How To Solve The 6s Challenge
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- Опубліковано 2 сер 2024
- Thanks to Tyler Cenko, and Caio Cerqueira from Brazil, for suggesting this wonderful number puzzle! I had a lot of fun solving this. Can you make 6 from 3 copies of the same number, where the number ranges from 0 to 10? You can use common mathematical operations, but you cannot introduce any new digits (so the cube root is not allowed), and you must have an equality (this is not a trick question with the "not equal" sign). This is a great exercise for building mathematical number sense! See the video for the many solutions.
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Nearly 2 million views in one year! Thank you!
Hi, Love your videos!
Does 8-(8/8)!= 6 work? I just learned what factorial means from the video so im not sure
you deserve it
I was thinking factorial can't be applied cause factorial uses other no... Like 3! = 3*2*1
.
Why??
You cant use sqrt() either. That's the same as 2root.
i smart 2+2+2 = 6
said it before he started working them out
dang dude, b-b-but how? is it even possible?
J4M13 b -but, how? This g - guy is hackin!
did stephen hawking wake up
@@morveoteki2719 2×2+2
If cube root is not allowed, square root should not be either, as it introduces a 2.
No it doesn't.
@@Stubbari yes it does
@@Stubbari Just because the 2 isn't written out, doesn't mean it isn't there. What do you think "square" means in "square root".
@@DarthBil1 So if there is no written digit then there's no digit. What's so confusing about that?
Inverse isn't the right word, but it's late and I can't remember the right word for what I'm thinking. I'll get back to it later.
I was messed up when you said that we weren’t allowed to use cube roots, because in my head that meant I wasn’t allowed to use square roots either. As soon as I realized I was allowed to use square roots I figured out the ones I was missing pretty quickly.
We can use cube and square roots but of the same number for you are solving.
Then how did he use a square root for 3s?
@@king_of_the_sun4897 he may use a square root because the operation is square by default, but no other root because that would require introducing a new number
@@redaipo the operation is the second root of 3*3
Could do the cubed root of 3 cubed. That uses only threes and all three of them to get 3.. never mind I'll be going...
Wait, I meant the 6th root of 6 to the 6th power! Yea! that's what I meant...
Very interesting. Initially I was able to figure out the answer only to the easier ones. But when you explain it, I can calculate it , as you do and understand it, because of the good way you explain it. Thanks!
I got the 2, 2, 2 thing in under a second maybe I am a prodigy
Lol!
Dude your going to get the nobel prize in mathematics
@@reidarkollstrm5218 let me finish college without failing math first
🤣🤣🤣
i also did the 6 6 6 in 5 seconds
I found a pretty solution for 10 10 10.
You can calculate 10 × 10 + 10 = 110, which is 6 in binary representation!
that is a nice answer but i don’t think that actually counts as a proper solution
well-
10 + 10 + 10
2 + 2 + 2
another binary one
There are 10 types of people in the world: those that understand binary and those that don't.
I don't think mixing bases would count.
at first it was like impossible, but after you start explaining it, it was like new ideas ware automatically coming into my mind, and I was getting the trick.
Awesome video!!!
In the first case I've actually used Cos(0) instead of 0! My other solutions were similar to yours. That was some really pleasant math here ;) Thanks for the video!
Cos(0) isn't a common mathematical operation
@@hugh.g.rection5906 tbh i've used the cosine function much more that factorials
@@hugh.g.rection5906 i think it is because he didn't restrictions in detail on this
@@hugh.g.rection5906prove it
@@hugh.g.rection5906 Ask Mathologer, I guess it is super-common to him?
Alternate title: how make three numbers equal 3 and then apply a factorial operation
yes i solved almost all that didnt include ! except 8 that was hard one tho, don't mind coz im newbie in math...
@@Nightmare_Developer How did u solve 1 withouts factorial. Would u mind sharing
@@SorakuteeYT umm no i said i solved all that didn't include !(factorial) except 8 coz i dont even know what ! is
@@Nightmare_Developer i mean the one with 1.
1 1 1=6
I cant solve it without factorial
@@SorakuteeYT he's saying he did all of the equations that DON'T require factorial. 1 1 1 requires a factorial, meaning he didn't do 1.
If you can't use "³√", using "√" should also be against the rules because it is just the short form of "²√".
And 4 is a short form of 1+1+1+1. With your locig the whole problem is impossible.
@@Stubbari you're not making any sense.
@@imonsanyal Radical symbol doesn't have a digit "2" in it.
N:th root is written with a corresponding digit.
This js enough "√" you don't need to add a digit "2" to make it square root.
With your logic 2+2+2=6 is just a shorter version of 1+1+1+1+1+1=6 which includes 6 new digits.
@@Stubbari I know... but it is the short form of *²√* and the rule clearly states that you can't introduce any new digits. *√* and *²√* are the same thing and hence cannot be used.
