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 $oxed{9}$.

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 $oxed{9}$.

["Solution: Sum of the First 12 Even Numbers After 2020 and Remainder When Divided by 11", "Calculating the sum of the first 12 even numbers immediately following 2020 is both a straightforward arithmetic exercise and a great example of applying sequence summation formulas. In this article, we break down the solution step-by-step and explain how to find the remainder when this total is divided by 11.", "---", "### Identifying the Sequence", "After 2020, the next even numbers are:", "2022, 2024, 2026, ..., up to 2044.", "This forms an arithmetic sequence where:\n- First term ( a_1 = 2022 )\n- Common difference ( d = 2 )\n- Number of terms ( n = 12 )", "The 12th term ( a_{12} ) is:", "[\na_n = a_1 + (n - 1)d = 2022 + (12 - 1) \cdot 2 = 2022 + 22 = 2044\n]", "---", "### Sum of the Arithmetic Sequence", "The sum ( S ) of an arithmetic sequence is given by:", "[\nS = \frac{n}{2} (a_1 + a_n)\n]", "Substituting the values:", "[\nS = \frac{12}{2} (2022 + 2044) = 6 \ imes 4066 = 24396\n]", "---", "### Finding the Remainder When Dividing by 11", "We now calculate:", "[\n24396 \div 11\n]", "To find the remainder efficiently, instead of full division, we compute:", "[\n24396 \mod 11\n]", "Using modular arithmetic:", "1. Break down ( 24396 ) into manageable parts or use alternating sums method for divisibility by 11, but for accuracy, we directly compute:", "[\n11 \ imes 2216 = 24376\n]", "[\n24396 - 24376 = 20\n]", "Thus, the remainder is 20. Since we want the mod 11 remainder:", "[\n20 \mod 11 = 9\n]", "---", "### Final Answer", "The remainder when the sum ( 24396 ) is divided by 11 is:", "[\n\boxed{9}\n]", "This solution demonstrates not only the sum formula but also practical modular arithmetic to efficiently determine remainders in number theory."]

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