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Home » Now, we find the time $ t $ such that $ P(t) = 16200 $ starting - Crosslake

Now, we find the time $ t $ such that $ P(t) = 16200 $ starting - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ â $ k(4) + 2k(4) = 0 + 0 $ â $ 3k(4) = 0 $ â $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1

📌 eq 0 $. Contradiction? Wait, from $ k(2) = 0 $, check $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + k(-1) - 2k(-1) = 1 - k(-1) $. Also $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $? No: $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $. So $ k(0) = 0 $. Then $ 0 = 1 - k(-1) $ â $ k(-1) = 1 $. Then $ x = -1, y = -1 $: $ k(-2) = 2k(-1) - 2k(1) = 2(1) - 2(1) = 0 $. $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + 1 - 2(1) = 0 $, consistent. Now $ x = 2, y = -1 $: $ k(1) = k(2) + k(-1) - 2k(-2) = 0 + 1 - 0 = 1 $, matches. No contradiction. Thus $ k(2) = 0 $. Final answer: $ oxed{0} $.

📌 Question: Find the remainder when $ x^5 - 3x^3 + 2x - 1 $ is divided by $ x^2 - 2x + 1 $.

📌 Solution: Note $ x^2 - 2x + 1 = (x - 1)^2 $. Use polynomial division or remainder theorem for repeated roots. Let $ f(x) = x^5 - 3x^3 + 2x - 1 $. The remainder $ R(x) $ has degree < 2, so $ R(x) = ax + b $. Since $ (x - 1)^2 $ divides $ f(x) - R(x) $, we have $ f(1) = R(1) $ and $ f'(1) = R'(1) $. Compute $ f(1) = 1 - 3 + 2 - 1 = -1 $. $ f'(x) = 5x^4 - 9x^2 + 2 $, so $ f'(1) = 5 - 9 + 2 = -2 $. $ R(x) = ax + b $, so $ R(1) = a + b = -1 $, $ R'(x) = a $, so $ a = -2 $. Then $ -2 + b = -1 $ â $ b = 1 $. Thus, remainder is $ -2x + 1 $. Final answer: $ oxed{-2x + 1} $.Question: A plant biologist is studying a genetic trait that appears in every 12th plant in a rows of crops planted in a 120-plant grid. If the trait is expressed only when the plantâs position number is relatively prime to 12, how many plants in the first 120 positions exhibit the trait?

📌 Solution: We are to count how many integers from 1 to 120 are relatively prime to 12. This is given by Eulerâs totient function applied to the set of positions, but since the condition is relative primality with 12, we compute $ \phi(12) $, the number of integers from 1 to 12 that are coprime to 12, and then multiply by the number of full cycles in 120.

📌 First, factor 12:

📌 $ 12 = 2^2 imes 3 $.

📌 \phi(12) = 12 \left(1 - rac{1}{2}

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