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
Aha! (spoiler)
