𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
前往频道在 Telegram
📈 Telegram 频道 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹 的分析概览
频道 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹 (@gate2027updates) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 14 369 名订阅者，在 教育 类别中位列第 13 986，并在 印度 地区排名第 28 834 位。
📊 受众指标与增长动态
自 невідомо 创建以来，项目保持高速增长，吸引了 14 369 名订阅者。
根据 27 七月, 2026 的最新数据，频道保持稳定运转。过去 30 天订阅人数变化为 570，过去 24 小时变化为 25，整体触达仍然可观。
- 认证状态： 未认证
- 互动率 (ER)： 平均受众互动率为 17.45%。内容发布后 24 小时内通常能获得 3.52% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 2 506 次浏览，首日通常累积 505 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 3。
- 主题关注点： 内容集中在 ace, wallah, notes, lectures, pdfs 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新（最新数据采集于 28 七月, 2026），频道始终保持新鲜度与高覆盖。分析显示受众积极互动，使其成为 教育 类别中的关键影响点。
14 369
订阅者
+2524 小时
+1737 天
+57030 天
帖子存档
Q) The average of 5 consecutive odd numbers is 27. The largest number is:
Repost from 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
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Q) A's salary is 25% more than B's salary. By what percent is B's salary less than A's salary?
IITM GATE 2027 Website is Live !!https://gate.iitm.ac.in/ https://gate.iitm.ac.in/ You can check the updated syllabus directly from the website
Q) Let A be a singular Hermitian matrix with an eigenvalue λ. The minimum eigenvalue of the matrix A² is:
Q) In how many ways can 5 people be seated around a circular table?
Q) In how many ways can 5 people be seated around a circular table?
Q) Statement: "The eigenvalues of a unitary matrix can take only 1 or -1."
Q) Let X be an eigenvector of matrix A. If B = P⁻¹AP, then the eigenvector Y of matrix B corresponding to the same eigenvalue is:
Q) If A is a singular square matrix of order n (where n >= 2), then Adj(A) is:
Q) If A is a 3x3 matrix with eigenvalues 1, 2, and 3, then what is the value of det(A^2 + A + I)?
Detailed Explanation1. How do we get a zero? A trailing zero is created every time you multiply a 2 and a 5 (because 2 × 5 = 10). In 100! (which is 100 × 99 × 98 × ... × 1), there are plenty of even numbers, meaning we have more than enough 2s. Therefore, the number of trailing zeros is determined entirely by how many 5s are hiding inside the numbers from 1 to 100. 2. Counting the 5s (Step 1): First, we count every number that is a multiple of 5 (5, 10, 15, 20... up to 100). Formula: 100 ÷ 5 = 20 So, there are 20 numbers that give us at least one 5. 3. The Trap (Why the answer isn't 20): Some numbers contain more than one 5. Take the number 25, for example. 25 is 5 × 5 (it has two 5s). The numbers 25, 50, 75, and 100 all have an "extra" 5 hiding in them. Our first step only counted one of them! 4. Counting the extra 5s (Step 2): To count those extra 5s, we divide 100 by 25 (which is 5 squared). Formula: 100 ÷ 25 = 4 So, there are 4 extra 5s. (Note: We don't need to check for 125 (5 cubed) because 125 is bigger than 100). 5. Final Total: Add them together: 20 (from the multiples of 5) + 4 (from the multiples of 25) = 24 total 5s. Because there are exactly 24 fives (and plenty of twos to pair them with), there will be exactly 24 trailing zeros at the end of 100!.
Q) How many zeros are there at the end of 100 factorial?
Q) How many zeros are there at the end of 100 factorial?
Q) If it takes 1 hour to boil 1 egg in a pot, how long to boil 4 eggs together?
