All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Nine forever (Posted on 2015-09-30)
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.

Comments: ( Back to comment list | You must be logged in to post comments.)
 Aha! (spoiler) | Comment 1 of 2
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

 Posted by Steve Herman on 2015-09-30 14:59:06

 Search: Search body:
Forums (0)