The thing is sqrt can be used without using any digits, so even though they have the same method like finding the nth root, you did not necessarily used a new number on the square root ,got me?
This challenge is a classic one in Brazil we learn as younglings. Nice to see it here!
Wow! This is your best yet. I love your videos.
I figured out
6 6 6 = 6
We cannot use not equal symbol
But can use = since not any rule for this
6 = 6= 6= 6
😂😂😂🤭🤭🤭🤣🤣🤣
Cheesy
Cheeky
Cheesy
Cheeky
What does it mean
Cheeky ckeesy whatever
Damn the maths developers really need to nerf this new "factorial" boost
They said they will in update 3.14
@@gjproductions9337 finally
@@gjproductions9337 Mmmm! Pumpkin 3.1416.
noooo! now I can magically make almost any equation just work
I actually used multifactorials for 2 of the solutions. Surprising that that was the first thing that came to mind for me. And yet I still used cosine for the 0s.
I followed the same path and worked out the first six pretty easily, but it never occurred to me to use the factorial before giving up and watching the solution. I think I could have got there eventually, but I’m not sure I ever would have thought how to work out the 888 one. That’s some Inception level stuff! Kudos to those who worked them all out
Edit: first seven, not six… got stumped by the use of factorials. Should have questioned why the answer was always 6!
so what was your solution to 000 and 111?
No I mean the first 7 he solved, because I followed the same path, not the first 7 sets in numerical order. So like, 222, 333, 444, 555, 666, 777 and 999 I could work out, but I got stumped by 000, 111, 888 and 101010, because I never thought to use factorial
Got it
The answer wasn't always 720
i got all of them except 10 10 10
Hi Presh.
Even though this video is five years old, it is my first view.
Again, you've created high-quality, educational content.
I'm a senior citizen, and I STILL enjoy your videos.
Thank you.
this is a high quality good viewer that we need to protect at all costs
(0!+0!+0!)!=6
For all the rest:
(sgn(x)+sgn(x)+sgn(x))!=6
It is not true
@@asgarrahmani939 it is true + we can say for every one of them including zero:
(sgn(x)!+sgn(x)!+sgn(x)!)!=6
If your unfamiliar with the sign function you should jnow it returns one if x is positive and minus one if x is negative (and zero if zero)
@@dorondaniel318 interesante!
Kumar Saurav or for 6 6 +6 =12 12-6=6
Kumar S / Because Presh doesn't ask for only natural numbers and you use a function that you have to explain ( you make a new "logic rule" ) then we can make another explanation ( rule, requirement, etc ) :
let's define de novo the following :
0=1, 1=2, 2=3....., (n=n+1)
where n={ 0,1,2,3,4...n+1 }
and then your solution is a general solution.
the only part i got confused was at the dont's rule.
it says it doesnt allow the introduction specific to ³√ but it still using the √ which in my understanding is the same as 2√
@Ro Bert Yeah because CR introduces a new digit "3" while SR doesn't.
@@Stubbari You don't have to write it but in SR is number 2 so it's a problem in rules.
@@videopoetic7101 If you don't write it then it doesn't get introduced. Simple as that.
Or do you know what digits I introduce here: " "?
It said digit, not number. So it is allowed since he wrote the square root. If that rule applied to everything, then multiplication wouldn’t be allowed either because that technically counts as this example: 2 x 2 x 2 = 8 (2 + 2 + 2 + 2 = 8) And that does add another number, but the rules said digits so it is allowed.
point is, it’s literally the same, the 2’s implied thus you dont have to actually write a number
I fact-checked this and this is absolutely correct. Good job! 👍
What a fantastic way of teaching operations. It is inspired.
Fun fact:
You can do this for arbitrary n, using ONLY addition, division, and trig identities.
That's right, without using square root OR factorial.
All you need is patience. And possibly a mental disorder. Thankfully, I have both!
One of the trig identities you all likely learned in High School, nestled in with the arcus functions (acos, asin, and atan) is:
cos(atan(x)) = 1/√(1+x²)
Therefore, using sec(x) = 1/cos(x), we get:
sec(atan(x)) = √(1+x²)
You may see where I'm going with this.
We know, for any n, that
(n+n)/n = 2
Therefore,
sec(atan((n+n)/n)) = √(1+2²) = √(5)
and
sec(atan(sec(atan((n+n)/n))) = √(1+5) = √(6)
We could continue this until we hit √(9) = 3 and use a factorial to get to 6. But if we've gone this far, do we *really* need to use a factorial? After all, √(36) is right there.
Therefore, for arbitrary n, I propose the solution that:
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan( (n+n)/n )))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) = 6
Plug it into Wolfram|Alpha. You'll see it works.
And pedants like me can avoid using √ :)
Bro 🤯👍
You’re the type of guy that does his homework on time
😳😳
can someone tell me if this guy is legit or trolling?
Who tf even are you?
