**psst** Are you enjoying Scam School? You've gotta check out our *brand-new series* "The Modern Rogue" at ruclips.net/user/modernrogue. If you've seen it, reply and tell me what you think. If not, then get on over there. I'll bet you a dollar you love it. -Brian
Only thing I can say about this one is that google calculator doesn’t like the sqrt(sqrt part. It gives an error. But then, it also gives an error for sqrt(8+8).
The rules were that you're able to use any (and he does emphasize 'any') mathematical symbol to complete the equation. All he says is that you need to make the statements correct. Besides, if they're using something obscure like factorials to solve it, then no reason to not use the ≠ sign.
I thought they couldn't use square roots as it has a implicit 2... But I could do all the numbers from 1 to 7 without sqrt =) the 4 was the most difficult. (4-(4/4))! = 3! = 6
thought so too. once you allow sqrt and factorial you sure gotta allow operators like floor and ceiling functions. this would enable you to solve this puzzle for any real number. here's a quick "algorithm" for [x∈ℝ]: *add/subtract until you have one number *take the absolute value *sqrt until 2 floor *factorial e.g. lets say x = -20 => ceiling{sqrt{sqrt{|-20-20-20|}}}! = 6 please note: there is a square bracket notation for the gauss' floor and ceiling functions. therefore you do not need to write "floor{x}"; so you are not violating the rules of not adding any letters or numbers. if you can find any flaw in my reasoning i'd be glad if you pointed it out! also: thank you brian brushwood for sharing all those tricks with us! scam school rocks :P
I got all of 1-7 pretty fast, it wasn't until I watched a bit of the video and realized that you consider sqrt to be fair game because many math nerds would consider that introducing the number .5 (by raising to that power), this game gets completely solvable for all numbers if you allow for picking any base you want or if you have 4 of the same number equally 6. You also can get 0's into ones by using log or natural log.
Yeah I was bothered by that as well. 8 can be done with semifactorials, but that is a very obscure symbol, one that kind of feels like cheating, since it's so obscure virtually no calculator that I'm aware of actually supports it, aside from specialized math programs for a computer. I'm not sure if 9 can be done at all without square root though. Edit: well it can be done, but it is *_ADVANCED_* and while it uses a common mathematical operation. In fact this method works for all numbers aside from zero.
I solved this before the hints. It was fun and challenging, but I have a hard time believing nobody has solved this in front of him before. I guess if I had a couple beers before trying this, my attention span would’ve given out too quickly lol
Couldn't you just take a derivative of all the numbers since the derivative of a constant is 0 and then go from there? So something like {(d/dx5)!+(d/dx5)!+(d/dx5)!}! = 6, would work for each row of numbers, just replace 5 with that rows number
LOL! I noticed that *all* the problems can all be solved the following way: ((c')!+(c')!+(c')!)! = 6 This will be true for *any* constants. They don't even need to be identical to one another hehe. :)
My solution was to draw a slash through the equals sign to change it into not-equals, and no matter what other symbols I drew, NOT-equal would make the statement true.
I square rooted a square root, I don't see how thats illegal, for the 8's anyway. Got all of them except the freakin 1's and 0's lol (never learned about factorials :( ) 8 - (sqrt(sqrt(8+8)))
I'm not a mathematician so I have no idea, but is a double square root the same as taking the square root of a square root? Seems like two different processes instead of one, and a square root isn't a number its a symbol? I googled it and even google calculator gives the square root of square root as simply that.
isnt using square root forbidden, I mean square root is that simbol with the small 2 infront of it ( which we dont write but we know its there). English is not my first language and Idk alot of math terms but I'll do my best. when using the same simbol and putting a small 3 infront we get a (third root?) out of it, so when they used a square root they basically added a new number which was not allowed. If you're not allowed to raise to the power of 2 then why should you be able to use its inverse function which is sqare root? I hope I explained my question good.
implied numbers are ok.you just can't write in numbers. for example to do a cube root (would have been useful for 8) you would have to actually write in the 3 over the radical symbol
evilsamar except that you are wrong. because there is always an implied exponent on any number. it is implied that any number without a square root sign has an exponent of 1. you don't have to write the 1 in because it is implied. therefore, you can actually solve the puzzle. By your reasoning, using factorial would not be allowed either because it implied the multiplication by other numbers. you can raise the number to an exponent of 1/2 because there is an operator that implies it without writing the numbers. if there were a cube root operator, it would also be legal. the trick to the puzzle for over thinkers like yourself, is recognizing that, Hey, I can cheat by writing a 1/2 exponent without actually writing the 1/2 because there is a universally accepted operator that does it for me. to take your reasoning further, multiplication should not be allowed because it truly only means that you add numbers together. 3x3 is really 3+3+3. Whoops, you can't write in numbers! Guess you can't multiply. believe what you'd like, but the reality is, to perform a square root, you do not have to write the 2 above the symbol which means that using square roots is totally compliant with the stated rules of the puzzle. It is a legal move.
meh! So what if there is an implied 1 as an exponent, it is alrady set in the puzzle. The puzzle would be completely the same if there was an exponent of 1 written on any number. As for multiplication you dont really pre imply any numbers with it. Multiplication has to be between two numbers to be solved while the square root doesn't because the 2 is implied. I'm not saying the puzle is wrong just that the rules aren't clear enough
Which logically leads me to an elegant solution for ANY number you choose! (Click "read more" for solution) 12 12 12 =6 13 13 13 =6 ...etc.. X X X = 6 |{X,X,X}|! =3! =6 *Note the cardinality of a set tells you how many members are in the set, and is denoted by the |...| symbols. Eg. the result of the cardinality of set {11,11} will give 2.
the cardinality of set {11,11} is not 2. It is 1 since every element of a set have to be unique. However making it (|{X}|+|{X}|+|{X}|)! will make is work i guess.
Rather than cardinality you could view it as a vector (x, x, x) and taking the modulus. Might just be exactly what you meant though but in these words I actually understand it myself :P Genius solution however!
But there is not 2 elements in the set {11,11}, there is only 1 element which is 11 Take a look at this operation; {11,11} - {11}=∅ (the empty set) by definition. Therefore {11,11}⊂{11} * Then {11} - {11,11} = ∅ by definition Therefore {11} ⊂ {11,11} ** And * and ** tell us {11} = {11,11} So if |{11,11}| =2 and |{11}|=1 and since 2=! 1 gives us a contradiction. Your thinking is actually really well imo
Every single one of them can be solved the same way. The derivative of any constant is 0. (If you can't remember your calculus, every number not multiplied by a variable is a constant) 0! is 1. 1+1+1=3 and 3! is 6. All of them are solved the same way, which makes this puzzle exciting to solve, but boring once its solved. Here, I solved it: oi60.tinypic.com/2i90cva.jpg
+Daniel Bundrick Problem is writing the letters is a grey area considering the rules established. Given the discussion about integrals, they may not be allowed.
