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

Home > Just Math
Possible or not? (Posted on 2015-01-17) Difficulty: 4 of 5
Prove or disprove the following:
For any integer number N there exists at least one integer number M, such that the decimal presentation of M*N needs only two distinct digits.

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.)
Warming up Comment 6 of 6 |
It helps to have:

N = n0 + 10n1 + 100n2 + ...
M = m0 + 10m1 + 100m2 + ...

where the undercase ms and ns are integers between 0 and 9. Then, the last two digits of U = NM are given by

u0 = n0m0 mod 10   and 
u1 = n0m1 + n1m0 + floor(n0m0/10) mod 10

It is good practice to look into the circumstances under which m0 and m1 can be chosen so that u1 = u0. This implies:

m1n0 = n0m0 - m0n1 - floor(n0m0/10) mod 10

Now, if n0 is relatively prime to 10 (n0 = 1,3, or 7) then we have the fortuitous circumstance that, by varying m1, we can reach all possible values mod 10, so we are also completely free in how we pick m0..

  Posted by FrankM on 2015-02-14 13:15:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information