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 |
587 x 999 | = 587 x (1000 - 1) |
= 587 x 1000 - 587 x 1 | |
= 587000 - 587 | |
= 586413. |
(112 x 54) = 112 x | ❨ | 10 | ❩ | 4 | = | 112 x 104 | = | 1120000 | = 70000 |
2 | 24 | 16 |
⟹ x + 3699 + 1985 = 31111 + 2047
⟹ x + 5684 = 33158
⟹ x = 33158 - 5684 = 27474.
72519 x 9999 | = 72519 x (10000 - 1) |
= 72519 x 10000 - 72519 x 1 | |
= 725190000 - 72519 | |
= 725117481. |
(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 |
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