Question

How many pairs of X and Y are possible in the number 763X4Y2, if the number is divisible by 9 ?

Answer 1

Answer:

possible pairs are 11

Step-by-step explanation:

How many pairs of X and Y is possible in the number f 763X4Y2 if the number is divisible by 9?

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11 pairs are possible.

Explanation

Rule of Divisibility by 9 : The sum of digits of Number should be divisible by 9 then only the number itself will be divisible by 9.

e.g.

Number 729

Sum of Digits = 7+2+9 = 18 (as 18 is divisible by 9, Number 729 will be divisible by 9)

Here,

The given number is 763X4Y2

sum of number = 7+6+3+X+4+Y+2 =22+X+Y

Now, X can be any number between 0 - 9

also Y can be any number between 0 - 9

so, Possible values of X+Y is any number between 0 -18

Hence, Sum of digits (22+X+Y) will be between 22(22 + 0 where both X and Y is 0) and 40 (when both X and Y is 9).

Now, in the range of 22 and 40, only 27 and 36 are divisible by 9.

Hence, X+Y must be 5 or 14.

Thus the possible combinations are

For, X+ Y = 5

X=0, Y =5

X=1, Y=4

X=2, Y=3

X=3, Y=2

X=4, Y=1

X=5, Y=0

For X+Y=14

X=5, Y=9

X=6, Y=8

X=7, Y=7

X=8, Y=6

X=9, Y=5

Answer 2

11 pairs of X and Y are possible in the number 763X4Y2, if the number is divisible by 9

Given:

Number 763X4Y2 is divisible by 9  

To Find:

Number of possible pairs of X  and Y

Solution:

A number is divisible by 9 if sum of its digits is divisible by 9

Find sum of digits of 763X4Y2

7 + 6 + 3 + X + 4 + Y + 2

= 22 + X + Y

27 and 36  are two possible sum divisible by 9

Case 1:  Sum is 27

=> X + Y = 5

Possible Pairs of X and Y  

(X , Y)  = (0 , 5) , ( 1 , 4) , ( 2 , 3) , ( 3 , 2) , ( 4, 1) , ( 5 , 0)

6 pairs

Case 2:  Sum is 36

=> X + Y = 14

Possible Pairs of X and Y  

(X , Y)  = (5 , 9) , ( 6 , 8) , ( 7 , 7) , ( 8 , 6) , ( 9, 5)

5 Pairs

Total Pairs = 11

Hence 11 pairs of X and Y are possible in the number 763X4Y2, if the number is divisible by 9