Find the smallest value of a so that 42a48b (a > b) is divisible by 11.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e 5.
Divisibility Rule of 11 says that, if the difference between the sum of digits at an even place and the sum of the digits at an odd place is divisible by 11 then the given number must be divisible by 11.
Required difference = ( 8 + a + 4) - (b + 4 + 2)
⇒12 + a - b - 6
⇒ a - b + 6
At a = 5 and b = 0 the given condition satisfy.
The smallest value of a will be 5.