I made a math puzzle

>>6709835

Can we use each symbol more than once?

>>6709837
Of course.

>>6709839

Cool. For some reason I thought I was still on /tg/.

(0*(1+2+3+4+%+^+8+9))+7

>>6709842 = 7

(7*(9+2))+(0*(1+3+4+5+6+8)) = 77

>>6709842
The numbers must stay in order. You should be able to draw them into the image. Nice approach though.

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got the first:

0 * 1 * 2 * 3 * 4 + 5 - 6 + 7 - 8 + 9 = 7

>>6709845

((((8+2)*(9+1))+6+4+1)*7)+(0*(3+5))=777

>>6709842
cant do that m8

>>6709847

Oh, right. Makes it very different. I'll go back and try again

0*(1*2*3*4*5*6)+7*abs(8-9)=7

>>6709849
Correct. The first one is solved. It's the easiest one though.

My solution was a bit more complex:
0 + 1 + 2 + (3 - sqrt(4) - 5) + 6 + 7 - (8 - sqrt(9)) = 7

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>>6709849
also got the second

0 * 1 * 2 + 3 - 4 + ( 5 + 6 ) * 7 - 8 + 9 = 77

0*(1+2+3+4)+5*abs(6-7)+8*9=77

>>6709860
Second one solved. You're good, man. But 777 and 7777 are a bit harder.

My solution was this:
(0! + (1 + 2)! + 3 - 4 + 5) * (6 + 7 + 8) / sqrt(9)

>>6709845
Faggot u can jumble numbers. They should be in same order as in Op picture.
>Autistic faggot

>>6709868
Calm down kid, we're not on /b/.

-(0!+1+2*3)+4!*5*6+7*8+9=777

>>6709899
Dubs don't lie this time. Third one solved.

My solution was this:
(0 * 1 * 2 + 3) * (sqrt(4) + 5) * (6 * 7 - 8 + sqrt(9))

Now the final boss must be defeated.

Finally got the 777.
Here is the answer. Simpler than I expected:
(0 + 1 + 2)*(3 + 4 + 5*6)*7*(-8 + 9)=3*37*7*1=777

>>6709924
That works. You found the key that makes it easy: prime factors. I used them in the creation process too.

>>6709927
Okay. I'm quite proud of this one because by luck, it looked to me that 7776 was full of prime factors and after trying, it really was. Which made the whole thing easier.
0! + (1 + 2)^(3 + Sqrt[4])*Sqrt[5 + 6 - 7]^(8 - Sqrt[9])=1+(3^5)*(2^5).
GG.

>>6709939
Correct. Last one solved.

My solution is this:
abs(0! * (1 + 2 - 3)! + 4! - 5! - 6) * 7 * (8 + sqrt(9)) = 101 * 7 * 11 = 7777

Share the puzzle with people who can solve them if you want.

>>6709939are you retardet, pow() is not allowed

>>6709953
It is allowed. The order of numbers is kept and you can write it in one line.

>>6709953
>All mathematical symbols can be used but you can not add numbers.
So.... Are you retarded or simply illiterate?

>>6709952
This was obviously me. I don't know why the tripname disappeared.

-(0 - 1 + 2 * 3!) * (sqrt(4) + (5 - 6! - 7 + 8 + 9) ) = 7777

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>>6709965
Not really.

>>6709958
>All mathematical symbols can be used
>It works with the use of only these symbols: + - * / () ! abs() sqrt().

OP wasn't clear on the rules, obviously. Don't be a dick.

>>6709970
I was clear. You CAN use everything but you only NEED these.

>>6709968
you didn't enter it correctly

>>6709972
The calculator reformated the brackets but I entered it the way you posted it. Look at it and you see it's the same.

>>6709970
>are you retardet
Hmm.... He was being a dick first?

>>6709971
if you had said exactly that in OP it would have been clear.
it's not difficult to interpret OP in multiple ways as written. At least 2 people did (myself and >6709953)

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>>6709972
Here. My CASIO calculator gives me the same result.

>>6709976
I'm sorry for the misunderstanding. I thought I was clear enough.

>>6709973
my fault, i copied my formulka wrong from the paper
>-(0 - 1 + 2 * 3!) * (sqrt(4) + (5 - 6! - 7 + 8 + 9) ) = 7777
should be
-(0 - 1 + 2 * 3!) * (sqrt(4) - (5 - 6! - 7 + 8 + 9) ) = 7777

>>6709982
That gives -7777. Remove the minus at the beginning and it's correct.

7777 was the highest number I have a solution for. I couldn't find one for 77777. Its factors are 7*41*271 or 287*271.
I figured out that 0 - 1 + 2 * 3! * 4! = 287 but I couldn't get the rest to 271.

>>6709991
I am the guy from >>6709924 and >>6709939, and I was able to solve it.
My answer is:
0+(1+2)!!*3!!/4!*5-6*7!+8+9

No time to try the next one because it's probably more difficult than this one, but I may try at night.

>>6710008
And by the way, by !! I don't mean the double factorial function http://en.wikipedia.org/wiki/Double_factorial , I mean the factorial function applied twice... I should have wrote (3!)!.
Anyway,
0+((1+2)!)!*(3!)!/4!*5?6*7!+8+9=0+720*720*5/24-6*7!+8+9=108000-30240+17=77760+17=77777.

>>6710015
This works. Amazing, dude. I focussed on the prime factors too much, I guess.

((0102)x3+4+5)x6-7*(8-9)=7
0+1-2+3x4-(5+6)+7x(8+sqrt(9))=77
0+1-2+3x4x5+6!-7+8-sqrt(9)=777
0!+(1+2)x(3!)!/sqrt(4)+5x6)x7x(-(8-9))=7777
thanks was fun

>>6710035
what i realised with the last one was that its easier to find x in x*7*(8-9) giving x=1 or 11 or 111 and so on depending on what solution youre looking for. this limits your number pool to 0,1,2,3,4,5,6 and you immediately gain the factor of 7 you need, so this game simplifies to 0 1 2 3 4 5 6 = 1 and so on