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.

The right answer of first question is 11

B is the correct option