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Matt71
 4 years ago
Using the digit 7 five time, place the basic arithmetic signs in between the digits to obtain a total of 540
Matt71
 4 years ago
Using the digit 7 five time, place the basic arithmetic signs in between the digits to obtain a total of 540

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Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0I know, it's been driving me nuts

pratu043
 4 years ago
Best ResponseYou've already chosen the best response.0can you join two 7s together to form 77?

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Nah, you gotta put the signs in between each one

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0You're going crazy, aren't you?

Stom
 4 years ago
Best ResponseYou've already chosen the best response.0this question is right r u sure?

Stom
 4 years ago
Best ResponseYou've already chosen the best response.0because if it is not then bad things will happen

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Yeth, I'm pretty sure it's the question

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0And now everyone's thinking...

nikhil389
 4 years ago
Best ResponseYou've already chosen the best response.0It would be fast to do with computer code, any one good with programming ?

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0certainly question of the week

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8if it doesn't get solved here put it in the metamath section for future reference

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0I've really been trying this for a while and I think it's messed up anyone else think it's impossible?

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0as in, it's total overload here

jimmyrep
 4 years ago
Best ResponseYou've already chosen the best response.2i'm not sure if this fits the rules but 77 x 7 + (7 / 7) gives 540

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8+  / and * only, right? no squaring or anything?

Ishaan94
 4 years ago
Best ResponseYou've already chosen the best response.0concatenation isn't arithmetic right? nice work jimmyrep btw

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0jimmyrep, if no one can find a solution without the placing two 7s together, I'll use your solution

pratu043
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think you can do it without joining two 7s together, jimmy rep has the right answer.

jimmyrep
 4 years ago
Best ResponseYou've already chosen the best response.277 is probably not allowed

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0jimmy rep yehh 77 not allowd

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0It's probably not, but if it isn't I think it's impossible. I've been over this for over two hours, can't do anything to solve it. Yes, two hours, that's how obsessed I am with this question.

jimmyrep
 4 years ago
Best ResponseYou've already chosen the best response.2these types of question are obsessive!

pratu043
 4 years ago
Best ResponseYou've already chosen the best response.0you can't do it because 540 is not divisible by 7.

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0I KNEW IT! All that time wasted, I COULD'VE BEEN PLAYING TETRIS!

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Sweet baby turtles, I can't believe I missed that fact

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8that doesn't make a difference, you can make four 5's into 26 even though 26 is not divisible by 5 5*5+(5/5)

moyo
 4 years ago
Best ResponseYou've already chosen the best response.0in jimmy's example, he used 7/7 to add 1...

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, yeah, then my time HASN't been a complete waste

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8I would think so... it's just multiplication

moyo
 4 years ago
Best ResponseYou've already chosen the best response.0i was thinking maybe there was an order of operations trick to solving it or something... grasping at straws :)

shadowfiend
 4 years ago
Best ResponseYou've already chosen the best response.0You don't need them though, order of operations takes care of the problem above.

jimmyrep
 4 years ago
Best ResponseYou've already chosen the best response.2yes  i didn't need the parenthesis

sandra
 4 years ago
Best ResponseYou've already chosen the best response.0It all depends on if digits can be stacked without operations (is 77 valid?)  in which case 7*77+7/7 works as jimmyrep mentioned

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0question y u no make sense? ლ(ಠ益ಠლ)

shadowfiend
 4 years ago
Best ResponseYou've already chosen the best response.0I feel like concatenation is needed here though. I'm not sure there's another way to grow the number fast enough while reserving 7/7 to make the difference of 1 you need to get to 540 itself.

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0this is certainly a crap question because 540 is not divisible by 7 and i advise ppl to stop wastin time on it

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8I already pointed out that the fact that 540 is not divisible by 7 makes no difference, read above

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8the problem is sitting in metamath now, so it can be viewed in the future by someone who can figure this out, or prove it to be impossible.

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Sooooooo, you can't get a straight answer right?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8not here, not now...

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, well, guys I don't want to completely waste all your time

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0So, if everyone please get together with the plan and just randomly give everyone you can medals!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8Like I say, check the metamath section in the future, someone will probably either find the answer or prove it impossible... eventually.

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, to everyone, for all the help

Stom
 4 years ago
Best ResponseYou've already chosen the best response.0what an awesome way to waste time

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Well, medals or wasted time right? Might as well

Stom
 4 years ago
Best ResponseYou've already chosen the best response.0but to prove it impossible without a computers help is going to be interesting

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8moneybird has a knack for this kind of thing, but he's not around.

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, but it's gonna take a looot of time and luck

Matt71
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, something about koalas that makes him as smart as he is

Stom
 4 years ago
Best ResponseYou've already chosen the best response.0why r ppl still on this question, u u ppl stil solving it?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.8I'm thinking it is impossible: adding 7/7=1 is the only way I can see to get to a number that is not divisible by 7, which only leaves three 7's left over, and the most we can make that into is 7*7*7=343 taking 540(7/7)=539 and 539/7=77 hence it seems jimmyrep's answer is the only one possible, but I don't think this is a true proof that it is impossible.

Janie29
 4 years ago
Best ResponseYou've already chosen the best response.0I tried a bunch of different ways, and I can't find an answer. Now I have a Chem Final in 45min., and this problem is all I'll be thinking of!
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