How many pairs of X and Y are possible
in the number 763X4Y2, if the number
is divisible by 9?
(a) 8 (b) 9
(c) 10 (d) 11
7. If 8947A56B1 is divisible by 9, where
B is an odd number. Find the sum of
all possible value of A?
अगर 8947A56B1, 9 से विभाज्य है, जहा B
एक विषम संख्या है। A के सभी संभावित मानों
का योग ज्ञात करो
a) 26 b) 27
c) 30 d) 36 8. Find the remainder

1. 11 option (d) to be divisible by 9 the sum of numbers should be a multiple of 9. The sum of numbers is 22+ x+ y. The total sum can either be equal to 27 or 36. Let's take the first option : 27. Therefore x+y should be equal to 5. 6 pairs are possible. They are : (0,5) (5,0) (1,4) (4,1) (2,3) (3,2). Now the second option 36. Therefore x+ y should be equal to 14. 5 pairs are possible. (5,9) (9,5) (7,7) (8,6) (6,8). Therefore 11 such pairs are possible. ie. 6+5. Hence option D.
Same logic applies to your second question as well.

B is the correct option

The right answer of first question is 11