(xn + 1) will be divisible by (x + 1) only when n is odd.
(6767 + 1) will be divisible by (67 + 1)
(6767 + 1) + 66, when divided by 68 will give 66 as remainder.
Let the sum be Rs. x. Then,
C.I. = | ![]() | x | ![]() | 1 + | 4 | ![]() | 2 | - x | ![]() | = | ![]() | 676 | x | - x | ![]() | = | 51 | x. |
100 | 625 | 625 |
S.I. = | ![]() | x x 4 x 2 | ![]() | = | 2x | . |
100 | 25 |
![]() | 51x | - | 2x | = 1 |
625 | 25 |
x = 625.