+evknucklehead I totally agree with you, but also kinda disagree. I agree that it's kinda arbitrary that cubed roots are not allowed, but square roots are, and only because the 2 is implied. However, cubed root means raising the number to the 1/3 power. The puzzle would be very easy indeed if you could just raise the number to whatever is needed to make the number equal to 6. So I guess it's good that those numbers aren't allowed to be used. Also, he specifically said that integrals are allowed, so my d/dx gets a pass either way. :)
+Daniel Bundrick I thought Brian said integrals weren't allowed because you'd have to express the limits using numbers that you otherwise have no access to. Though, d/dx is pretty clever. :D
It is a valid mathematical symbol. Accoring to all the ruled you've defined in this chanlenge it seems to me as an acepateble and non-cheating way to complete the puzzle.
Im at 8:32 but could you just turn the equals sign into a non-equivalent symbol then do whatever you wanted to the number becuase no matter the combinations (unless its correct itd be wrong) itll be right
+Avinavi he did say in the beginning "you cant do cube roots because that would involve drawing a 3" which implies that square roots are fine because they dont involve drawing a 2
I'm new to this whole factorial stuff so correct me if needed, but isn't 8! = 40320? So another factorial would make it immeasurably big, right? Like 40320 × 40319 × 40318 and so on..?
My math teach in 8th grade, about 18 years ago would make us do something similar to this. Each day we would go into class he would put the date up on the board. For example today is 5/22/2014. We had to make a math equation that was true, and we could use any math symbols, had to use an equal sign wherever we wanted, but the numbers had to stay in that order. I got so good at this due to a fascination I have with numbers, that he would start to limit me to 10 equations because I could easily fill the board and it took him too long to get through them all. For instance: 5 2 2 2 0 1 4 could be (5 x 2) + (2 + 2 + 0) = 14 This was fun to watch them struggle. The answer I came up with was (√((8 / 8) + 8))! I paused the video at 12:01 to write this. I swear I didn't cheat. The other answer, if it is allowed would be to not use a square root, but a cube root. The cube root of 8 is 2. The math then simply becomes: (cube root 8) + (cube root 8) + (cube root 8) = 6 I believe you wouldn't allow that though, because the symbol on a calculator for cube root is 3√, but on calculators that have cube root, the square root symbol is shown as 2√. I'd call that a technicality though, because it's not fair that they didn't decide on a better symbol for it. Great video though guys. I liked this a lot.
Wow just finished watching the whole thing.. for 8 I did this (after already solving with inverse-tetration (which is basically a different way of what I'm about to write but whatever)... 8+8 is 16.. square root of 16 is 4, square root of 4 is 2, 2-8 is negative 6 and absolute value of -6 is 6... |sqrt(sqrt(8+8))-8|
This is cool, my high school math teacher/ chess club advisor taught us this riddle back in 2012. It's interesting to think this was so obscure since it became a almost viral puzzle to show all our friends that year.
Alan Shea no, Square root is the same as taking the power of 1/2. so a sqrt(sqrt(8+8)) =((8+8)^(1/2))^(1/2)...........to simplify the powers you multiply them, 1/2*1/2 is 1/4 so =(16)^1/4 =2 and 8-2=6 you write sqrt of sqrt, but it's really a fourth root.both notations are valid, but one does not require you to write a number.
Eight was easy lsqrt(sqrt(8 + 8)) - 8l. The square root of 8+8 or 16 is 4. the square root of 4 is 2. 2-8 is -6 take the numeric value of -6 and you get 6.
kinda miffed you used Square Roots given that you can't use Squares (except for 2) and Cubes (except for 3) and that Square Roots are just to the Power of 1/2 but this is coming from the same person that did 8-8/8-(i*i*i*i) so
The square root is allowed in this puzzle, since the symbol doesn't involve writing any additional numbers. Any roots above this are out because you'd also have to include another number when writing the symbol.
I didn’t get this the first time I watched it and had to watch for the answers. I gave this riddle to the girl I’m talking to and she solved them all without any help within minutes... I think I’m in love
When you said stopped and do the solutions I got these. (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 This is where I got stumped because I thought √ was ²√ (which would involve another number) so I watched into it until I saw that √ was fine and then I got these {[√(8 + 8)]! / 8}! = 6 √(9) * √(9) - √(9) = 6
I got it. Did all but the last three before letting the video run, paused at the commercial and got 0 and 1. And then paused again when you put 8 on the separate page to get the last solution, though slightly different than you showed in the video.
Random idea: I just put a "/" over the "=" sign. It would be much easier to make all functions TRUE. (I know that is not the point of the puzzle but kinda fun idea)
Very clever. I saw your comment stating that you found an expression for every real number. But I saw in a glimpse that you used ln and sqrt. So I started trying to figure out an expression like yours, and when I found one, I came to check your's and we had the same expression. (I guess it works for all complex numbers, except for 0).
Martijn Courteaux You're right, it won't work on zero, neither on negative numbers. If you use |x| though, which is still correct according to the rules, it will work to all negatives. Zero does not work however, as ln(0) is not defined as e^x never gets to 0 and lim(x->0) e^x tends to negative infinity.. Another solution to zero, as zero factorial is not that common, cos(0) also results into 1, add all and a factorial and you're done as well. Complex numbers will definately not work as you will never get rid of i this way.
Well, lim(x->0) (exp(x)) = exp(0) = 1. An exp(x) goes towards -∞ if for example x = lim(a->∞) (a+i*π). And it works for the whole complex plane except for 0. log/ln is defined for the whole complex plane except for 0. (Screenshot in maple: i.imgur.com/xLuGkAl.png)
Started a stop watch, within a 2 minutes had 2, 3, 4, 5, 6, and 7. Took another minute to realize that 9 was the 3 solution with, took 2 more minutes to figure out 1, realized almost instantly that 0 was the same as 1, took the entire rest of the time to finally get 8. Total elapsed time 11:30, over half of it spent getting 8. 0! =1 or cos(0) = 1 sqr(9) = 3 8 was the tricky 1. 8 - sqr( sqr( 8 + 8) ). This wasn't easy, but it wasn't that hard. ! means factorial as in 3! = 3 x 2 x 1 0! happens to be defined to be 1. I'm using Sqr(x) to mean Square Root of x, (which, if I could put the symbol here, would not have any other number on it) ( 0! + 0! + 0! ) ! = 6 ( 1 + 1 + 1) ! = 6 (2 + 2 + 2) = 6 (3 * 3 - 3) = 6 Sqr(4) + Sqr(4) + Sqr(4) = 6 5 + 5/5 = 6 6 * 6 / 6 = 6 7 - 7/7 = 6 8 - sqr( sqr( 8 + 8) ) = 6 sqr(9 ) * sqr(9) - sqr(9) = 6
Tony Reno I did the same as you for all except 4. 8 took me a while but I came up with the same solution as you. My solution for 4 is ((4!)/(4+4))!=6 4!=24 4+4=8 24/8=3 3!=6 I think I took the harder solution for that one.
