OSSSC / OSSC MATH AND REASONING
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𝐀𝐧𝐲 PROMOTION RELATED CONTACT OWNER @BUNNYTAPAS Join our YouTube Channel: https://shorturl.at/Ulwjw 🔴 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐟𝐟𝐚𝐢𝐫𝐬 @CURRENT_AFFAIRS_REGULARLY 🔴𝐄𝐧𝐠𝐥𝐢𝐬𝐡- @BLACKBOOK_ENGLISH
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网络:📈 Telegram 频道 OSSSC / OSSC MATH AND REASONING 的分析概览
频道 OSSSC / OSSC MATH AND REASONING (@pinnacle_math_reasoning) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 11 535 名订阅者，在 教育 类别中位列第 17 102，并在 印度 地区排名第 33 457 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 11 535 名订阅者。
根据 02 十月, 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 147，过去 24 小时变化为 0，整体触达仍然可观。
- 认证状态： 未认证
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- 帖子覆盖： 每篇帖子平均可获得 2 377 次浏览，首日通常累积 488 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 4。
- 主题关注点： 内容集中在 ʜᴇʀᴇ, cgle, cgl, mcqs, mcq 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“𝐀𝐧𝐲 PROMOTION RELATED CONTACT OWNER @BUNNYTAPAS
Join our YouTube Channel: https://shorturl.at/Ulwjw
🔴 𝐂𝐮𝐫𝐫𝐞𝐧𝐭 𝐚𝐟𝐟𝐚𝐢𝐫𝐬 @CURRENT_AFFAIRS_REGULARLY
🔴𝐄𝐧𝐠𝐥𝐢𝐬𝐡- @BLACKBOOK_ENGLISH”
凭借高频更新（最新数据采集于 03 十月, 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。
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𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients are 1, 11 and 2, respectively. What is sum of the two numbers ?
Ans. (a) : 1149 – 827 = 322
1310 – 1149 = 161
1310 – 827 = 483
Largest desired number
= H.C.F of 322, 161 and 483
= 161 ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
Which is the largest number that divides 827, 1149 and 1310 to leave the same remainder in each case?
Ans. (c) 156 – 6 = 150
181 – 6 = 175
331 – 6 = 325
Required number
= HCF of 150, 175 and 325
= 25 ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
What is the greatest number by which when 156, 181 and 331 are divided, the remainder is 6 in each case?
Ans. (a) :
∵Length = 15 m 17 cm = 1517 cm. Breadth = 9m 43 cm = 943 cm.
On taking HCF of 1517 cm and 943 cm H.C.F = 41
Then the least number of tiles placed ( 1517 × 943 ) / (41 × 41 )
= 37 × 23 = 851 ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
What is the least number of square tiles required to pave the floor of a room 15m 17cm long and 9m 43cm broad?
Ans. (d): LCM of the number 6, 7 and 8 = 168
The largest number of three digits · 999
On dividing 999/ 168 the Remainder is 159.
∴ 999 – 159 = 840
The number 840 is completely divisible by 168. ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
The largest three digit number that is exactly divisible by 6, 7 and 8 is:
Ans. (b) :
∵ The LCM of 56, 57 and 58 is K.
∴ 59 is a prime number.
L.C.M of 56, 57, 58 and 59
= 59 × K = 59K ✅✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
If the Least Common Multiple of 56, 57 and 58 is K, then what will be the Least Common multiple of 56, 57, 58 and 59?
Ans. (b) :
Let the three numbers be 3x, 4x and 5x respectively.
LCM of all three numbers
= 3×4×5×x = 60 x LCM = 1800
∴60 x = 1800
⇒ x = 30
∴ Second number = 4x= 4×30= 120 ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
The proportion among three numbers is 3:4:5 and their LCM is 1800. Then the second number is:
Ans. (b) Given, 5 digit number · 538xy This number is divisibe by 3, 7 and 11 Hence, the number will be a multiple of (L.C.M) 3,7 and 11.
∴ L.C.M of 3, 7 and 11 = 231
538xy = 53823
∴ x = 2
y = 3
∴ x²+ y²= 2² + 3² => x2 + y2 = 13✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
If the 5-digit number 538xy is divisible by 3, 7 and 11, then the value of (x² + y²) is:
Ans. (d) L.C.M of 2, 3, 4, 5, 6 and 7 =420
Hence x = 420 k + 1
According to the condition
The value of x lies between 2000 and 2500
On putting k = 5
∴ x = 420 × 5 + 1
x = 2101
∴ The sum of the digit of x = 2 + 1 + 0 + 1 = 4 ✅
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
Let x be the least number of 4 digits that when divided by 2, 3, 4, 5, 6 and 7 leave a remainder of 1 in each case. If x lies between 2000 and 2500, then what is the sum of the digits of x?
Ans. (a) :
L.C.M of 15, 18 and 36
15 = 3 × 5
18 = 2 × 3 × 3
36 = 2 × 2 × 3 × 3
LCM = 180
Required number = 180 k + 9
=> 180 × 6 + 9 = 1089
Hence, on putting k = 6, the number is divisible by 11
𝐎𝐃𝐈𝐒𝐇𝐀 𝐕𝐈𝐑𝐓𝐔𝐀𝐋 𝐋𝐈𝐁𝐑𝐀𝐑𝐘
What is the least number which when divided by 15, 18 and 36 leaves the same remainder 9 in each case and is divisible by 11 ?
