72519 x 9999 | = 72519 x (10000 - 1) |
= 72519 x 10000 - 72519 x 1 | |
= 725190000 - 72519 | |
= 725117481. |
⟹ x + 3699 + 1985 = 31111 + 2047
⟹ x + 5684 = 33158
⟹ x = 33158 - 5684 = 27474.
4500 x x = 3375 ⟹ x = | = | 3 | |
4 |
The smallest 5-digit number = 10000. 41) 10000 (243 82 --- 180 164 ---- 160 123 --- 37 --- Required number = 10000 + (41 - 37) = 10004.
Unit digit in 358 = Unit digit in [(34)14 x 32]
= Unit digit in [Unit digit in (81)14 x 32]
= Unit digit in [(1)14 x 32]
= Unit digit in (1 x 9)
= Unit digit in (9)
= 9
Unit digit in (795 - 358) = Unit digit in (343 - 9) = Unit digit in (334) = 4.
So, Option B is the answer.
∴Sum = | n | (a + l) | = | 50 | x (51 + 100) = (25 x 151) = 3775. |
2 | 2 |
(2n + 3)2 - (2n + 1)2 = (2n + 3 + 2n + 1) (2n + 3 - 2n - 1)
= (4n + 4) x 2
= 8(n + 1), which is divisible by 8.
5 | x z = 13 x 1 + 12 = 25 -------------- 9 | y - 4 y = 9 x z + 8 = 9 x 25 + 8 = 233 -------------- 13| z - 8 x = 5 x y + 4 = 5 x 233 + 4 = 1169 -------------- | 1 -12 585) 1169 (1 585 --- 584 --- Therefore, on dividing the number by 585, remainder = 584.
This is an A.P. in which a = 10, d = 5 and l = 95.
tn = 95 ⟹ a + (n - 1)d = 95
⟹ 10 + (n - 1) x 5 = 95
⟹ (n - 1) x 5 = 85
⟹ (n - 1) = 17
⟹ n = 18
∴Requuired Sum = | n | (a + l) | = | 18 | x (10 + 95) = (9 x 105) = 945. |
2 | 2 |
35423 317 x 89 = 317 x (90 -1 ) + 7164 = (317 x 90 - 317) + 41720 = (28530 - 317) ----- = 28213 84307 - 28213 ----- 56094 -----
Here a = 3 and r = | 6 | = 2. Let the number of terms be n. |
3 |
Then, tn = 384 ⟹ arn-1 = 384
⟹ 3 x 2n - 1 = 384
⟹ 2n-1 = 128 = 27
⟹ n - 1 = 7
⟹ n = 8
∴ Number of terms = 8.
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