Tony, I used the same one for 8 as you. Elchi, I'm confused by your solution for 8. Might be clearer with brackets. Looks to me like you're doing [sqrt(8+8)]!/8=4!/8=3, not 6. You could stick another factorial on at the end though, ie. {[sqrt(8+8)]!/8}!
Took me 5mins, the trick being to use (pi +pi)/pi = 2 if we call that sum S then 0+0+0+S+S+S=6 1xS + 1xS +1xS = 6 8+8-8-S=6 the other numbers as demonstrated are simple (arguably pi is a number but you did say mathematical symbols are allowed) Loved the guys use of factorals !!! Great puzzle, keep them coming
Don't know if this counts as cheating but the way I figured it out in less than 10 seconds is put a slash through is equal sign making it not equal then put an addition sign in between each number only don't do it with the 2s
I think I might have done this back when the video first came out, but ran into the video again and puzzled it out on paper (sqrt stands for square root): (0!+0!+0!)! = 6 (1 + 1 + 1)! = 6 2 + 2 + 2 = 6 3! + 3 - 3 = 6 sqrt(4) + sqrt(4) + sqrt(4) = 6 5 + 5 : 5 = 6 6 + 6 - 6 = 6 7 - 7 : 7 = 6 (sqrt(8 + 8)! : 8)! = 6 sqrt(9)! + 9 - 9 = 6 The 8s were the trickiest.
The dude didn't give them time to think with all the talking.... they breezed through the fist 4 and if given some minutes to think they would've thought of factorial, but the dude kept going "OOOOOOOK This is as far as you're gonna get... blah blah... I'll give you a hint" lol
if pi is allowed as not a number, as it is reference as a symbol, all are easy by using (pi/pi) =1 or (e/e0=1 to add or subtract 1on any question. only problem both pi and e are irrational numbers and would probably count as cheating.
+Eugenio Garza That's exactly what I thought, but then I was afraid I was somehow cheating. I resolved it to √(8+8)! / 8 = 6, which I still think is a bit simpler than what they did.
+Arta Soral the square root of 8+8 which equals to 16 is 4, if you square root that number you get 2, at that point all that's left is simple subtraction
This one has been so much fun on the last family get-together. I owe you some beers guys ! btw.. do you know the song "bamboozled" ? nice slick blues song?
Tudor Barbu The zero before a decimal is traditional, and not really mathematically necessary since you don't usually use it in any calculations. Sticklers for rules and elementary school math teachers would still call bullshit though, lol.
Alex .Taylor No, it is not. I'm not sure how to explain it except try counting in binary on your fingers. You might be confused by this, but look at it this way: 0 0 0 0 0 0 0 0 0 0 Each 0 represents 1 finger, if you were to start counting it would look like this(use your fingers, 0 represents your fingers being closed, 1 represents your finger standing up): 1: 0 0 0 0 0 0 0 0 0 1 2: 0 0 0 0 0 0 0 0 1 0 3: 0 0 0 0 0 0 0 0 1 1 4: 0 0 0 0 0 0 0 1 0 0 And so on, and so on. So, if we have 111 in my system, that would look like this: 0 0 0 0 0 0 0 1 1 1 Now, we know that the first 1 represents a 4 because 4 in binary is written with 100, the second 1 represents a 2, because 2 in binary is written with 10, and the third 1 represents 1, because 1 in binary is just 1. 4+2+1=7 I hope you understand my explanation, I tried my best to use correct grammar, I am Swedish, so English is not my first language :)
+Akshay Aradhya He ended up with 8 - 2, why isn't it valid? 8 - sqrt(sqrt(8+8)) = 8 - sqrt(sqrt(16)) = 8 - sqrt(4) = 8 - 2 = 6. It follows all the rules set at the beginning and DJT even said that there are many solutions.
**psst** Are you enjoying Scam School? You've gotta check out our *brand-new series* "The Modern Rogue" at ruclips.net/user/modernrogue. If you've seen it, reply and tell me what you think. If not, then get on over there. I'll bet you a dollar you love it. -Brian
Scam School looks like I owe you a dollar. It was great 👍🏾
I freaking love The Modern Rogue even more than Scam School, which is saying something.
8 is the easiest. cube root 8 + cube root 8 + cube root 8 = 6
8-√√8+8))
Awesome
the eighth one could also be expressed as:
8 - √(√(8 + 8)) = 6
8 - √(√(16)) = 6
8 - √(4) = 6
8 - 2 = 6
which to me is much simpler
Cube root all 8's and add up those numbers
Roko Palada you can’t use the cube root
You could do:
floor(sqrt(8)) + floor(sqrt(8)) + floor(sqrt(8)) = 6
@@rokopalada571 Lol, I found that 8 immediately with cube root. 1 and 0 are the hardest; you'll have to know factorials to solve them.
@@somebodyelse5784 It starts off saying that you can't use any operation that requires another number -- he specifically states you can't cube root
There is much easier way to solve for 8s, without factorials: (8 - sqrt(sqrt(8 + 8)))
Came to the comments to type this
@@OwenCampbellMoore same
Only thing I can say about this one is that google calculator doesn’t like the sqrt(sqrt part. It gives an error. But then, it also gives an error for sqrt(8+8).
Math is beautiful. So many options
Also wouldn’t that technically be cubed root? Which is against the rules. No writing another number. Like squares or cubes
13:04 That poor dude trying to high five LOL!
I am a bad friend to Roberto.
the guy in the end got left hanging disgustingly
Aadil SAAGamer tl vgvgb
the
what
I saw that too!
I know right....cringe
@@guillermoronga1403 are you stroke
To solve the eights, I was going to to have:
8 + (8 / 8) = 6
...and then turn the paper upside-down.
9 = (8 / 8) + 8
Damn, that's smart as fuck
issue is they have to equal 6 as said in the start however if he didn't then you would be the smartest human on the fucking planet
Yep, doesn't work tho
You can just cube root the 8s to get 2s and then add them to make 6
Add a square root (sqrt ()) and a factorial (!) somwhere. 9 year old
Does '6=6=6=6' work?
Outstanding move
yea
Modern problems require modern solutions.
Wow
This is how I did it:
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
'≠' means does not equal. So, for example, '1 + 1 + 1 ≠ 6' is a correct statement because that equals 3, not 6.
