** Prove** that if the digits of

**N**, read from left to right, form a strictly increasing sequence, the sum of the digits of

**9*N** is always

**9 (nine)**.

You are requested to formally **prove**, not to verify.

Well, this stumped me at first. I was looking forward to seeing somebody prove this unlikely (and not quite believable) result. And then I had a flash of insight.

Multiplying by 9 is the same as multiplying by (10 - 1)

Let's say the number is abcd.

Then 9 * abcd = abcd0 - abcd

Because the digits are strictly increasing, only the unit digits involve "borrowing".

The resulting digits after multiplying by 10 and subtracting the original number are:

a, (b-a), (c-b), (d-1-c), and (10 - d).

And the sum of these digits = 9.

I have proved it for 4 digit numbers, but this same proof method can be used for any length number ( i.e, 1 through 9 digits).

q.e.d