Required numbers are 102, 108, 114, ... , 996
This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then,
a + (n - 1)d = 996
102 + (n - 1) x 6 = 996
6 x (n - 1) = 894
(n - 1) = 149
n = 150.
12345679 x 72 | = 12345679 x (70 +2) |
= 12345679 x 70 + 12345679 x 2 | |
= 864197530 + 24691358 | |
= 888888888 |
Let x be the number and y be the quotient. Then,
x = 357 x y + 39
= (17 x 21 x y) + (17 x 2) + 5
= 17 x (21y + 2) + 5)
Required remainder = 5.
Clearly, 4864 is divisible by 4.
So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3.
P = 1.
When n is odd, (xn + an) is always divisible by (x + a).
Each one of (4743 + 4343) and (4747 + 4347) is divisible by (47 + 43).
Given Exp. | = -84 x (30 - 1) + 365 |
= -(84 x 30) + 84 + 365 | |
= -2520 + 449 | |
= -2071 |
Let x = 296q + 75
= (37 x 8q + 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.