Vault Boy the
game is to make all the numbers equal to six
The rules were that you're able to use any (and he does emphasize 'any') mathematical symbol to complete the equation. All he says is that you need to make the statements correct. Besides, if they're using something obscure like factorials to solve it, then no reason to not use the ≠ sign.
There's always someone who thinks they're clever by saying ≠
The easiest way to handle the problem is add an exclamation point before the equals making it an accepted notation for not equals.
I never would have thought of factorials.
But I did 8 :D
sqr(sqr(8+8))=2
Then you do 8-2 and it's done :D
nice!
What I eventually got was 8-(8cos~1+8cos~1)=6
I did floor(sqrt(8))+floor(sqrt(8))+floor(sqrt(8))
NudelGames I have no idea what you just did :P What is floor?
Well.... that's an option
I watched this 3 years ago just watching it again Lol I love it
i watched this video at least 4 years ago and i just learned what factorial is about 2 weeks ago in school
I just saw this comment 3 years after posting, time flies, you should watch again
@@KhaosDCrab man I watched this in like eighth grade the first time and I’m graduated 😭 time really does fly
@@bryant7369 I feel ya bro, I watched this at 13, now im 20 out of school and in the workforce, well played young me stealing my youth XD
I thought they couldn't use square roots as it has a implicit 2... But I could do all the numbers from 1 to 7 without sqrt =)
the 4 was the most difficult. (4-(4/4))! = 3! = 6
Tadashi Mori impressive!
thought so too. once you allow sqrt and factorial you sure gotta allow operators like floor and ceiling functions. this would enable you to solve this puzzle for any real number.
here's a quick "algorithm" for [x∈ℝ]:
*add/subtract until you have one number
*take the absolute value
*sqrt until 2 floor
*factorial
e.g. lets say x = -20 => ceiling{sqrt{sqrt{|-20-20-20|}}}! = 6
please note: there is a square bracket notation for the gauss' floor and ceiling functions. therefore you do not need to write "floor{x}"; so you are not violating the rules of not adding any letters or numbers.
if you can find any flaw in my reasoning i'd be glad if you pointed it out!
also: thank you brian brushwood for sharing all those tricks with us! scam school rocks :P
WOW TADASHI MORI IS HERE
Vispi Dhala =)
4+4-2=6
I got all of 1-7 pretty fast, it wasn't until I watched a bit of the video and realized that you consider sqrt to be fair game because many math nerds would consider that introducing the number .5 (by raising to that power), this game gets completely solvable for all numbers if you allow for picking any base you want or if you have 4 of the same number equally 6. You also can get 0's into ones by using log or natural log.
Yeah I was bothered by that as well.
8 can be done with semifactorials, but that is a very obscure symbol, one that kind of feels like cheating, since it's so obscure virtually no calculator that I'm aware of actually supports it, aside from specialized math programs for a computer.
I'm not sure if 9 can be done at all without square root though. Edit: well it can be done, but it is *_ADVANCED_* and while it uses a common mathematical operation. In fact this method works for all numbers aside from zero.
There are multiple solutions for all of them, even for 8, no need to use factorials to solve 8: 8-sqrt(sqrt(8+8)=6
Your sentence: "There are *multiple* solutions *for all of them* " is false. BTW, nice your solution about 8s!
Can't you use a not equals sign?
not if you want a beer from me. :-D
But it's clever-er than all of those math geeks.
I'm clever-er than any math geek for one reason: I'm a *_NERD_*
Jangwoo Lee NERDs FOR THE WIN!
WHOOOOOOOOOOOOO!!!
Scam School Who said I wanted alcohol; a cup of juice will do.
I solved this before the hints. It was fun and challenging, but I have a hard time believing nobody has solved this in front of him before. I guess if I had a couple beers before trying this, my attention span would’ve given out too quickly lol
Just a thought for the zeros, but before you mentioned factorials, I was thinking of using cosine, as cos(0)=1, if I remember correctly.
Couldn't you just take a derivative of all the numbers since the derivative of a constant is 0 and then go from there? So something like
{(d/dx5)!+(d/dx5)!+(d/dx5)!}! = 6, would work for each row of numbers, just replace 5 with that rows number
13:03 That awkward high five moment...
LOOK AT 13.05 THE GUY WANTED TO GIVE A HIGH FIVE BUT NO ONE LOOKS AT HIM...LOL
LOL! I noticed that *all* the problems can all be solved the following way:
((c')!+(c')!+(c')!)! = 6
This will be true for *any* constants. They don't even need to be identical to one another hehe. :)
Nice work.
Sorry, what does x' do?
A Casual German how does it work with 2 and higher
Meis smart It finds the derivative of the constant (equals to zero)...
Kurt Mueller The derivative of any non-complex number equals zero. You end up solving it just as you would do zero.
My solution was to draw a slash through the equals sign to change it into not-equals, and no matter what other symbols I drew, NOT-equal would make the statement true.
haha i thought of that too!
I square rooted a square root, I don't see how thats illegal, for the 8's anyway. Got all of them except the freakin 1's and 0's lol (never learned about factorials :( )
8 - (sqrt(sqrt(8+8)))
For the 8s you could also use floor functions.
A double sqrt is a cube root which is illegal though.
I'm not a mathematician so I have no idea, but is a double square root the same as taking the square root of a square root? Seems like two different processes instead of one, and a square root isn't a number its a symbol? I googled it and even google calculator gives the square root of square root as simply that.
Chris Adams it's not a cube root more like (x)^1/4
((sqrt(8 + 8))! / 8)! = 6
isnt using square root forbidden, I mean square root is that simbol with the small 2 infront of it ( which we dont write but we know its there). English is not my first language and Idk alot of math terms but I'll do my best. when using the same simbol and putting a small 3 infront we get a (third root?) out of it, so when they used a square root they basically added a new number which was not allowed. If you're not allowed to raise to the power of 2 then why should you be able to use its inverse function which is sqare root? I hope I explained my question good.
implied numbers are ok.you just can't write in numbers. for example to do a cube root (would have been useful for 8) you would have to actually write in the 3 over the radical symbol
I thought the exact thing, I was under the impression that sqrt was not allowed. Since it is raising the number to a power of 1/2.
