 numbers might not satisfy the divisibility condition. However, any selection of 1004 numbers from the first 2005 natural numbers must have at least one pair where one number is divisible by the other. Hence, the least value of ( k ) ensuring the condition is: [ boxed{1004} ]

answer:# Problem Analysis 小明和小梅通过摸球游戏比赛，每人摸5个球，根据球的颜色计分，红球为-3分，白球为0分，黄球为2分。比赛结束后，他们将各自摸到的球所代表的分数相加，分数较高者获胜。 # Solution （1）计算小明和小梅的得分，判断谁获胜 - **小明的得分计算**： 小明摸到的球分别为：红球、黄球、红球、白球、红球。 代入分数计算得分：-3 + 2 + (-3) + 0 + (-3) 计算过程：-3 + 2 - 3 + 0 - 3 = -7 因此，小明的得分为 -7。 - **小梅的得分计算**： 小梅摸到的球分别为：黄球、黄球、白球、红球、红球。 代入分数计算得分：2 + 2 + 0 + (-3) + (-3) 计算过程：2 + 2 + 0 - 3 - 3 = -2 因此，小梅的得分为 -2。 - **比较得分，判断获胜者**： 比较小明和小梅的得分：-2 > -7 因此，小梅获胜。 boxed{text{小梅获胜}} （2）若比赛规则改为和的绝对值较大者获胜，判断谁获胜 - **小明的得分绝对值**：|-7| = 7 - **小梅的得分绝对值**：|-2| = 2 - **比较得分绝对值，判断获胜者**： 比较小明和小梅的得分绝对值：7 > 2 因此，小明获胜。 boxed{text{小明获胜}}

answer:To determine when Justin got home from baseball practice, we need to calculate the total time he needed to complete his dinner, homework, and chores. Dinner: 45 minutes Homework: 30 minutes Cleaning room: 30 minutes Taking out the trash: 5 minutes Emptying the dishwasher: 10 minutes Now, let's add up these times: 45 + 30 + 30 + 5 + 10 = 120 minutes Justin needed a total of 120 minutes (or 2 hours) to finish everything before the movie. Since the latest time he could start was 6 pm and he needed 2 hours to complete his tasks, we can calculate the time he got home from baseball practice by subtracting the 2 hours from 6 pm. 6 pm - 2 hours = 4 pm Therefore, Justin got home from baseball practice at boxed{4} pm.