try this math explain any general rule? Any number get subtracted from its reverse will be divisible by 9?

Question

(1234-4321)/9=-343 which is an iteger
(52322-22325)/9= should be an integer, try.

explain?

Answer 1

Let's start with an easy case of a 3 digit number 'abc'.

The value of 'abc' is:
a*100 + b*10 + c

And the value of the reverse 'cba' is:
c*100 + b*10 + a

Subtracting these two equations you get:
a*100 + b*10 + c - c*100 - b*10 - a

Simplifying you get:
99a - 99c

You can factor out a 9 to get:
9(9a - 9c)

Since a and c are integers, then 9a - 9c will be an integer. And since you factored out a 9, this means that the difference is always divisible by 9.

You can do this for more digits too:

Let's say you had 6 digits 'abcdef'
100000a + 10000b + 1000c + 100d + 10e + f

The reverse is 'fedcba'
100000f + 10000e + 1000d + 100c + 10b + a

Subtracting you get:
99999a + 9990b + 900c - 900d - 9990e - 99999f

Again factor out a 9:
9(11111a + 1110b + 100c - 100d - 1110e - 11111f)

So it is divisible by 9.

The rule follows regardless of the number of digits which you can prove using x0, x1, x2, x3, etc...

Answer 2

Pick a 3-digit number in which the first and last digits differ by more than one, reverse this number (531 becomes 135) and subtract the smaller from the larger, add this number to the reverse of itself. I'll bet your answer is 1089.

I guess we have to stipulate that the first and last digits must differ by more than one, so we ever don't end up with a zero as the left digit, which would complicate matters. Here is our number (pretend that x>z): 100x+10y+z. We reverse and subtract: 100(x-z-1)+90+(10-x+z). The reverse of this is 100(10-x+z)+90+(x-z-1). Adding, we get 1089.

1089 is divisible by 9