Rodolfo Baez It' simple mathematical logic. You and I are right, everyone who disagrees is wrong. *top gun high five*
evilsamar except that you are wrong. because there is always an implied exponent on any number. it is implied that any number without a square root sign has an exponent of 1. you don't have to write the 1 in because it is implied. therefore, you can actually solve the puzzle. By your reasoning, using factorial would not be allowed either because it implied the multiplication by other numbers. you can raise the number to an exponent of 1/2 because there is an operator that implies it without writing the numbers. if there were a cube root operator, it would also be legal.
the trick to the puzzle for over thinkers like yourself, is recognizing that, Hey, I can cheat by writing a 1/2 exponent without actually writing the 1/2 because there is a universally accepted operator that does it for me.
to take your reasoning further, multiplication should not be allowed because it truly only means that you add numbers together. 3x3 is really 3+3+3. Whoops, you can't write in numbers! Guess you can't multiply.
believe what you'd like, but the reality is, to perform a square root, you do not have to write the 2 above the symbol which means that using square roots is totally compliant with the stated rules of the puzzle. It is a legal move.
meh! So what if there is an implied 1 as an exponent, it is alrady set in the puzzle. The puzzle would be completely the same if there was an exponent of 1 written on any number. As for multiplication you dont really pre imply any numbers with it. Multiplication has to be between two numbers to be solved while the square root doesn't because the 2 is implied. I'm not saying the puzle is wrong just that the rules aren't clear enough
Yay, I did it, and 5 of mine were subtly different from yours. I love these types of puzzles.
Which logically leads me to an elegant solution for ANY number you choose! (Click "read more" for solution)
12 12 12 =6 13 13 13 =6 ...etc.. X X X = 6
|{X,X,X}|!
=3!
=6
*Note the cardinality of a set tells you how many members are in the set, and is denoted by the |...| symbols. Eg. the result of the cardinality of set {11,11} will give 2.
hah, right !
the cardinality of set {11,11} is not 2. It is 1 since every element of a set have to be unique. However making it (|{X}|+|{X}|+|{X}|)! will make is work i guess.
Rather than cardinality you could view it as a vector (x, x, x) and taking the modulus. Might just be exactly what you meant though but in these words I actually understand it myself :P Genius solution however!
But there is not 2 elements in the set {11,11}, there is only 1 element which is 11
Take a look at this operation; {11,11} - {11}=∅ (the empty set) by definition.
Therefore {11,11}⊂{11} *
Then {11} - {11,11} = ∅ by definition
Therefore {11} ⊂ {11,11} **
And * and ** tell us {11} = {11,11}
So if |{11,11}| =2 and |{11}|=1 and since 2=! 1 gives us a contradiction.
Your thinking is actually really well imo
you don't need factorial for the 8s: 8 - sqrt(sqrt(8+8)) = 8 - sqrt(sqrt(16)) = 8 - sqrt(4) = 8 - 2 = 6
I went the same way, 0 and 1 took me the longest. 8 came straight to me.
Every single one of them can be solved the same way. The derivative of any constant is 0. (If you can't remember your calculus, every number not multiplied by a variable is a constant) 0! is 1. 1+1+1=3 and 3! is 6. All of them are solved the same way, which makes this puzzle exciting to solve, but boring once its solved.
Here, I solved it: oi60.tinypic.com/2i90cva.jpg
+Daniel Bundrick Problem is writing the letters is a grey area considering the rules established. Given the discussion about integrals, they may not be allowed.
+evknucklehead d/dx just means "derivative of". It isn't actually a number or variable. A lot of mathematical functions use letters as symbols.
Same could be said of the 3 on cube roots, but those were explicitly forbidden. But, whatever.
+evknucklehead I totally agree with you, but also kinda disagree. I agree that it's kinda arbitrary that cubed roots are not allowed, but square roots are, and only because the 2 is implied. However, cubed root means raising the number to the 1/3 power. The puzzle would be very easy indeed if you could just raise the number to whatever is needed to make the number equal to 6. So I guess it's good that those numbers aren't allowed to be used.
Also, he specifically said that integrals are allowed, so my d/dx gets a pass either way. :)
+Daniel Bundrick I thought Brian said integrals weren't allowed because you'd have to express the limits using numbers that you otherwise have no access to. Though, d/dx is pretty clever. :D
This is how I solved the 8:
log( 8% / 8 ) + 8
8% = 0.08
0.08 / 8 = 0.01
log( 0.01 ) = -2
-2 + 8 = 6
Wow, people have found so many ways!
i did 8 - sqrt (sqrt (8+8))
= 8 - sqrt (sqrt (16))
= 8 - sqrt (4)
= 8 - 2
= 6
still valid right?
+Dominic Jonnathan Yu
I got to 8 and 2 the same way but didn't see the 8-2 and instead did
(sqrt(8+8)!/8)!
(4!/8)!
(24/8)!
(3)!
+Dominic Jonnathan Yu - Yep, that's how I figured it as well. And for 0's I did cos(0) = 1.
Yeah, me too. Just roots. Couldn't figure out how to use factorial with 8's
Nice
My dad is insanely good at stuff like this, he did this faster than I could run an 800m run. Takes me 1:54 to run 800m. My dad did this in 1:52.
Yeah, this was too easy. Even if you solved it in the intentional way it was meant and didn’t just create “not equal to” signs.
Was I the only one to use semifactorials? XD
(8!!)/(8*8)=6
I think it's quite elegant ;)
Yeah, there were way too many options to solve these that it shouldn’t have stumped anyone with knowledge of ! let alone knowledge of !!
For the 8s i came up with 8 - (sqrt(sqrt(8+8)))
I ADded a degree sign to the zero and then made cos0degrees=1, then did the 1+1+1 factorial thing. Are we allowed to do that?
"Not equal" is a math sign too... Just cross off the equal sign and they'll be all correct
Dino Prašo If that's a solution, then I suppose tearing up the puzzle and saying "puzzle? I don't see any puzzle," would be one, too.
Scam School ahh good point!
It is a valid mathematical symbol. Accoring to all the ruled you've defined in this chanlenge it seems to me as an acepateble and non-cheating way to complete the puzzle.
Dino Prašo I agree
Scam School I believe he has just outscamed you
8-√(√(8+8)) =6
8+8=16
Square-root of 16=4
Square-root of 4=2
8-2=6
True.
Exactly as I did. but is it allowed?
Aritra Ghosh Yeah, of course it is!! :D
It's way simpler than the one with the !.
what the balls... this video still plays in the video player from 2013
Im at 8:32 but could you just turn the equals sign into a non-equivalent symbol then do whatever you wanted to the number becuase no matter the combinations (unless its correct itd be wrong) itll be right
When you note down a square root, aren't you technically writing in a 2?
+Avinavi yeah, since you're technically raising it to the (^1/2)
I was kinda confused
+Avinavi no you are just doing a tick symbol
+Avinavi he did say in the beginning "you cant do cube roots because that would involve drawing a 3" which implies that square roots are fine because they dont involve drawing a 2
yes it is just understood like the one in front of every number
+medusa T. Square roots have a 2 in them. In the same way as the sentence "Go over there" has a subject.
Is it posible to use trigonometrical fuctions?