Square root is controversial for all these kind of quest, as it's square 2.
my thoughts exactly - I threw it out the window right off the bat, silly me :D And I am not nearly confident enough to toy around with goniometry to get there that way
Agrees I too thought it was out of bounds and despite having the solutions using it still believe thusly
@@JossWainwright bruh, do you scan this comment section for this one question?
if so- *i shall follow you kid*
The rule should clarify you can't introduce other numbers, but numberless symbols are allowed. Like √ .
@@JossWainwright Good point.
I learned what a factorial is because of this video!I love learning new things!
For the 8's, I did (8!!)/(8 * 8), because 8!! = 8*6*4*2 = 384, and 384/64 = 6. I definitely brute-forced the 10's with some somewhat dubious logic, doing (10!!!!!!! + 10!!!!!!!)/10 to get (30 + 30)/10 = 6 XD
just another one for 8 8 8, for a sample:
(square(8 + 8)) = 4; 4! = 24
24/8 = 6)
seems pretty easy, once come up with, and effective here)
though, the ones in t' video r too, surely)
Cheers
@@yehor_ivanov 24/8 = 3 not 6. You'd have to say:
square(8+8) = 4
-> 4! = 24
-> 24/8 = 3
-> 3! = 6.
@@yehor_ivanov say sqrt(x) not square(x). It almost makes it look like you're saying (x)^2.
Just this,
8+8=16 aight?
16 root=4
4 root=2
8-2=6
You used 4 10’s
I never even thought of factorial as one of the operations I could use, but I'm happy to say as soon as you brought it up, the ones I was struggling with fell into place in my brain!
As soon as I saw this, I thought 0 0 0 would be (0!+0!+0!)! bruh
*888 is hardest*
Me : 6 CIRCLES!!!
😂😂😂
😂
Keerthi Mahesh genius👍👍👍
Ehm.... Yes?
888 was easy
1. third root of 8 is 2
2. add them up
3. ....
4. profit!
Great solution... your way of thinking is just awesome.
this is one of the few puzzles i could actually do and it was actually pretty fun
Theoretically, you could find the derivative of every single number, which gets you to 0 0 0, and then factorial each of them to get 1 1 1, then add them together to get 3 and factorial it to get 6. This could work for all of the problems.
For example,
4 4 4 = 6
(d/dx 4) (d/dx 4) (d/dx 4) = 6
0 0 0 = 6
0! 0! 0! = 6
1 1 1 = 6
(1+1+1)! = 6
3! = 6
Since the derivative of any constant is always 0, we can use this approach on any number to get 6, hence we could have solved all of the problems in the same exact way.
haaaaaaaaa, this is the equivalent to the infinite money glitch hahhahah
Bravo!
невероятно, я восхищен
Big Brain Time
I resorted to something like this to solve 8. I used delta function to convert each number to zero. Then I realized it works for all but 0.
Was able to do all of them except 0, I forgot that 0! = 1. This is a fun puzzle exercise.
Thanks for creating this!
3:29 (9 + 9)/ sqr(9)
I did that too.
Sorry im late to the comment section
You cannot use sqr(9)
@@nicolasrozenberg5209 do you even know what sqr() is?
Of course that equals six, because sqr(9) = 3, and (9 + 9) = 18, and 18 / 3 = 6, and square root doesn’t require a number.
@@faradaykhaleesi877 Sorry, I didn't watch the video. It shouldn't have been used though, because sqr() is not an operator, it is a function that represents the root of index 2 of a certain number. And that involves implicitly using number 2. The task is not explained correctly
The whole is about solving to 3 and taking its factorial.
@@leinfare Wow!
@@leinfare Yup
Some people might say sqrt symbol should not be allowed, as it is really introducing ²√ or ^0.5. And by extension as shown in the "8" example you are allowing all root powers of 2. I wonder if some of these would be solvable in that case?
A nice one, thanks Presh
One more solution with zeros (it's mine before watching this video):
(cos(0) + cos(0) + cos(0))! = 6
@@nikizhu78 it's factoriel
@@nikizhu78 Factorial.
(log(10 * 10 * 10))! = 6
@@nikizhu78 не позорь русских
@@nicholasjarrett4480 not log but lg
lol nice try. square root is technically a short hand for 2√ ... and the suggestion of simple math operation is also quite misleading. Most of people would exclude factorial.
And logarithms, since (log10+log10+log10)!=6
Steven Song take ln0 so
I know for e ;)
Most people wouldn't know about factorials
so what do we call a root symbol alone and how does it work
Yah Root 2 is basically X ^ (1/2)
For 10, i did [Log( 10 ) + Log ( 10 ) + Log ( 10 ) ] ! = 6. Usually log is base 10, so no new digits
Nice, or we can use use ln or even sqrt directly if we introduce ceiling or flooring
@Able89535 yeah but I don't like to use floor or ceiling because they annot be described with regular math.
