psi20 Posted June 10, 2005 Share Posted June 10, 2005 I made up this new card game and the first version goes like this. You get a deck of cards and remove the jacks, queens, and kings. Now you have Aces and 2 to 10. Aces are worth 1 and the numbers are worth themselves. Shuffle the cards and pick 5 cards. Now with those five cards, make an equation. You can use + - x / ( ) powers and roots. For example, if you picked up 9,5,1,2,7 you could make the equation (1*9+5)/2 = 7 If you picked up 9,5,6,3,3 you could make the equation 3*3^(6-5) = 9 The game gets harder--for me at least-- with the face cards in them. It's even harder (perhaps impossible) with some sets of 4 cards. But all the 5 card sets I've drawn can be made into equations. This set of numbers gave me a hard time. The puzzle is to make an equation with 3,4,7,8,10. Link to comment Share on other sites More sharing options...
Guest pupil Posted June 11, 2005 Share Posted June 11, 2005 i have one for ya psi; 2 2 2 3 - 8 + 7 + 10=4 i wasnt sure how to type powers. I havent had a comp for very long, so i improvised. Link to comment Share on other sites More sharing options...
Guest pupil Posted June 11, 2005 Share Posted June 11, 2005 Crap. Didnt work. The 2's are to go above and to the right of the 3, 8, and 7.....LOL! Any tips on typing the powers would be appreciated. Sorry for my noobness. Link to comment Share on other sites More sharing options...
psi20 Posted June 11, 2005 Author Share Posted June 11, 2005 I don't know how to do powers either, except the old 3^2 - 8^2 + 7^2 + 10 = 4 But you can't do that because you didn't draw any twos. If you're going to use powers, both base and exponent must be drawn. Link to comment Share on other sites More sharing options...
Ducky Havok Posted June 11, 2005 Share Posted June 11, 2005 I got it. The (cuberoot of 8)*7-4=10 (the cuberoot uses the 3) edit: I added parenthases to lessen confusion Link to comment Share on other sites More sharing options...
ydoaPs Posted June 11, 2005 Share Posted June 11, 2005 that would need an ace:[imath]8^{\frac{1}{3}}*7-4=10[/imath], right? Link to comment Share on other sites More sharing options...
Ducky Havok Posted June 11, 2005 Share Posted June 11, 2005 no, I meant just the cube root of just 8 (2) and he said we could use roots Link to comment Share on other sites More sharing options...
ydoaPs Posted June 11, 2005 Share Posted June 11, 2005 i missed the part about roots Link to comment Share on other sites More sharing options...
psi20 Posted June 11, 2005 Author Share Posted June 11, 2005 Yeah that works. I've been playing this game for a day, and I haven't come across a set which couldn't be done. (using 5 cards of ace through 10, I was stuck on sets of 4's) Is there a set which can't be put into an equation? Link to comment Share on other sites More sharing options...
psi20 Posted June 28, 2005 Author Share Posted June 28, 2005 Now I'm having trouble with 2,2,5,5,12. Anyone got an equation for them? Link to comment Share on other sites More sharing options...
jcarlson Posted June 29, 2005 Share Posted June 29, 2005 Now I'm having trouble with 2,2,5,5,12. Anyone got an equation for them? does sqrt require a 2 to use? or can we just sqrt something? if it doesnt, then sqrt(5*5) * 2 + 2 = 12 Link to comment Share on other sites More sharing options...
psi20 Posted June 30, 2005 Author Share Posted June 30, 2005 Yeah, sqrt needs the extra 2. Link to comment Share on other sites More sharing options...
psi20 Posted June 30, 2005 Author Share Posted June 30, 2005 I'm going to make a computer program to figure it out. Hope I'm done by tomorrow. Link to comment Share on other sites More sharing options...
psi20 Posted July 1, 2005 Author Share Posted July 1, 2005 Oh, heh, didn't see this one. 12^(2-2) = 5/5 Link to comment Share on other sites More sharing options...
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