General Questions
Practice and master this topic with our carefully crafted questions.
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| Given Exp. = | (1000)9 | = | (103)9 | = | (10)27 | |
| 1024 | 1024 | 1024 |
| Given Exp. = 35 + 15 x | 3 | = 35 + | 45 | |
| 2 | 2 |
20 x x = (64 + 36)(64 - 36) = 100 x 28
 x = | 100 x 28 | = 140 |
| 20 |
(22 + 42 + 62 + ... + 202) = (1 x 2)2 + (2 x 2)2 + (2 x 3)2 + ... + (2 x 10)2
= (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102)
= 22 x [12 + 22 + 32 + ... + 102]
 | Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) |  | |
| 6 |
| = |  | 4 x | 1 | x 10 x 11 x 21 |  |
| 6 |
= (4 x 5 x 77)
= 1540.
Required numbers are 10, 15, 20, 25, ..., 95
This is an A.P. in which a = 10, d = 5 and l = 95.
tn = 95 a + (n - 1)d = 95
10 + (n - 1) x 5 = 95
(n - 1) x 5 = 85
(n - 1) = 17
n = 18
 Requuired Sum = | n | (a + l) | = | 18 | x (10 + 95) |
| 2 | 2 |
| Given Exp. | = a2 + b2 - 2ab, where a = 287 and b = 269 |
| = (a - b)2 = (287 - 269)2 | |
| = (182) | |
| = 324 |
91 is divisible by 7. So, it is not a prime number.
132 = 4 x 3 x 11
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 
11,3,4 (/)
396 11,3,4 (/)
462 11,3 (X)
792 11,3,4 (/)
968 11,4 (X)
2178 11,3 (X)
5184 3,4 (X)
6336 11,3,4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
Clearly, 97 is a prime number.