"!" Is just broken in mathematics 💀 bro almost solo'd the secret quest
He said the word "factorial" and I knew I was in over my head, haha
Don’t be intimidated! Factorial is *extremely* easy to understand. It’s literally just the multiplication of every whole number before x as well as x itself. Ex: 6! = 6 x 5 x 4 x 3 x 2. That’s it lmao, you’ll grasp it quickly.
@@is1hair Yeh, I went and googled it, thanks
@@is1hair thank u bro .. it help me a lot.. and prevent to Google it?!!!
in my opinion i dont think he factorial should have been alowed as it is not deemed as a common mathematical concept. No one is going to be like "oh yeah I have 6 factorial dollars"
@@NERONRR Can you solve them without factorials?
To use 10 10 10 there is another solution
(log(10*10*10))!
Awesome dude!
That's an innovative solution!
@20 Subs Before Tomorrow? I think, he mean lg (log with base 10)
@20 Subs Before Tomorrow? A piece of trunk.
What is log? Baby don't hurt me~
For 10 10 10 = 6, I did: ( log(10) + log(10) + log(10) ) ! = 6
That was amazing
I must admit, I fell in love with your videos and solving problems. It's challenging but fun! Just watching your videos without being able to solve the problem makes me realize different approaches to solving these puzzles. You're a great teacher too, it's fairly easy to understand your resolvement, even in harder puzzles.
Literally the only two I’m smart enough for:
2+2+2=6
And
6+6-6=6
😂🤣😂
Same
wow same lmao
5/5+5=6
kingmaker (5^2 + 5)/5
Me too lmao
I have a neat solution for every positive integer N of the N N N = 6 problem. My solution is (log(sqrt(N), N*sqrt(N)))! = 6, where log is the logarithm, for example log(10, 1000) = 3.
Took me 10 minutes to come up with this, amusing puzzle by the way.
That's brilliant and deserves more attention than it's got so far!
What an awesome and elegant solution!
I love it when someone just kills a puzzle.
I hope this counts!
brilliant: it gives allways 3! (log(sqrt(N), N*sqrt(N)))! = log(N*sgrt(N))/log(sgrt(N)=(log(N)+log(sgrt(N))/(log(sgrt(N)=(log(N)/log(sgrt(N)+1)! =(2*log(N)/log(N)+1)!=3! . You are jenius men!
That was fun!
3 is truly wielding that exclamation point like a mighty blade today
for 999 just flip the nines upside down
the 6 on the right also get up side down
Nein nein nein, that won't work.
Enin Enin Enin
Nothing worked
(9+9)/√ 9
Whoaaaaaaaaa there brother maybe we can relax
Reading that first equation very excitedly
“Zero! Plus zero! Plus zero! (!!!)”
@@user-SG717 what's 230 - 220 x 0.5?
You probably wouldn't believe me, but the answer is 5!
@@user-SG717 just because you're right doesn't mean I'm wrong
@@user-SG717 it's 5!
@@user-SG717 5! = 120
challenge: do the ones with square roots without them, as square roots technically involve other numbers like cube roots do
@@JossWainwright huh? If Cube root isn't allowed then square root shouldn't be. Just because you can get away without writing the " ² " dosent mean it isn't there, you just don't write it because it's implied
@@JossWainwright when we first learned square root we would write the 2. At some point in math they dropped the two. It is there, you just can't see it. It is still introducing a new digit
@kaderen8461 👍 finally, someone's thinking what i'm thinking
@@H0uxdubxston exactly
@@JossWainwright So I can write a square root, but say it’s implied as a cube root and then cube root now counts?
we also had this challenge but with another number in 6th grade, and we weren't allowed to use factorials or square roots since factorials are just like introducing numbers with 3! as 3x2x1 except for 1! which doesn't really do anything. Square roots are like cube roots, but a two. However this challenge is almost unsolvable without square roots and factorials. This challenge/puzzle is so fun to do and I like your explanations :)
not sure I agree with factorial introducing new numbers, I would then argue that '+' introduces new numbers as putting it between two '2' introduces a '4'
@@Archy_The-WizardFactorials literally turn 0 into 1. How is that not introducing a new number?
each time you showed the solution I was like "how didn't I think of that?!!" haha
Same with me😂😂
In the instructions you said you cannot introduce any new digit, but can you explain how the factorial is not an introduction of new digits.. cause 3!= 3×2×1
@@aubreyundi
Sure, both introducing ! and square root are actually introducing new digits. But since there is a code for those which does not imply WRITING the digit, it's considered valid. It annoyed me too at first, but then, i realized there probably was no other way and that the problem needed that "trick".
I fully understand you, the problem lies in the phrase "do not introduce new digit". Many people would immediately understand that "code without explicit digit, even if implicit, are fine". Others, like you and me, would immediately consider "implicit digits aren't allowed either, thus square root and factorial can't be allowed".
I'm actually pretty sure most of the people are in the second situation.
