Find the greatest value of b so that 30a68b (a>b) is divisible by 11.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 3
Divisibility by 11:
The difference between numbers at even positions to the number at odd positions is a multiple of 11.
⇒ 30a68b = (3 + a + 8) - (b + 6 + 0)
⇒ a + 11 - b - 6
⇒ 5 + a - b = 0
This case is not possible.
⇒ 5 + a - b = 11
⇒ a - b = 6
⇒ a = 9 so b = 3