9548 16862 = 8362 + x + 7314 x = 16862 - 8362 ----- = 8500 16862 -----
∴ (4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11;
(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11;
(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11;
(4x - 9y) = (4 x 5 - 9 x 1) = 11, which is divisible by 11.
Now, 1397 = 11 x 127
∴ The required 3-digit number is 127, the sum of whose digits is 10.
Given Exp. | = a2 + b2 - 2ab, where a = 287 and b = 269 |
= (a - b)2 = (287 - 269)2 | |
= (182) | |
= 324 |
Then, x2 = (6q + 3)2
= 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
Thus, when x2 is divided by 6, then remainder = 3.
= 460 x (4 x 85)
= (460 x 340), which is divisible by 10.
Then, | n | [2a + (n - 1)d] = 1800 |
2 |
⟹ | n | [2 x 6 + (n - 1) x 6] = 1800 |
2 |
⟹ 3n (n + 1) = 1800
⟹ n(n + 1) = 600
⟹ n2 + n - 600 = 0
⟹ n2 + 25n - 24n - 600 = 0
⟹ n(n + 25) - 24(n + 25) = 0
⟹ (n + 25)(n - 24) = 0
⟹ n = 24
Number of terms = 24.
Given Exp. = 666 x | 1 | x | 1 | = 37 |
6 | 3 |
By hit and trial, we find that 114345 is divisibleby 11 as well as 9. So, it is divisible by 99.
Required H.C.F. = | H.C.F. of 9, 12, 18, 21 | = | 3 |
L.C.M. of 10, 25, 35, 40 | 1400 |
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