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

Home > Numbers
Tricky Ten Digit Number (Posted on 2006-07-31) Difficulty: 3 of 5
Find a number ABCDEFGHIJ, with all its digits different, such that:
  • A, C, E, G, and I are odd
  • HIJ is a multiple of BCD
  • GH is a multiple of AB
  • HIJ/BCD equals GH/AB

No Solution Yet Submitted by Yosippavar    
Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 7 of 11 |

If GH/AB = HIJ/BCD = 2

Then A & C would be 1,3,5,7 and G & I would be 3,5,7,9

B & D would be 6,8 & J & H would be 2,6 Which is not possible

If GH/AB = HIJ/BCD = 3

Then A = 1,3 & G = 3,9 but B has to be 2 hence H has to be 6.

D has to be 8 & J has to be 4 and as regards C & I following possibilities exist

a) If C is 1 then I = 5

b) If C is 3 then I = 1

c) If C is 5 then I = 7

d) If C is 7 then I = 3

e) If C is 9 then I = 9

Possibilities b), d) & e) are not possible. And possibility c) will make H odd hence possibility a) would be true.

Then A has to be 3 & G has to be 9

the remaining odd & even numbers E=7, F=0

So ABCDEFGHIJ is 3218709654

Have not tried possibility of 4 but that may not be possible.

 

 

 

 

 


  Posted by CMAS23 on 2006-08-09 01:08:48
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 (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information