If 608xy0 is divisible by both 3 and 11, the nonzero digit in the hundred’s place and ten’s places, respectively, are:
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 8 and 5.
Divisibility rule of 3: If the sum of all digits is divisible by 3, the number is divisible by 3.
Divisibility rule of 11: If the difference between the sum of the digits at the odd places and the digits at the even places is either 0 or multiple of 11, the number is divisible by 11.
According to the given question,
(6 + 8 + y) - (0 + x + 0) = 11
⇒ x = 3 + y
Now, sum of all digits = 6 + 0 + 8 + x + y + 0
Putting the value of x, 14 + 3 + y + y = 17 + 2y
Here, 28 + 2y must be divisible by 3.
Possible values of y = 2, 5, 8
Now, values of x = 5, 8, 11(not possible)
Possible pairs of x and y = (2 and 5); (5 and 8)
According to the given options, required answer is (8 and 5).