Logic Q

[Shotokan_Karate]

[QUOTE]Originally posted by Amina-DZ

Welcome back ...couldn't resist huh!

Now now I had no idea where to start with this problem! How did you solve it?

Thanks dear rival! Tell me how I could've resisted huh?!

Yes same as Phylay! Try and stop when all digits are different! But I'm thinking if there is another way too, only I don't want to cause nightmares!

[phylay]

Originally posted by Amina-DZ

Originally posted by phylay
Lol, No didn't dream about it or so I think (remember I forget my dreams )
the idea was to do a^3 starting from a = 22 and stop when all the digits are different

Yeah I remember, but still strange to find the answer at 7am

OK so it's about trying till you get the answer! Fair enough!
I was looking to find a mathematical pattern to solve it!

sure, that would have been a sexier solution...

[Shotokan_Karate]

[QUOTE]Originally posted by Amina-DZ

OK, don't tell me you can see an obtuse triangle in my picture!! Seeing is believing

Nah! Seeing can be deceiving! No optical illusions thanks!

[Amina-DZ]

I see! It's a possible solution I suppose. But I'm still thinking about another way of doing it!!

I don't like to rely on hit 'n miss too much...


[Edited by Amina-DZ on 1st September 2005 at 02:50]

[Amina-DZ]

Originally posted by Shotokan_Karate
Nah! Seeing can be deceiving! No optical illusions thanks!

C'mon now! Can you see the obtuse triangle??

[Amina-DZ]

OK If you divide the triangle in the middle, you end up with either 2 right angles (line A) or one acute and one obtuse triangle (line B or C) (i.e. back to the starting point with the new obtuse triangle)

This is not the way to do it then!

[Amina-DZ]

They all look acute to me!