Example: (cos(0)+cos(0)+cos(0))!=6
david guerra not bad!
sure, that's an alternate solution for 0's
Scam School ...I paused at 3:43...What if I just drew a line through the equals signs?
I was thinking that.
This video encouraged be to get an interest in maths which is my best subject in my senior year. Thank you scam school
The most compact approach with 8's using the double factorial:
(8!!)/(8*8) = 6
Gytax0 whoa.
I'm new to this whole factorial stuff so correct me if needed, but isn't 8! = 40320? So another factorial would make it immeasurably big, right? Like 40320 × 40319 × 40318 and so on..?
Ylijohtaja Lintunen
with double factorials, you basically skip every other number. So 8!! is (8*6*4*2) = 384. And 384/(8*8) = 6
Ohh! Okay! That's cool! :)
ok for 8 it s a more simple way sqrt(8+8)-8=-6
My math teach in 8th grade, about 18 years ago would make us do something similar to this. Each day we would go into class he would put the date up on the board. For example today is 5/22/2014. We had to make a math equation that was true, and we could use any math symbols, had to use an equal sign wherever we wanted, but the numbers had to stay in that order. I got so good at this due to a fascination I have with numbers, that he would start to limit me to 10 equations because I could easily fill the board and it took him too long to get through them all. For instance:
5 2 2 2 0 1 4 could be (5 x 2) + (2 + 2 + 0) = 14
This was fun to watch them struggle.
The answer I came up with was (√((8 / 8) + 8))!
I paused the video at 12:01 to write this. I swear I didn't cheat.
The other answer, if it is allowed would be to not use a square root, but a cube root.
The cube root of 8 is 2. The math then simply becomes:
(cube root 8) + (cube root 8) + (cube root 8) = 6
I believe you wouldn't allow that though, because the symbol on a calculator for cube root is 3√, but on calculators that have cube root, the square root symbol is shown as 2√. I'd call that a technicality though, because it's not fair that they didn't decide on a better symbol for it.
Great video though guys. I liked this a lot.
^Spoiler Alert!
I posted my (the correct) answer in this message.
what a cool idea!
just draw a line in to the equal sign and make it not equal I think ?
could work, since it is mathematical XD
boo! Everyone thinks of that.
*****
probaly everyone skipped over that!
*****
BURN THE WITCH!
***** This needs more likes.
Brian: There's no way you're gonna get the 0's. In fact, I'm gonna-
Right Guy: Challenge accepted.
Laughter all around
It's easy use factorial and addition. Basic facts:
0!=1
1!=1
2!=2
3!=6
4!=24
Try 5!.
5!=?
13:03 guy on right TOTALLY IGNORED!
What about 8-sqrt(sqrt(8+8)) ?
Wow just finished watching the whole thing.. for 8 I did this (after already solving with inverse-tetration (which is basically a different way of what I'm about to write but whatever)... 8+8 is 16.. square root of 16 is 4, square root of 4 is 2, 2-8 is negative 6 and absolute value of -6 is 6... |sqrt(sqrt(8+8))-8|
think about how many numbers you would have drawn to do this
tyler edwards |(√√(8+8))-8|
This is cool, my high school math teacher/ chess club advisor taught us this riddle back in 2012. It's interesting to think this was so obscure since it became a almost viral puzzle to show all our friends that year.
yay i found out 8 all by myself.... after all the hints. but still, i is happy
You don't need factorials for 8 8 8 = 6
You can take 8 - square root (square root (8 + 8)) = 6
8 - square root(square root(16))
8 - square root(4)
8 - 2 = 6
wouldn't that be a cubed root requiring you to write a number?
Alan Shea no, Square root is the same as taking the power of 1/2. so a
sqrt(sqrt(8+8))
=((8+8)^(1/2))^(1/2)...........to simplify the powers you multiply them, 1/2*1/2 is 1/4 so
=(16)^1/4
=2
and
8-2=6
you write sqrt of sqrt, but it's really a fourth root.both notations are valid, but one does not require you to write a number.
Alan Shea You can write another square root without a numer like square root(square root (16))
the square root of 16 is 4
psychicalligator and if you take the square root of the square root, you'll get 2 and 8-2 and that's = 6
8 - sqrt( sqrt(8+8) ) = 6
This took me 4 days to solve. If it wasn't for your challenge, I would have just waited for the answers. Well done.
nice! congrats.
i got the other 7 on my own, just couldn't find anything for 000 and 111
ROFL there are 10 not 9. Good job on getting seven of the but I don't know how you did it if you can't count to ten lol.
Eight was easy lsqrt(sqrt(8 + 8)) - 8l. The square root of 8+8 or 16 is 4. the square root of 4 is 2. 2-8 is -6 take the numeric value of -6 and you get 6.
Rafi Lerman Their way does work but the way Marc said it I personally think is easier.
Came up with the same thing as you but hey, they said there were multiple answers.
Or you could just do 8 - sqrt(sqrt(8+8)) = 6.
kinda miffed you used Square Roots given that you can't use Squares (except for 2) and Cubes (except for 3) and that Square Roots are just to the Power of 1/2
but this is coming from the same person that did 8-8/8-(i*i*i*i) so
The square root is allowed in this puzzle, since the symbol doesn't involve writing any additional numbers. Any roots above this are out because you'd also have to include another number when writing the symbol.
Lmao, that answer for 8 made my day
I didn’t get this the first time I watched it and had to watch for the answers. I gave this riddle to the girl I’m talking to and she solved them all without any help within minutes... I think I’m in love
When you said stopped and do the solutions I got these.
(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
This is where I got stumped because I thought √ was ²√ (which would involve another number) so I watched into it until I saw that √ was fine and then I got these
{[√(8 + 8)]! / 8}! = 6
√(9) * √(9) - √(9) = 6
Exact same thing happened to me. I didn't think sqrts would be cool, like would cube roots? Would log, natural log or even log base 2?
c - c - c ≠ 6 for all the answers gets you free beer for life using the rules provided
I thought the same thing wondered if that was technically legal
i got a much easier answer for the 8s
8-sqrt(sqrt(8+8))=6
***** well done!
I did like this too!
That was my solution as well.
Me too
Same solution for me as well. 8 came naturally when I started thinking about square roots. 0 and 1, I couldn't figure out on my own however.
I got it. Did all but the last three before letting the video run, paused at the commercial and got 0 and 1. And then paused again when you put 8 on the separate page to get the last solution, though slightly different than you showed in the video.
Easy...
Put a line through the "equal sign" to make it ≠.
You can do it in all the rows, and do any sort of math you wish.
I think 8-(sqrt(sqrt(8+8)))=6 is right too. The calculator said so. Haha. it's a fun challenge. 😆
so is √8! + √8! - √8! hardest one has 3 different answers xD
I see what you were going for haha, but the equation has to equal 6, not be greater or less than. Good one though lol.
