Aptitude - Numbers - Discussion

Divisibility by 3 condition is sum of all the digits are divisible by 3.
Divisibility by 11 condition is diff of sum of odd and even place digits should be equal to zero.

Let the missing digits be x and y.
Then given value becomes 476xy0.
For divisible by 11: (4+6+y)-(7+x+0)=0
=> 10+y=7+x.
=> y = (7-10)+x.
=> y = x-3 ------> (1).

Now for divisibility by 3: 4+6+7+x+y+0 => 17+x+y.
=> 17+x+x-3 => 14+2x should be divisible by 3,
now try substuiting values:
x=1: 16 not div by 3,
x=2: 18 div by 3,
x=3: 20 not div by 3.
.
.
x=5: 24 div by 3.

So for x=2
=> y= -1 which is not valid.
=> for x= 5
=> y=2. then 10+y =7+x
so the answer 2 and 5.
It will work for 8 and 5.

How can come x=2 or 8?

Please explain.

Thanks @Tanya.

Thanks @Tanya Mazumdar.

Thank you @Tanya Mazumdar.

Thank you @Tanya.

Please can anyone explain from (0 + x + 7) - (y + 6 + 4) =x - y - 3. Must be either 0 or 11?

5 and 2 are not there in the answers given. Hence 8 and 5 selected.

By solving through divisibility method we can aslo get value X = 5 and Y = 2. Is it correct?

Just try hit and trial method to solve it, by putting the values in the missing places.