𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
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📈 Analytical overview of Telegram channel 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹
Channel 𝗕𝘁𝗲𝗰𝗵 & 𝗚𝗮𝘁𝗲 𝗦𝘁𝘂𝗱𝘆 𝗺𝗮𝘁𝗲𝗿𝗶𝗮𝗹 (@gate2027updates) in the English language segment is an active participant. Currently, the community unites 14 365 subscribers, ranking 13 992 in the Education category and 28 875 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 14 365 subscribers.
According to the latest data from 26 July, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by 565 over the last 30 days and by 20 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 16.74%. Within the first 24 hours after publication, content typically collects 3.52% reactions from the total number of subscribers.
- Post reach: On average, each post receives 2 401 views. Within the first day, a publication typically gains 505 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 3.
- Thematic interests: Content is focused on key topics such as ace, wallah, notes, lectures, pdfs.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 27 July, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
14 365
Subscribers
+2024 hours
+1597 days
+56530 days
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| Date | Subscriber Growth | Mentions | Channels | |
| 28 July | +4 | |||
| 27 July | +25 | |||
| 26 July | +20 | |||
| 25 July | +14 | |||
| 24 July | +26 | |||
| 23 July | +23 | |||
| 22 July | +27 | |||
| 21 July | +38 | |||
| 20 July | +11 | |||
| 19 July | +24 | |||
| 18 July | +16 | |||
| 17 July | +17 | |||
| 16 July | +21 | |||
| 15 July | +25 | |||
| 14 July | +35 | |||
| 13 July | +14 | |||
| 12 July | +18 | |||
| 11 July | +25 | |||
| 10 July | +5 | |||
| 09 July | +15 | |||
| 08 July | +24 | |||
| 07 July | 0 | |||
| 06 July | +25 | |||
| 05 July | +36 | |||
| 04 July | +12 | |||
| 03 July | +20 | |||
| 02 July | +13 | |||
| 01 July | +20 |
Channel Posts
Q) The average of 5 consecutive odd numbers is 27. The largest number is:
| 2 | ✅ GATE 2027 ECE , EE , IN Preparation Channel
✏️Short Notes | Important Formulas | Previous Year Concepts | Quick Revision Material
⚡️Useful for last-minute revision and concept clarity.
⚡️Join now and prepare smart 📚
🔗@Gate_ECE_EE_IN_Short_Notes
🔗@Gate_ECE_EE_IN_Short_Notes
🔗@Gate_ECE_EE_IN_Short_Notes | 846 |
| 3 | Q) A's salary is 25% more than B's salary. By what percent is B's salary less than A's salary? | 925 |
| 4 | 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 | 2 310 |
| 5 | Q) The unit digit of 1! + 2! + 3! + ... + 100! is: | 2 881 |
| 6 | Q) Let A be a singular Hermitian matrix with an eigenvalue λ. The minimum eigenvalue of the matrix A² is: | 2 954 |
| 7 | Q) In how many ways can 5 people be seated around a circular table? | 2 735 |
| 8 | Q) In how many ways can 5 people be seated around a circular table? | 1 |
| 9 | Q) Statement: "The eigenvalues of a unitary matrix can take only 1 or -1." | 2 586 |
| 10 | 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: | 2 216 |
| 11 | Q) If A is a singular square matrix of order n (where n >= 2), then Adj(A) is: | 2 241 |
| 12 | Q) If A is a 3x3 matrix with eigenvalues 1, 2, and 3, then what is the value of det(A^2 + A + I)? | 2 254 |
| 13 | Detailed Explanation
1. 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!. | 2 184 |
| 14 | Q) How many zeros are there at the end of 100 factorial? | 2 168 |
| 15 | Q) How many zeros are there at the end of 100 factorial? | 1 |
| 16 | Q) How many months have 28 days? | 2 476 |
| 17 | Q) If it takes 1 hour to boil 1 egg in a pot, how long to boil 4 eggs together? | 2 665 |
| 18 | Q) The argument of j² is: | 3 356 |
| 19 | 220 Days Left For GATE 2027 | 3 431 |
| 20 | Give Reactions for more polls👇👇 | 3 423 |
