Permalink Submitted by Anonymous on December 21, 2012

No divisibility rules or 99 times table is needed for the proof.

If the number is 100a + 10b + c, and a>c then
100a + 10b + c
- 100c + 10b - a
= 100*(a-c) + (c-a) -- or, since c-a is negative
= 100*(a-c-1) + 90 + (10+c-a)
This shows that the middle digit is 9.
Then adding digits:
100*(a-c-1) + 90 + (10+c-a)
+ 100*(10+c-a) + 90 + (a-c-1)
= 1000 - 100 + 180 + 9 = 1089

## conditions and proof for 1089

No divisibility rules or 99 times table is needed for the proof.

If the number is 100a + 10b + c, and a>c then

100a + 10b + c

- 100c + 10b - a

= 100*(a-c) + (c-a) -- or, since c-a is negative

= 100*(a-c-1) + 90 + (10+c-a)

This shows that the middle digit is 9.

Then adding digits:

100*(a-c-1) + 90 + (10+c-a)

+ 100*(10+c-a) + 90 + (a-c-1)

= 1000 - 100 + 180 + 9 = 1089