∴Each one of (4743 + 4343) and (4747 + 4347) is divisible by (47 + 43).
Soln:
(56*Q)+29 = D -------(1)
D%8 = R -------------(2)
From equation(2),
((56*Q)+29)%8 = R.
=> Assume Q = 1.
=> (56+29)%8 = R.
=> 85%8 = R
=> 5 = R.
(489 + 375)2 - (489 - 375)2 | =? |
(489 x 375) |
Given Exp. = | (a + b)2 - (a - b)2 | = | 4ab | = 4 |
ab | ab |
= 324 x 3 x 4 x 10
= (324 x 4 x 30), which is divisible by30.
(17200 - 1200) is completely divisible by (17 + 1), i.e., 18.
⟹ (17200 - 1) is completely divisible by 18.
⟹ On dividing 17200 by 18, we get 1 as remainder.
[ | Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) | ] | |
6 |
= | ❨ | 20 x 21 x 41 | - | 10 x 11 x 21 | ❩ |
6 | 6 |
= (2870 - 385)
= 2485.
∴ x = 2.
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 |
Given Exp. | = -84 x (30 - 1) + 365 |
= -(84 x 30) + 84 + 365 | |
= -2520 + 449 | |
= -2071 |
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.
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