Let (x + 1) be completely divisible by the natural number N. Then,
(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
Given Exp. | = -84 x (30 - 1) + 365 |
= -(84 x 30) + 84 + 365 | |
= -2520 + 449 | |
= -2071 |
Let the required fraction be x. Then | 1 | - x = | 9 |
x | 20 |
∴ | 1 - x2 | = | 9 |
x | 20 |
⟹ 20 - 20x2 = 9x
⟹ 20x2 + 9x - 20 = 0
⟹ 20x2 + 25x - 16x - 20 = 0
⟹ 5x(4x + 5) - 4(4x + 5) = 0
⟹ (4x + 5)(5x - 4) = 0
x = | 4 |
5 |
∴ x = 2.
∴Each one of (4743 + 4343) and (4747 + 4347) is divisible by (47 + 43).
3251 Let 4207 - x = 3007 + 587 Then, x = 4207 - 3007 = 1200 + 369 ---- 4207 ----
Sn = | n | [2a + (n - 1)d] | = | 45 | x [2 x 1 + (45 - 1) x 1] | = | ❨ | 45 | x 46 | ❩ | = (45 x 23) |
2 | 2 | 2 |
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035.
Shorcut Method:
Sn = | n(n + 1) | = | 45(45 + 1) | = 1035. |
2 | 2 |
Clearly, 653xy is divisible by 10, so y = 0
Now, 653x0 is divisible by 9.
So, (6 + 5 + 3 + x + 0) = (14 + x) is divisible by 9. So, x = 4.
Hence, (x + y) = (4 + 0) = 4.
∴ 100 is not a prime number.
√101 < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
∴101 is a prime number.
Hence 101 is the smallest 3-digit prime number.
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