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Home » Solución: El problema es una permutación de un multiconjunto. - GetMeFoodie

Solución: El problema es una permutación de un multiconjunto. - GetMeFoodie

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 A = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6}

📌 = \sqrt{21 \cdot 8 \cdot 42} = \sqrt{7056} = 84

📌 The altitude to a side is $ h = rac{2A}{ ext{base}} $. To find the longest altitude, use the shortest side, since altitude is inversely proportional to the base. The shortest side is 13, so:

📌 h_{ ext{longest}} = rac{2 \cdot 84}{13} = rac{168}{13}

📌 Final answer: $ oxed{rac{168}{13}} $Question: What is the remainder when the sum of the first 12 even numbers after 2020 is divided by 11?

📌 Solution: The first 12 even numbers after 2020 are $2022, 2024, \ldots, 2044$. The sum of an arithmetic sequence is $S = rac{n}{2}(a_1 + a_n)$. Here, $n = 12$, $a_1 = 2022$, $a_n = 2044$. Thus, $S = rac{12}{2}(2022 + 2044) = 6 imes 4066 = 24396$. Dividing 24396 by 11: $11 imes 2216 = 24376$, so the remainder is $24396 - 24376 = 20$. However, $20 \mod 11 = 9$. Therefore, the remainder is $oxed{9}$.

📌 Question: What is the least common multiple of 2048 and 1024?

📌 Solution: The prime factorization of 2048 is $2^{11}$, and 1024 is $2^{10}$. The LCM of two numbers is the product of the highest powers of all primes present. Thus, $ ext{LCM}(2048, 1024) = 2^{11} = 2048$. The answer is $oxed{2048}$.

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