Harshit Singh
Last Activity: 3 Years ago
Dear Student
Let ‘a’ be placed here
Sum of its digits at odd places = 9 + 3 + 2
= 14Sum of its digits at even places = 8 + a + 9
= 17 + a
Difference = 17 + a–14
=3 +a
The difference should be 0 or a multiple of 11, then the number is divisible by 11
If 3 + a = 0
a = -3
But it cannot be a negative
Taking a closest multiple of 11 which is near to 3
It is 11 which is near to 3
Now, 3 + a = 11
a = 11–3
a=8
Therefore the required digit is 8
Thanks