However most of those still are able to switch back to the first, by realizing the problem is impossible without this assumption.
Don't be too extreme on implicit vs. explicit. There is not good solution. Both are always possible, and can always be extreme (go too far).
If you assume implicit should be always the rule, then you can never solve anything. Because if you keep pushing in that direction, you never have a satisfying "proof". Nothing can be proved expect "cogito ergo sum", and stricly applying implicit rules means everything else is irrelevant. They you can just go back to bed and die.
It's true with the other exageration. Explicit is ALWAYS possible, you can invent a new way of writing, a new code, which would make the exercice always super easy. But then everything becomes irrelevant too.
Sure playing around the definition and the limits of a problem is fine... but first you need to accept the limits and definition, solve the problem WITHIN this conditions, and ONLY THEN you can play around with the rules and try to bend them.
The big problem here isn't mathematics. It's understand what (most of) people would agree too.
Not being able to understand what most of the others do or think is actually very frequent. It's also unfuriating, frustrating, because everybody around think you are odd while you did nothing wrong. Your interpretation is just different.
It's fine being different, keep at it.
But keep in mind that human beings progress A LOT by sharing knowledge and understandings. Thus what you need to do is to work on that too (and probably first). Then you can be different AND able to understand others, play with them, and learn from them.
@@aubreyundi Technically multiplication is adding new digits because it's repeated addition
@@aubreyundi a*b=a*a*a... b times
ah yes, maeth
You always have solution in mathematics when u stuck.
"Let's assume 0 0 0 = 6"
Hahah xd
😂😂😂
You are Savage until I am thug bro
this is AMAZING
If the floor function is used, it will be valid for all positive integers. If the absolute value function is also used, it will also be valid for negative integers. Since we are using the floor function, it will be valid for all real numbers. It will even work with complex numbers. Here are some examples:
floor(sqrt(11))*floor(sqrt(11))-floor(sqrt(11))=6
(|-1|+|-1|+|-1|)!=6
floor(pi)*floor(pi)-floor(pi)=6
(|i|+|i|+|i|)!=6
Or you use the Cardinality, and it does not even have to be Numbers at all. Like (|{X}|+|{X}|+|{X}|)!=6 for all complex Numbers or anything else.
Actually that's true, repeatedly take the square root and floor in the end to reduce the problem to 1+1+1 and solved for any number.
6:16 can you use (cos(0) + cos(0) + cos(0))! = 6 ? I´m not adding any digits, but I don´t know if in this case you can do it...
good idea!
cos is not operation it is FUNCTION
@@preyunknown1820 oh sugar lumps yur right. However, it was a nice try.
I think it should count, but, welp, rules are... rules I guess? (I think it should count xd)
@@jachpi1080 yeah nt tho
The 8’s one is actually really easy once you realize that if you can make it equal to nine you’re good:
8+8/8=9. Square root and factorial
So it is {√[8+(8/8)]}!
Technically a square root also requires adding a new number to the equation. Mathematicians just agreed to not put a number “2” near the root symbol to save some time. But it’s not fair to say “you can’t use new numbers as in the 3 root” and then use square root yourself. Seems cheap.
“Cannot use sqrt 3 because it introduces a new number “
_ignores the fact that square root uses a 2_
|{x,x,x}|! = 6 for all integers x between 0 and 10. Fight me.
Actually x can be any number :D
Smart! :)
Underrated comment. This is the best solution.
So... Is " |{....}| " The math notation for counting an array?... Please correct me if I'm wrong...
@@restablex Braces { } denote a set. Elements in a set are separated by commas, so {1, 2, 3} is the set containing the elements 1, 2, and 3. The absolute value sign | | here is called the cardinality in set theory, and evaluates to the number of elements in a set. So |{3, 5, 27}| = 3. And then lastly we take the factorial of 3 which gives 6.
The coolest part about this solution is that the elements can be absolutely anything. They don't even have to be numbers! |{duck, chicken, goose}|! = 6.
Edit: As mina86 pointed out, technically a set cannot have duplicates of the same element. However I believe we could consider the numbers to be a sequence, which allows duplicates.
@@timduffy8935 thanks. So, cardinality is 3 even when the element is repeated? Just want to be sure that |{a,a,a}| is 3 and not 1.
square rooting introduces a 2 into the equation. sqrt shouldn't be allowed
True
But when you write the symbol of a square root you dont really write the digit 2 down, so technically it still works
Well, it is a little tricky because I can not see any "two" in the square root symbol. Quite clever, indeed.
I don’t think you could do it with numbers 8, 9, or 10 without the sqrt
Mike Disney if it can not be done with out it then it can not be done. The floor method someone else mentioned is beyond my comprehension
1 video can teach me more than 100 days of school
I never thought there would be a complicated way to ask "What's 8 - 2?"
