This is an A.P. in which a = 102, d = 6 and l = 996
Let the number of terms be n. Then tn = 996.
∴ a + (n - 1)d = 996
⟹ 102 + (n - 1) x 6 = 996
⟹ 6 x (n - 1) = 894
⟹ (n - 1) = 149
⟹ n = 150
∴ Number of terms = 150.
=Unit digit in { 292915317923361 x 17114769 }
= (1 x 9) = 9
Thus, when 2n is divided by 4, the remainder is 2.
This is an A.P. in which a = 1, d = 1, n = 45 and l = 45
∴Sn = | n | (a + l) | = | 45 | x (1 + 45) = (45 x 23) = 1035 |
2 | 2 |
Required sum = 1035.
854 x 854 x 854 - 276 x 276 x 276 | =? |
854 x 854 + 854 x 276 + 276 x 276 |
Given Exp. = | (a3 - b3) | = (a - b) = (854 - 276) = 578 |
(a2 + ab + b2) |
∴ | a + b | = | 12 | ⟹ | ❨ | 1 | + | 1 | ❩ | = | 12 |
ab | 35 | b | a | 35 |
∴ Sum of reciprocals of given numbers = | 12 |
35 |
= Unit digit in [(42)896 x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)317 = Unit digit in (5)317 = 5
Unit digit in (341)491 = Unit digit in (1)491 = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.
= (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102)
= 22 x [12 + 22 + 32 + ... + 102]
[ | Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) | ] | |
6 |
= | ❨ | 4 x | 1 | x 10 x 11 x 21 | ❩ |
6 |
= (4 x 5 x 77)
= 1540.
8796 x 223 + 8796 x 77 | = 8796 x (223 + 77) [Ref: By Distributive Law ] |
= (8796 x 300) | |
= 2638800 |
2133 → 9 (X)
2343 → 12 (/)
3474 → 18 (X)
4131 → 9 (X)
5286 → 21 (/)
5340 → 12 (/)
6336 → 18 (X)
7347 → 21 (/)
8115 → 15 (/)
9276 → 24 (/)
Required number of numbers = 6.
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