Question

The difference between a multiple of 9 and the number formed by reversing the order of its digits is always a multiple of 9 . Why?​

Answer 1

Answer:I think the question needs a bit of modification. It should be a positive integer having two or more digits.

So all the answers in this thread are all perfect. Hence I am going to try explaining this a bit more easily.

I shall show few proofs for numbers having more than one digits.

Case 1:

Consider a two digit number. Suppose it has digits  x  in tens place and  y  in unit place.

So the number is  10x+y  

Now it's reverse is  10y+x  

So the difference will be  (10x+y)−(10y+x)=9(x−y)  

So it is clearly indicative of the fact that it is divisible by 9.

Case2:

Consider a three digit number. Suppose it has  x  in hundredth place  y  in ten’s place and  z  in units place.

So the number is  100x+10y+z  

So the reverse of the number will be  100z+10y+x  

Therefore the difference will be  (100x+10y+z)−(100z+10y+x)=9(11x−11z)  

Hence this too is a multiple of 9.

Thus you can show for all such different digits numbers where the difference will always leave a multiple of 9 from which you can take it common to prove the statement.

Note: Here the sign is insignificant that which number we subtract from the other. So we generally take the mod value. Otherwise there will just be a (-ve) sign but that won't affect its divisibility by 9.

I hope I was able to clear this and that I am correct in my approach.

Any suggestions shall be appreciated!

Thank you!

Step-by-step explanation:

Answer 2

Answer:

9(y-x)

Step-by-step explanation:

let X and y be theones and tens digitof a multiple of 9

the number multiple will be 10y+x

after reversing the number 10x+y

now ATQ,

(10y+x) - (10x+y)

10y+x - 10x+y

9y-9x

=9(y-x)

so it clearly indicate that it is multiple of 9 and always multiple of 9