You never explained in detail what was allowed or gave an example. I had no idea what I could do.
funny enough you can't use cubic root of a number but can use the square? under what logic? lul
@@SkullDraker he said it. You can't involve new numbers. You don't need numbers to do a square root since you don't write the 2.
@@prenomenomine9355 but square root is power to 1/2.. basically new number is there.
@@prenomenomine9355 yes cause it's implicite, you don't NEED to write cause it is known, like that the sun is hot...
@@SkullDraker any operation can be expressed with more digits. +3 is just like +1+1+1
3:30 NEIN NEIN NEIN
Xd
Michael Foley kein Problem
😂 Guter Witz!
geil
GameJunk geiler typ 😬😁
Cool video!
So happy to say I found them all without cheating ! The 8 one was the hardest
"common mathematical functions that don't introduce new digits"
>Uses factorial, extremely uncommon, albeit simple function
>Uses square root but cube root and exponents are not allowed
By saying "common" I think he referred to "known"..
And about the factorial being legal but exponents not, I think its because when you exponent a number you can manipulate the exponent itself
(for ex: 2,3 etc.) but when using the factorial you cant do that because its a fixed function which only depends on the number you put factorial on
and in this problem the numbers are fixed.
For the root being legal and not the cube, I think thats because of the root being a basic function and using cube or anything else is just
manipulating the basic function to be a different one.
Thats my opinion though..
Exactly my thoughts, and seems I'm far from the only one.
John Jose cube root is not a manipulation of square root. Rooting a number requires a digit. It’s like saying that exponent 2 is a manipulation of exponent 1, it’s not, it’s just a different power exponent.
@@jaakezzz_G you didnt get my point.. my point is that factorial us a fixed function, just like adding or substracting or multiplying.
In this example you need to use functions that you cant change them.
Exponents however, you can change the exponent however you like.. 2,3,5 etc. Same for root you can do sqrt, cube or anything else but factorial depends on the number you do factorial on, you cant change the factroial function to work in a different way as you desire.
The one that stood out to me was taking a square root of the square root, that’s basically what he said was against the rules
nice, although one could argue that sq root 'theoretically' introduces the digit 2 into the equations. it is a defined symbol of sq root. Same way the 3rd root is not necessarily an introduction of the digit 3, it's a mere symbol, but a good puzzle nonetheless :)
Actually this argument wins, since sq root of any number is that number raised to the power 1/2
You could say the same with the factorial symbol, you are adding numbers and multiplying them even though you don't see them. I think that the puzzle's goal is to only have 3 visible numbers, and make a 6 out of it
@@JossWainwright Tbh, it is totally "I wanna catch you with the rules" BUT the 2 is simply there as much as 3 is in the cubic one. The cubic sign and the square sign are just 2 signs that indicate 2 different functions. Either both or none. Otherwise, you are just being a ahole as the puzzle giver and trying to be a smartass. (Which I think is still wrong because square root still means 2 whether you write it or not - you included another integer.)
@@JossWainwright I read milions of posts in the chain above. You're incorrect. Its simply that. You dont write it because of redundancy but its there whether you like it or not. Also, by proving you mean you will say to me that its exactly how you think it is.
@@JossWainwright Whatever you say. The number 2 is assumed always there if you dont write it. It becomes invisible 2. The fact the video rules out 3√x which is x^(1/3) [the number 3 you can take like a part of the sign, not as an integer because the whole 3√x is a number, not just the 3] but leaves √x which is x^(1/2) is just trying to be a smartass and not working. The 2 is there just you dont write it to save time because everyone understands what you mean. End of story.
Thanks to this, I made my own challenge inspired by this, and I have completed it. It is basically X X X = Y, where X and Y are both values between 0 and 10 including 0 and 10
2×2×2=8
Omg that's amazing! You did every number from 0 all the way up to 10!
That's all numbers up to 3628800, wow! That must have taken months.
Gotta be careful with punctuation around math.
@@MrEscape314 I only took 2 months. Try solving 0 0 0 = 4. Is is hard.
@@thedudehimself69420 I'm trying to figure out other hard ones like 999 999 999 = 1234..
I was trying to be silly cause you said you did all the numbers from 0 up to 10!
I mean for example 10 * 10/10 = 10 is the highest one I did. Also I did ones that included 11 and 12 as X and Y values as well. 12 * 12/12 = 12
A good part of the solutions relies on the student knowing the basic laws of mathematics and what it is that the student is allowed to do with math to solve these problems.
Ayyy best solution for 0
( cos(0) + cos(0) + cos(0) )!
Edit: Somebody already did this solution :(
👍👍👍awesome solution dude, perfect .
0 to the 0 power added three times then factorial.... = 6
Your incorrect. cos (0) = 1, so ( cos (0) + cos (0) + cos (0) ) = 3
@@snirpleinad8592 he wrote "!" in the end. So its (1 + 1 + 1)! = 3*2*1 = 6.
