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Variant

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Lepton

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  1. Wow, I'm dumb. I'd considered doing exactly that and found it wrong... but I must have just DID it wrong lol. Probably because of order of operations... again.
  2. Oh wow... I never even considered putting something like what you did for x. I'm DUH! Heh. Also can't believe I forgot order of operations... shows how much 20 year old math I actually have to use (a.k.a NONE)
  3. Yeah, I went back and edited my post. So I guess there is no solution within the rules. I'm not satisfied saying that... so here goes more wasted paper...
  4. (3+3)x3-3/3=29 becomes 6x3-3/3=29 I meant this to be 6*5*3-3/3=29 not 6*5*3-(3/3)=29 18x-3/3=29 But I don't know where this comes from: 18x-3/3=29 becomes 18x-1=29 6*5 = 30 30*3 = 90 90 - 3 = 87 87 / 3 = 29 nevermind, I see you said order of operations. I guess it can't work that way.
  5. Ahh, but it IS a trick question. Due to the nature of the way the symbols work, you only need to look at four possibilities to eliminate them all. (3-3) = 0, you'll get nowhere with that. (3/3) = 1, can't get very far with a 1 either. (3x3) = 9, you used your multiply, you have one add to use and it's downhill from there. (3+3) = 6, used your add already, 3 x 6 = 18, downhill from there. You simply can't get a high enough number with any of them for the rest to work, you will always end up with a smaller number if you continue the problem. But if you use x as a variable, this is possible: (3 + 3)x3 - 3 / 3 = 29, where x = 5 6*5*3 = 90 90 - 3 = 87 87 / 3 = 29
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