Subject: Some 6 equations
Posted by: jolana
Date: Nov 30 11
Do any maths operations to solve these equations. (It was sent to me by my math teacher colleague and I´d like to share, it´s fun):
1) 1 1 1 = 6
2) 2 2 2 = 6
3) 3 3 3 = 6
4) 4 4 4 = 6
5) 5 5 5 = 6
6) 6 6 6 = 6
7) 7 7 7 = 6
8) 8 8 8 = 6
9) 9 9 9 = 6
8) 8 8 8 = 6
9) 9 9 9 = 6
Posted by: jolana
Date: Nov 30 11
Do any maths operations to solve these equations. (It was sent to me by my math teacher colleague and I´d like to share, it´s fun):
1) 1 1 1 = 6
2) 2 2 2 = 6
3) 3 3 3 = 6
4) 4 4 4 = 6
5) 5 5 5 = 6
6) 6 6 6 = 6
7) 7 7 7 = 6
8) 8 8 8 = 6
9) 9 9 9 = 6
8) 8 8 8 = 6
9) 9 9 9 = 6
17 replies. On page 1 of 1 pages. 1
Jakeroo 
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2) (2+2)x2 = 6 OR 2+2+2 = 6 3) (3x3)-3 = 6 5) (5/5)+5 = 6 6) (6/6)x6 = 6 7) 7-(7/7) = 6 Okay, I got the easy ones, up to the rest of you math wizzes to get the rest lol Reply #1. Nov 30 11, 6:47 PM |
jolana
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Oh, how I hate the changes of my Czech keyboard in the title! It´s "I´d like to share, it´s fun." It´s the apostrophe that makes the problem. Btw, did you know that we, Easterners, have "y" instead of "z" and vice versa? Reply #2. Nov 30 11, 6:49 PM |
| jolana
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Jakeroo, almost right, I would say, (2+2)x2 = 8:) Never mind. The second choice is correct:) Reply #3. Nov 30 11, 6:52 PM |
| Jakeroo
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oops typo: first entry for "2" should read (2x2)+2 not 2x2x2 (duh lol) Reply #4. Nov 30 11, 6:52 PM |
geniusonwheels
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Can we use roots like (third root 8) + (third root 8) + (third root 8) = 6? Reply #5. Nov 30 11, 6:53 PM |
| jolana
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You can use powers and roots, of course. Reply #6. Nov 30 11, 6:55 PM |
| jolana
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In the end, I should post the correct answers, I hope to find the signs on my keyboard:) Reply #7. Nov 30 11, 6:56 PM |
toocool4u
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9) (9 + 9)/ (root 9) Reply #8. Nov 30 11, 9:30 PM |
| toocool4u
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4) 4 + 4 - (root 4) = 6 or (root 4) + (root 4) + (root 4) =6 Reply #9. Nov 30 11, 9:33 PM |
Juggernaut314
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Well square roots take care of 4 & 9 then: 4^(1/2) + 4^(1/2) + 4^(1/2) = 6 (9 + 9) / (9^(1/2)) = 6 Reply #10. Nov 30 11, 9:34 PM |
| Juggernaut314
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Thinking about this harder, if I can write in any power I want or any root beyond square root, there are an infinite number of solutions to these. Reply #11. Nov 30 11, 9:39 PM |
| Juggernaut314
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A clean one for 1: (1+1+1)! = 6 Reply #12. Nov 30 11, 9:41 PM |
lola0177
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I think any might hold in modular arithmetic Reply #13. Apr 05 12, 7:06 PM |
superfan123
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Here's #9: (9-(?9+?9))! In other words, take the square root of two 9s to get 9, 3, 3. Add 3+3 to get 6 and subtract 6 from 9 to get 3. Then factorial it. Reply #14. Oct 16 12, 3:07 PM |
| superfan123
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Ok, wait, apparently I can't post square roots on the boards, otherwise they turn up as ?s. Below is another version: (9-(SQ9+SQ9))!=6 SQ means "square root of" in terms of this equation. Reply #15. Oct 16 12, 3:09 PM |
Gil_Galad
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Juggernaut had it right. Bringing roots into this is practically cheating. Reply #16. Oct 17 12, 11:40 PM |
| superfan123
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I think so too... Reply #17. Oct 18 12, 1:49 PM |
17 replies. On page 1 of 1 pages. 1
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