If the number 476**0 is divisible by both 3 and 11, then in the hundredth and tenth places, the non-zero digits are, respectively:
A2 and 3
B3 and 2
C5 and 8
D8 and 5
Answer:
D. 8 and 5
Read Explanation:
Divisibility rule of 3 = If the sum of digits of a number is a multiple of 3, the number will be completely divisible by 3
Divisibility rule of 11 = If the difference of the sum of alternative digits of a number is divisible by 11 or it is 0, then that number is completely divisible by 11.
On considering option 8 and 5 ,The number is 476850
Sum of the digits = 4 + 7 + 6 + 8 + 5 + 0 = 30,which is divisible by 3
Sum of digits at odd places = 4 + 6 + 5 = 15
Sum of digits at even places = 7 + 8 + 0 = 15
Difference = 15 -15 = 0
this is divisible by 11 as well