Where are you getting this idea that "√8! + √8! - √8!" = 6...??
Please. Enlighten us.
nevermind, I'm not a smart person, I thought 8!=1+2+3 . . . +7+8, which equals 36 and well you get the idea, too bad there isn't a symbol for that
Turn those pluses 45 degrees and you've got it. :P
It is so simple for 8 too:
8 - (sqrt (sqrt( 8+ 8 )) = 6
Jagadish Uppala INDEED
Jagadish Uppala You arent allowed to double square root.. Thats cube root which was explained to not be allowed.
Rupenbritz double square root is not cube root. It's fourth root. Someone needs some math classes.
Rupenbritz double square root is not cube root.
Jagadish Uppala I figured the same solution before the video ended.
Now if only my school ever taught that. I didnt even know what factorial was till this show
I'm a bit of a math geek so I already knew what a factorial was. Wasn't too difficult from there.
Random idea: I just put a "/" over the "=" sign. It would be much easier to make all functions TRUE. (I know that is not the point of the puzzle but kinda fun idea)
ruclips.net/video/V8XTpCwicwE/видео.html
Clearly someone that hasn't watch any other Scam School episode involving equations
Tri Mai i did too
I thought that too. Only rule was do not draw numbers so drawing a / isn't against the rules.
had the same thought :P
Solution to all real numbers:
(ln(sqrt(x))+ln(x))/ln(sqrt(sqrt(x)))=6
Very clever. I saw your comment stating that you found an expression for every real number. But I saw in a glimpse that you used ln and sqrt. So I started trying to figure out an expression like yours, and when I found one, I came to check your's and we had the same expression. (I guess it works for all complex numbers, except for 0).
Martijn Courteaux
You're right, it won't work on zero, neither on negative numbers. If you use |x| though, which is still correct according to the rules, it will work to all negatives. Zero does not work however, as ln(0) is not defined as e^x never gets to 0 and lim(x->0) e^x tends to negative infinity..
Another solution to zero, as zero factorial is not that common, cos(0) also results into 1, add all and a factorial and you're done as well.
Complex numbers will definately not work as you will never get rid of i this way.
Well, lim(x->0) (exp(x)) = exp(0) = 1. An exp(x) goes towards -∞ if for example x = lim(a->∞) (a+i*π). And it works for the whole complex plane except for 0. log/ln is defined for the whole complex plane except for 0.
(Screenshot in maple: i.imgur.com/xLuGkAl.png)
Ah, that's what I meant with the exp.
I found a way to make it work for whole C. Just wrap every x between exp(exp(abs())).
No hints, hour and a half to solve them all... Can't wait to show this to all of my engineering classmates, they're gonna love it!
well done!
Started a stop watch, within a 2 minutes had 2, 3, 4, 5, 6, and 7. Took another minute to realize that 9 was the 3 solution with, took 2 more minutes to figure out 1, realized almost instantly that 0 was the same as 1, took the entire rest of the time to finally get 8. Total elapsed time 11:30, over half of it spent getting 8.
0! =1
or cos(0) = 1
sqr(9) = 3
8 was the tricky 1. 8 - sqr( sqr( 8 + 8) ).
This wasn't easy, but it wasn't that hard.
! means factorial as in 3! = 3 x 2 x 1
0! happens to be defined to be 1.
I'm using Sqr(x) to mean Square Root of x, (which, if I could put the symbol here, would not have any other number on it)
( 0! + 0! + 0! ) ! = 6
( 1 + 1 + 1) ! = 6
(2 + 2 + 2) = 6
(3 * 3 - 3) = 6
Sqr(4) + Sqr(4) + Sqr(4) = 6
5 + 5/5 = 6
6 * 6 / 6 = 6
7 - 7/7 = 6
8 - sqr( sqr( 8 + 8) ) = 6
sqr(9 ) * sqr(9) - sqr(9) = 6
Tony Reno For 8 there are a few solutions...Mine would've been sqrt(8+8)!/8=6
and for 9: (9+9)/sqrt(9)=6
for 4: (4/4+sqrt(4))!=6
Tony Reno I did the same as you for all except 4. 8 took me a while but I came up with the same solution as you. My solution for 4 is
((4!)/(4+4))!=6 4!=24 4+4=8 24/8=3 3!=6
I think I took the harder solution for that one.
I like that solution. No Square root needed.
Your solution is exactly the same as mine, except I overthought it and got Gamma(4) + 4 - 4 = 6 haha (since Gamma(n) = (n-1)!, where n is an integer).
Tony, I used the same one for 8 as you.
Elchi, I'm confused by your solution for 8. Might be clearer with brackets. Looks to me like you're doing [sqrt(8+8)]!/8=4!/8=3, not 6. You could stick another factorial on at the end though, ie. {[sqrt(8+8)]!/8}!
Guessing before hand, you just change draw a slash though the equals sign
Thats what I thought too :P Its really cool that there is a 'real' anwser to this puzzle (:
*IM BORED*
Took me 5mins, the trick being to use (pi +pi)/pi = 2
if we call that sum S then
0+0+0+S+S+S=6
1xS + 1xS +1xS = 6
8+8-8-S=6
the other numbers as demonstrated are simple
(arguably pi is a number but you did say mathematical symbols are allowed)
Loved the guys use of factorals !!!
Great puzzle, keep them coming
Don't know if this counts as cheating but the way I figured it out in less than 10 seconds is put a slash through is equal sign making it not equal then put an addition sign in between each number only don't do it with the 2s
lol what If a teacher gave you this for a test :P
that would NOT be awesome but I will give this to my math teacher XD and then lets see if she's smart enough to solve this puzzle XD
what tsskyx? I posted that comment...
I like the new hair :p
I think I might have done this back when the video first came out, but ran into the video again and puzzled it out on paper (sqrt stands for square root):
(0!+0!+0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3! + 3 - 3 = 6
sqrt(4) + sqrt(4) + sqrt(4) = 6
5 + 5 : 5 = 6
6 + 6 - 6 = 6
7 - 7 : 7 = 6
(sqrt(8 + 8)! : 8)! = 6
sqrt(9)! + 9 - 9 = 6
The 8s were the trickiest.
8 is simply 8 - sgn(8) - sgn(8)
sgn just returns 1 for positive numbers and -1 gor negative numbers. 🤡
Put + everywhere and then Draw a line trought equals.. Ex. 1+1+1 = 6 if you draw a line trought = it means it doesnt equals to and u solve it :)
so right :D
Csanád Kovácsházi hmm. clever. OR
6 + 6 - 6
5 (divided by) 5 + 5 or your way. Your way is better though.
who says u need addition? 000≠6 111≠6 222≠6 333≠6 444≠6 555≠6 666≠6 777≠6 888≠6 and 999≠6
dolfin rule your not supposed to do that
youre not supposed to do that
can you just do
1-1-1(not equal to sign)6 and do that for all
+War Hawks that wouldn't be solving the puzzle.