@@ck3908 ahh, not exactly as 0^0 is indeterminate form 🤷♂️
Wait isn't common operations just "+" "-" "/" "x" ? What's with the square roots and factorials? xD
Have you ever been to school?
i partially agree i think that square roots are ok but i don't think that factorials count as simple common operations
@@ThumbsTup Yeah.. have you??
@@LasTCursE69 sorry, I misunderstood the question for a sec, my mistake. And, yes, I have
@@JossWainwright It isn't about the solution of the puzzle.. It's about how they phrase the rules and the question..
I got all but 10, so I spent twenty minutes dicking around on my calculator until I learned that cos(10!) = 1, so I did (cos(10!) + cos(10!) + cos10!)! = 6. Your solution is much simpler and more elegant.
I got 8-logbase(sqrt(8),8) for 8s and
10-logbase(sqrt(sqrt(10)),10) for 10s.
Great video!
According to rule 1,you can use d/dx,and d/dx will make any constant 0,then you know what to do.
Favorite comment on this video
Good one bro😝😝😝
_Rice_
I don’t think calculus is allowed
wow thats kinda boring.
I would say it's a good trick to remember that 3! is because this means if any operation gets you to 3 you have solved the problem. For 10 10 10 and 8 8 8 I also used the fact that 9 is the square of 3 and so getting a result of 9 also immediately yields a solution. Finally any of these numbers can yield their own digit + or - 1 by simply addying or subtracting the quotient of the last two digits. And you have just gotten a rule for basically any of these numbers. ( For 0 0 0 you simply use the fact that you can reduce it to 1 1 1 and then 1+1+1 is 3, and thats all your numbers done in very few steps)
But square was not allowed I guess
He isn't squaring 3 in his example he is sqrting 9 to get 3 which then he factorials to get 6.@@XYZGarfieldZYX
I used the same approach but made a mental list along the way to knock off more variations. In the example below I am only going to match the previous number and not all the operations to get back to the original
6 = 3! = sqrt(9)
Then I also though of ways to manipulate the original equation: to get to these numbers
x/x = 1
x-x = 0
(x+x)/x = 2
There is a neat solution for the 8 8 8 using thermial. N thermial, noted n?, is defined to be 1+2+3+...+n, analogously to the factorial definition, but for sums instead of multiplication. Then 8? = 36 and one can easily do sqrt ((8+8-8)?) = 6
*You didn't say I couldn't introduce symbols for irrational constants...*
That means stuff like this technically _isn't_ cheating:
• round(3 + 3 + 3 - π) = 6
• round(3 + 3 + 3 - e) = 6
• round(9 × e - 9 - 9) = 6
• round(π(10 + 10) ÷ 10) = 6
for 10 10 10, i just took the log of 10 which =1, then added them up and took a factorial.
"Clever girl"
But log 10 (10) introduces a new number
@@createyourownfuture5410 not if you use natural logarithm.
@@createyourownfuture5410 Log base 10 is implied, same as the two in a square root. If you can use one, you can use the other.
@@freewing3964 I see. Log base ten can be written as lg in the same way as log base e can be written as ln.
In the future please try to be a bit more specific in ur directions because it was unclear which mathematical symbols were allowed
he said all common mathematical operations which do not explicitly introduce a new number is valid... I used logarithm to solve 10 10 10 = 6 as log (10) = 1
@@sondesobbaia1886 But why choose a base 10 logarithm? This is just a completely arbitrary base to choose.
Thunderbolt O because it is the only base that doesn’t require a number to be written in order to represent
@@sondesobbaia1886 log is not a common mathematical operation. The only ones are +×÷-. Square root is ambiguous because it's ^1/2, and there is no way that factorial is common. You will never use that outside of high school-college level math or above.
@@sondesobbaia1886 log(10) = 2.302... using the natural logarithm.
WHAT.THE.HECK.THIS.IS.GENIUS!
I have another way for 4 4 4 = 6. It is
4+4-√4.
Well done. BLOG IT
Same
does that read 4 plus 4 negative square root of 4?
@@valdehuesagoddessphoebej.1082 Yes.
Sphynx it’s minus not negative
For 8, I use the following equation: ((square root (8+8))!/8)! =6
explanation:
8+8 equal 16, square root of 16 is 4, factorial of 4 is 24, divided by 8 equal 3, and then factorial of 3 is 6
you can't use cube root, which means you shouldn't be able to use square root either
@@-.a No, dude. That would be the fourth root of x. You can only build (power of 2) roots out of square roots.
just use (8/8 + 8/8 + 8/8)! = 6
@@alexkelley8342 cant... because it only allow to use 3 digit not 6 like yours
8 is as easy as number 10, but in 8 you add instead of subtract (square root ( 8 + ( 8 / 8))! =6
I clicked without knowing what was The 6s Challenge and got a 6s ad.
"Rule: Can't use '=' symbol"
Me, an intellectual: Solution{6}
(n°+n°+n°)!=6 for any n-real number 😎🤓
Was stuck on a few until you mentioned factorial. I completely forgot about factorial!!!
