The least multiple of 11, which leaves a remainder of 8, when divided by 5, 3, 18 and 20 is :

Correct option is B

Given:
Least multiple of 11, which leaves a remainder of 8, when divided by 5, 3, 18 and 20 is :
Solution:
LCM(5, 3, 18, 20) = LCM(5, 3, 2×3², 2²×5) = 4×9×5 = 180.
The number N leaves a remainder of 8 when divided by these numbers. So, N = 180k + 8 for some integer k.
We need (180k + 8) to be divisible by 11.
180k + 8 ≡ 0 (mod 11)
Since 180 = 11 × 16 + 4, we can write:
4k + 8 ≡ 0 (mod 11)
4k ≡ -8 (mod 11)
4k ≡ 3 (mod 11) (since -8 + 11 = 3)
To find k, we can multiply by the modular inverse of 4 (mod 11). The inverse of 4 is 3 because 4 × 3 = 12 ≡ 1 (mod 11).
3 × 4k ≡ 3 × 3 (mod 11)
12k ≡ 9 (mod 11)
k ≡ 9 (mod 11)
The smallest positive integer value for k is 9.
Substitute k = 9 into N = 180k + 8:
N = 180(9) + 8 = 1620 + 8 = 1628.
The least multiple is: 1628