20050317, 07:49  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Kiddie problem.
I am a 3 digit number. If you take away the 1st digit the remaining no. is 1/5th of the original 3 digit no. . If you take away the next digit the remaining no. is 1/5th of the 2 digit no. What no. am I? Mally 
20050317, 12:16  #2 
May 2003
Belgium
2×139 Posts 
easy....
125 
20050317, 17:02  #3  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Kiddie problem
Quote:
Mally 

20050318, 22:55  #4 
Jun 2003
3^{2}×5^{2}×7 Posts 
Here is the solution
125 Last fiddled with by Wacky on 20050319 at 00:24 Reason: spoiler tag added 
20050323, 08:07  #5 
Sep 2002
3C_{16} Posts 
It's not really a 3 digit number... 000

20050324, 05:16  #6 
2^{5}×3×101 Posts 
125

20050324, 16:09  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Kiddie problem
Quote:
Was it guess work or you have a method? Mally 

20050324, 19:57  #8 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3·13·229 Posts 
How about [b].[/b]125

20050324, 20:28  #9 
Sep 2002
Database er0rr
5·701 Posts 
And ""12.5" assuming the that a half can be written as ".5""
(I don't think I have the blanking thingy sorted yet) 
20050324, 22:30  #10  
"Sander"
Oct 2002
52.345322,5.52471
29×41 Posts 
Quote:


20050324, 22:41  #11 
Sep 2002
Database er0rr
6661_{8} Posts 
Thanks
another the imaginary number 12.5i 
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