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

Home > Numbers
Duplicate Digit Determination (Posted on 2006-12-03) Difficulty: 3 of 5
If 2^P and 5^P start with the same (non-zero) digit for positive integer P, what is that digit? Can you prove it must be so?

See The Solution Submitted by Old Original Oskar!    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Another solution without logs | Comment 5 of 6 |

But this one is based on the log solution.

Consider 2^p to be expressed in scientific notation as a*10^m, and 5^p as b*10^n.  Their product, a*b*10^(m+n) must be a power of 10, 10^(m+n+1), so a*b = 10.

If a and b were exactly equal, they'd be sqrt(10), which begins with a 3.  If a were so low as exactly 3, b would be 3.333..., and when a got down to 2.999 (finite number of 9's, so just under 3), b would still start with 3 and there'd be a mismatch.  If a were as high as 3.99999, b would be2.5, a mismatch. Any further deviations would be further mismatches, so the only matching starting digit is 3.

  Posted by Charlie on 2006-12-04 09:11:08
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 (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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