As soon as you mentioned it, things became so much clearer...
Consider n as a generic Real number. If you take the limit of x/x when x tends to n, the result is always 1, whe then can get to the 1 1 1 = 6 case, and solve the puzzle for any number, even if it is negative, has several digits and decimals.
I don't know if logarithms and exponentials are allowed, if they are, I might have found a way to break this problem: we know that ln(e) = 1, and we can add or substract an unlimited amount of these, since they don't use any digits.
So 4 4 4 could be solved this way: 4+4-4+ln(e)+ln(e) = 6
This works with all integers not only the one from 0 to 10, and frankly you could do anything you want with the 3 numbers, as long as the result is an integer, and you balance it with the appropriate number of ln(e).
This however makes the problem useless and a lot less fun.
On 10 i did:
(lg 10 + lg 10 + lg 10)!
just another way to do it
@Govinda Solanki Vlogs how about (ceil(log(8))+ceil(log(8))+ceil(log(8)))! :O
If ypu exclude log, you should also exclude sqrt, both have a intrinsic number to the operation
@Bjjs you can use
When u include lg you automatic include log base 10. You can't include extra digits
@@RB-cl8tc and how about:
( sin(8!)! + sin(8!)! + sin(8!)! )! = 6
8! = 40320
sin(40320) = 0
0! = 1
1+1+1 = 3
3! = 6
voila, a hard way to solve it! :D
(I figured it out myself)
I don't understand the "you cannot introduce new numbers"
How is cube root introducing new numbers? Why can you use sqrt but not cube root?
you don't have to write the number "2" when you write the square root symbol, while you do have to write the number "3" when writing the cubic root symbol
But the square root is akin to raising the number to the power of (1/2)...
Matthew Wilson Its technically a trick cuz u dont write the number
That's why I say that this is more of a text puzzle than a math puzzle. sqrt is canonically written without a number superscript, while cube root is canonically written with one. That's all.
It is really strange that we can use sqrt but not cube root.
Damn wow, bro this was the hardest problem I saw here. I couldn't solve a single one except 2 2 2 = 6
Thanks for the video.
This ones are very amusing!
In fact, there's another trick to solve any equation of this type, N N N = 6, for any positive integer N: if we consider the "floor" function as an allowed one to use (defined as floor(x)=the biggest integer less than or equal to x, so for example floor(pi)=3 and floor(e)=2), we can always concatenate square roots of N until we have
1 < sqrt(sqrt(sqrt(...sqrt(N)))...) < 2 ,
hence floor(sqrt(sqrt(sqrt(...sqrt(N)))...))=1 and we have reduced the problem to "1 1 1 = 6", so adding up the ones and taking factorial, we're done:
( floor(sqrt(sqrt(...sqrt(N))...)) + floor(sqrt(sqrt(...sqrt(N))...)) + floor(sqrt(sqrt(...sqrt(N))...)) )! = 6
Wonderful, I think that is fair game since the floor function is commonly used. And your method also leads to a good mathematical question: why does repeatedly applying the square root lead to a number between 1 and 2? Alternately stated: why does the nth root of a number tend to 1? Here's one proof:
planetmath.org/limitofnthrootofn
That's right! Or even easier: given a fixed N we can always find n big enough so that 1
Interesting. One could do the same with the ceiling function and get everything into the form (2,2,2).
I also used the floor and the ceiling function but because when those equations were introduced to me I also thought that sqrt isn't allowed, I solved them with the e-function and its inverse.
Greetings from Germany :)
That's like the 4 4s problem, which can be done for any number using the same method. I forget the exact method, but there is a numberphile video on it.
Anyone else get 2+2+2 and then feel really proud of themselves?
Seb most of the comments are nerds lol
I got 2 × 2 + 2 and overcomplicated things.
Shut up, using bad language is prohibited.
Please refrain from such use of language, as it may result in ban
(SOORY UTUBE ME SMOLL)
2! ^ 2! + 2!
Wonderful task
We can use the function: (lnx+ln(sqrt(x))/ln(sqrt(sqrt(x))). It is always equal to 6 when x is not 0 or 1.
It's a shame such an elegant solution went unnoticed.
Beautiful, and this even doesn't depend on the base of the logarithm you choose, so (log(x) + log(sqrt(x)))/(log(sqrt(sqrt(x))) also works.
Very nice one! Just move one parenthesis to make it work: (lnx+ln(sqrt(x)))/ln(sqrt(sqrt(x)).
Should be three ending before the division.
So basically... just use factorials and you’re all good
Ye
I didn’t realize the point was to make an arithmetic statement. i just saw it as a puzzle to turn it into a true statement, so i drew a horizontal line through each problem, leaving the 6. 6 is 6 . a true mathematical statement. I honestly thought this was the solution and raced back here to see.