8+8/8=9 9 root=3 3! 'Drumroll please' =6
+Howard Ong absolute val 8- the sq root of the sq root of 8+8
+Howard Ong you can't use any other numbs than 8 for 8 8 8
I just did 8- sqrt[sqrt(8+8)]
8 - sqrt(4)
8-2 = 6
I did it! :) For 8 I did.. 8 - root(root(8 + 8))
Or you could use decimal points and the "round up/down to the nearest integer" symbols for trivial solutions
I found that on my own too. But I wouldn't have the patience to finish the whole thing IRL.
I just did (root 8+8:8)!
Works as well ;) bec 3! is 6
***** same i thought i was the only one!
I did the exact same way
I learnt this in an extension course in year 6 (Cube roots make 8 easy) the rest is just simple addition, division, square-roots and factorials.
The dude didn't give them time to think with all the talking.... they breezed through the fist 4 and if given some minutes to think they would've thought of factorial, but the dude kept going "OOOOOOOK This is as far as you're gonna get... blah blah... I'll give you a hint" lol
if pi is allowed as not a number, as it is reference as a symbol, all are easy by using (pi/pi) =1 or (e/e0=1 to add or subtract 1on any question.
only problem both pi and e are irrational numbers and would probably count as cheating.
another way to do the 8's is
8-(√(√8+8)
GUYS I GOT IT *!!!* He said you could use *ANY* operation you want so just add random opertions and change the equals sign to a "does not equal sign"
Hardest my ass. All you need to "know" is that you can use factorial and square root.
ceputza You're super smart.
I wasn't being sarcastic or a smartass. This was really one of the easiest ones because it had a no bullshit, straight math answer.
ceputza yup, got it in like 5 mins as well
but NOT addition, subtraction, multiplication(which you need for factorials), or division(which you need for square roots) ok... that makes sense
So, 8 is not as hard as it seems... 8-√(√(8+8))=6
Saw it too
+Eugenio Garza That's exactly what I thought, but then I was afraid I was somehow cheating. I resolved it to √(8+8)! / 8 = 6, which I still think is a bit simpler than what they did.
Brittany Adams NICE!!! I like that answer, extremely creative!
Whoops! My solution was actually wrong and would equal three. I don't know how I missed that, :P. Maybe someone else can amend it.
+Brittany Adams just add a factorial... I thought I saw it
I got the 8 without even knowing of factorial
+Joshua B well done!
+Joshua B how? like what was your solution?
+Arta Soral the square root of 8+8 which equals to 16 is 4, if you square root that number you get 2, at that point all that's left is simple subtraction
Joshua B o right ty, brain fart :)
no problem
This one has been so much fun on the last family get-together. I owe you some beers guys ! btw.. do you know the song "bamboozled" ? nice slick blues song?
That moment when you realise you did the 8 WAY too complicated 😅
{[Sqrt(8+8)]!/8}!
That's how I did it.
for 5 i thought of
5 + .5 + .5 = 6
lol
I'm not sure if that's correct.I think you have to use 0.5
Tudor Barbu The zero before a decimal is traditional, and not really mathematically necessary since you don't usually use it in any calculations. Sticklers for rules and elementary school math teachers would still call bullshit though, lol.
5/5+5
I did 5.5+.5=6
Ha I didn't know 0! was 1... I used (cos(0)+cos(0)+cos(0))! instead
Sorry, I think you're not allowed to write letters.
he said numbers not letters,
would pi be considered a number even though its a symbol? Because I did -(cos(8pi)) - (cos(8pi)) + 8 = 6
For the 8's could you do the cube roots of each 8 and then add them together go get 6 or is that not allowed?
I was thinking in binary 1 1 1 is just binary for 6
Alex .Taylor not bad!
Alex .Taylor i was thinking the same xD
Alex .Taylor 111 is 7 in binary..
Alex .Taylor No, it is not. I'm not sure how to explain it except try counting in
binary on your fingers. You might be confused by this, but look at it
this way:
0 0 0 0 0 0 0 0 0 0
Each 0 represents 1 finger, if you were to start counting it would look like this(use your fingers, 0 represents your fingers being closed, 1 represents your finger standing up):
1:
0 0 0 0 0 0 0 0 0 1
2:
0 0 0 0 0 0 0 0 1 0
3:
0 0 0 0 0 0 0 0 1 1
4:
0 0 0 0 0 0 0 1 0 0
And so on, and so on.
So, if we have 111 in my system, that would look like this:
0 0 0 0 0 0 0 1 1 1
Now, we know that the first 1 represents a 4 because 4 in binary is written with 100, the second 1 represents a 2, because 2 in binary is written with 10, and the third 1 represents 1, because 1 in binary is just 1.
4+2+1=7
I hope you understand my explanation, I tried my best to use correct grammar, I am Swedish, so English is not my first language :)
Unknown 111 = 7 110 = 6
another solution to 888: -( sqrt(sqrt(8+8)))+8
try the math. upvote so he'll see it
does 0 + 0 + 0 + (Pie + Pie + Pie + Pie + Pie +Pie) / Pie
count?
because if it works, then it's pretty easy
btw Copywright me
if that works then
8 - (8 / 8) - (Pie / Pie) = 6
I guess not but I'm still gonna try the 8
8 - V( V (8 + 8) )
Done haha
You can not add any numbers
*Pi btw lol
I used cos(0) to get 1 from zero. No funky definition rule needed.
Well, i used e^0
@@5tyxh You could use anything ^0, but you are not allowed to use any other digits, so you cannot use the 0, so that doesn't work.
@@paulwagorn6200 I dont use another digit. I just put an "e" in front of the zero. So my answer is : (e^0+e^0+e^0)! = 6
I got them all. But for 8 I got :
8 - (_/(_/(8 + 8))
The puzzle is German, so am I... Does that mean I cheated?
+Prerendered renderation I did that too
+Prerendered renderation Thats not even Valid
+Akshay Aradhya He ended up with 8 - 2, why isn't it valid?
8 - sqrt(sqrt(8+8)) = 8 - sqrt(sqrt(16)) = 8 - sqrt(4) = 8 - 2 = 6.
It follows all the rules set at the beginning and DJT even said that there are many solutions.
Kazutoification Oh okay thats sqrt I thought that slash was division
This is like math for a 5th grade azn kid. Not even challenging a bit. Exaggerate much?
Ellery Pham LESS BRAGGING, MORE ANSWERS!
Not really
Solved it. Different answers than they got, but equally as valid.