answer:To solve this problem, we follow a systematic approach to understand how the sexagenary cycle progresses over years. Given that the year 2023 is "Gui Mao", we aim to find what the year 3023 corresponds to in the cycle. 1. **Understanding the Cycle:** - The ten heavenly stems are labeled as 1 (Jia) through 10 (Gui). - The twelve earthly branches are labeled as 1 (Zi) through 12 (Hai). - The cycle pairs a heavenly stem with an earthly branch sequentially, forming a pair like (1, 1) for "Jia Zi", (2, 2) for "Yi Chou", and so on. 2. **Representing the Year 2023:** - The year 2023 is "Gui Mao", which corresponds to the 10th heavenly stem (Gui) and the 4th earthly branch (Mao). - Therefore, 2023 can be represented as (10, 4). 3. **Calculating the Difference:** - The difference between the years 3023 and 2023 is 1000 years. 4. **Determining the Heavenly Stem for 3023:** - Since there are 10 heavenly stems, we divide 1000 by 10, which equals 100 with no remainder. - This means the heavenly stem cycles through all 10 stems 100 times, returning to the 10th stem, which is Gui. 5. **Determining the Earthly Branch for 3023:** - Since there are 12 earthly branches, we divide 1000 by 12, which gives 83 with a remainder of 4. - Starting from the 4th earthly branch (Mao) and adding the remainder 4, we move 4 positions forward in the cycle of earthly branches. - Counting from Mao (4), we proceed to Chen (5), Si (6), Wu (7), and finally to Wei (8). 6. **Conclusion:** - Combining the results, the year 3023 corresponds to the 10th heavenly stem (Gui) and the 8th earthly branch (Wei). - Therefore, the year 3023 is the year of "Gui Wei". Hence, the final answer, encapsulated as requested, is boxed{text{Gui Wei}}.

answer:Let's denote the principal amount (the initial sum of money deposited) as P, the rate of interest as r, and the number of years it takes for the amount to reach Rs. 2420 as t. According to the problem, the rate of interest (r) is 10%, or 0.10 in decimal form. We are given that the amount (A) becomes Rs. 2420 in t years and Rs. 2662 in t+3 years. Using the compound interest formula: A = P(1 + r)^n For Rs. 2420 in t years: 2420 = P(1 + 0.10)^t ... (1) For Rs. 2662 in t+3 years: 2662 = P(1 + 0.10)^(t+3) ... (2) We can divide equation (2) by equation (1) to eliminate P and solve for t: (2662 / 2420) = [(1 + 0.10)^(t+3)] / [(1 + 0.10)^t] (2662 / 2420) = (1 + 0.10)^3 1.10 = (1 + 0.10)^3 Now, we calculate (1 + 0.10)^3: (1 + 0.10)^3 = (1.10)^3 = 1.10 * 1.10 * 1.10 = 1.331 Now, we can solve for the ratio: (2662 / 2420) = 1.331 1.10 = 1.331 Now, we can see that the ratio of the amounts for the two different time periods is equal to the compound interest factor for 3 years at 10%. This means that the amount of Rs. 2420 was reached exactly 3 years before the amount of Rs. 2662. Therefore, it takes t = boxed{3} years for the amount to reach Rs. 2420.

answer:# Problem Statement: Consider an n times m grid where m, n geq 2. This grid is referred to as a "corner." A "corner" is obtained by removing an (n-1) times (m-1) rectangular section from the grid, leaving a corner-shaped figure. Two players take turns coloring any non-zero number of adjacent cells that form a rectangular or square shape within the corner. A player loses if, after their move, the entire corner becomes colored. Determining the winning strategy for the first player when n or m is greater than or equal to 2. 1. **Identify the Initial Strategy:** First, determine the structure of the game. The first player will start by coloring cells positioned along the longest side of the corner. - Since the grid is n times m, let's assume n geq m. Therefore, color n - m + 1 cells along the longer side, starting from a corner. 2. **Remaining Structure Analysis:** After coloring n - m + 1 cells, let's analyze the remainder of the grid. - This initial move leaves the two equal strips along the smaller side, each of size (m-1). 3. **Symmetric Move Mechanism:** The first player then adopts a symmetric strategy: - For every move made by the second player in one strip, the first player mirrors that move in the other strip, ensuring symmetry. This replication continues until there's only one uncolored rectangle left larger than a single cell. 4. **Final Move Strategy:** The strategy then depends on the count of remaining single-cell rectangles: - If an odd number of single-cell rectangles remains, the first player colors any multiple-cell rectangle completely. - If an even number of single-cell rectangles are left, the first player leaves one cell uncolored. 5. **Ensuring the Second Player’s loss:** By following the above pattern, after each move by the first player, there must always be an odd number of single-cell rectangles left uncolored. This forces the second player to always color one of the last remaining single-cell rectangles, thereby losing the game: - Therefore, the first player ensures they force the last uncolored cell on the second player, resulting in the second player losing the game. # Conclusion: With the described strategy, the first player can ensure a win by consistently maintaining the asymmetric move repetitions and ending configurations that favor their victory. Thus: [ boxed{text{The first player wins}} ]

answer:To find the 20th digit to the right of the decimal point in the decimal expansion of (1/2)^20, we need to calculate the value of (1/2)^20. (1/2)^20 = 1 / 2^20 2^20 is 2 multiplied by itself 20 times, which equals 1,048,576. So, (1/2)^20 = 1 / 1,048,576 Now, to find the decimal expansion, we can perform the division. However, since we are only interested in the 20th digit to the right of the decimal point, we can use a shortcut. We know that the result will be a very small number, and since 2^20 is a large number, the decimal expansion will start with several zeros after the decimal point. The 20th digit to the right of the decimal point will be the first non-zero digit in the decimal expansion of (1/2)^20. This is because the value of (1/2)^20 is less than 1/10^19 but greater than 1/10^20. Therefore, the 20th digit to the right of the decimal point in the decimal expansion of (1/2)^20 is boxed{1} .