Ok all you have to do is add the symbols to make the following equations true. 0 0 0 = 6 1 1 1 = 6 2 2 2 = 6 3 3 3 = 6 4 4 4 = 6 5 5 5 = 6 6 6 6 = 6 7 7 7 = 6 8 8 8 = 6 9 9 9 = 6 The rules are simple you can use any math symbols you want but you cannot add any new numbers to the equation. Adding exponents is out because doing something like 3^2 (how you show 3 to the second for those not familiar with the format)would be out because it adds a "2". You can use square roots because there is no number in the symbol but can't use cube roots because it has a 3 in the symbol. You get it absolutely NO NEW NUMBERS. All other symbols are allowed. As an example I'll do 2 2 2 2= 6 could be 2+2+2=6 or 2x2+2=6 or 2^2+2=6 All of these are possible and as above some have several correct answers. If nobody gets them all I'll post the answers on my wall or something probably Saturday evening. ---------- Post added 9th May 2014 at 07:23 AM ---------- Oh almost forgot 2,3,6 are the easiest low hanging fruit. 5 and 7 are next followed by 4 and 9, but most people hit a wall with 0,1,8 they are possible but require some advanced math.
Sqrt (4) + sqrt (4) + sqrt (4) = 6. ---------- Post added 9th May 2014 at 06:06 AM ---------- (3*3) - 3 = 6 ---------- Post added 9th May 2014 at 06:09 AM ---------- 7 - (7÷7) = 6 ---------- Post added 9th May 2014 at 06:27 AM ---------- ( Arccos (1) - arcsin (1) )÷ arctan (1) = 6 ---------- Post added 9th May 2014 at 06:29 AM ---------- ( cos (0) + cos (0) + cos (0) )! = 6 ---------- Post added 9th May 2014 at 06:31 AM ---------- ( Sqrt(9) x sqrt(9) ) - sqrt (9) = 6.
(1 + 1 + 1)! = 6 (0! + 0! + 0!)! = 6 3 - 3 + 3! = 6 ln(3 / 3) + 3! = 6 [4 - exp(4 - 4)]! = 6 ---------- Post added 9th May 2014 at 10:59 AM ---------- I'm not sure whether I'm allowed to use the exp function (since it's e^... and e is a number). Anyway, here are more answers (figured out that factorials are very useful for this problem!): (4 - 4/4)! = 6 [sqrt(8 + 8/8)]! = 6
Good job quem. Yea when you figure out you need factorials this one solves itself, but most people don't know about them (at least not in the US).
(0!+0!+0!)! = 6 (1+1+1)! = 6 2+2+2 = 6 3x3-3 = 6 √4+√4+√4 = 6 (5/5)+5 = 6 6+6-6 = 6 7-(7/7) = 6 8-√√(8+8)) = 6 9 - (√√9)*(√√9) = 6
Well, since a solution has been posted, maybe there should be a new puzzle??? "The sum of three consecutive integers is also equal to their product. What are the three integers?" Hint: There are actually three sets of three integers that give a solution to this problem.
Seems like Minnie got them all. Anyway, another one for the original problem (Using resu's idea of two square roots): [sqrt(sqrt(4))]*[sqrt(sqrt(4))] + 4 = 6 @Redneck: Didn't know that! I live in Europe by the way. :icon_bigg
