Now, 1397 = 11 x 127
∴ The required 3-digit number is 127, the sum of whose digits is 10.
∴ (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.
9548 16862 = 8362 + x + 7314 x = 16862 - 8362 ----- = 8500 16862 -----
= 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.
∴Sum = | n | (a + l) | = | 50 | x (51 + 100) = (25 x 151) = 3775. |
2 | 2 |
(Place value of 7)-(face value of 7)
=7000-7=6993.
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