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

Home > Just Math
Nine forever (Posted on 2015-09-30) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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). 


  Posted by Steve Herman on 2015-09-30 14:59:06
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information