How To Solve The 6s Challenge

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.
Sources
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[@MindYourDecisions]

Nearly 2 million views in one year! Thank you!

[@panginoonn]

Hi, Love your videos!

[@luisalcocer5095]

Does 8-(8/8)!= 6 work? I just learned what factorial means from the video so im not sure

[@Lara-fg9vq]

you deserve it

[@prabeshgautam4112]

I was thinking factorial can't be applied cause factorial uses other no... Like 3! = 3*2*1 . Why??

[@davidpatterson9770]

You cant use sqrt() either. That's the same as 2root.

[@JM-po5hu]

i smart 2+2+2 = 6

[@JM-po5hu]

said it before he started working them out

[@pedromendz]

dang dude, b-b-but how? is it even possible?

[@tumeric3988]

J4M13 b -but, how? This g - guy is hackin!

[@iamhappythingy9258]

did stephen hawking wake up

[@giovannizottola106]

@@morveoteki2719 2×2+2

[@okramra]

I got the 2, 2, 2 thing in under a second maybe I am a prodigy

[@wahabbayor9090]

Lol!

[@reidarkollstrm5218]

Dude your going to get the nobel prize in mathematics

[@okramra]

@@reidarkollstrm5218 let me finish college without failing math first

[@spartalives]

🤣🤣🤣

[@-rheto-7837]

i also did the 6 6 6 in 5 seconds

[@jingusflorpus4274]

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.

[@aproplayer8581]

We can use cube and square roots but of the same number for you are solving.

[@king_of_the_sun4897]

Then how did he use a square root for 3s?

[@redaipo]

@@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

[@awesomeleozejia8098]

@@redaipo the operation is the second root of 3*3

[@MrEscape314]

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...

[@roblatour3511]

0 0 0 = 6 ; put the first zero on top of the 2nd zero; 8 0 = 6; move the remaining zero to the other side of the equals sign; 8 = 0 6; move the zero one last time so that overlays the six; 8 = 8

[@BlueCoreGamming]

If cube root is not allowed, square root should not be either, as it introduces a 2.

[@Stubbari]

No it doesn't.

[@DarthBil1]

@@Stubbari yes it does

[@DarthBil1]

@@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".

[@Stubbari]

@@DarthBil1 So if there is no written digit then there's no digit. What's so confusing about that?

[@DarthBil1]

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.

[@jakobf6165]

I found a pretty solution for 10 10 10. You can calculate 10 × 10 + 10 = 110, which is 6 in binary representation!

[@ethanpatch6840]

that is a nice answer but i don’t think that actually counts as a proper solution

[@txtp]

well- 10 + 10 + 10 2 + 2 + 2

[@txtp]

another binary one

[@BigMikeECV]

There are 10 types of people in the world: those that understand binary and those that don't.

[@Serai3]

I don't think mixing bases would count.

[@MrTacoLama]

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!

[@hugh.g.rection5906]

Cos(0) isn't a common mathematical operation

[@therealmaster9686]

@@hugh.g.rection5906 tbh i've used the cosine function much more that factorials

[@msncat]

@@hugh.g.rection5906 i think it is because he didn't restrictions in detail on this

[@rayaanansari4834]

@@hugh.g.rection5906prove it

[@swedishpsychopath8795]

@@hugh.g.rection5906 Ask Mathologer, I guess it is super-common to him?

[@eddeh0772]

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!

[@Vicandiers]

so what was your solution to 000 and 111?

[@eddeh0772]

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

[@Vicandiers]

Got it

[@bragesrensen9889]

The answer wasn't always 720

[@smaransure2234]

i got all of them except 10 10 10

[@connormorton665]

Alternate title: how make three numbers equal 3 and then apply a factorial operation

[@Nightmare_Developer]

yes i solved almost all that didnt include ! except 8 that was hard one tho, don't mind coz im newbie in math...

[@SorakuteeYT]

@@Nightmare_Developer How did u solve 1 withouts factorial. Would u mind sharing

[@Nightmare_Developer]

@@SorakuteeYT umm no i said i solved all that didn't include !(factorial) except 8 coz i dont even know what ! is

[@SorakuteeYT]

@@Nightmare_Developer i mean the one with 1. 1 1 1=6 I cant solve it without factorial

[@arcanesmemes]

@@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.

[@babitasaxena2369]

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 😂😂😂🤭🤭🤭🤣🤣🤣

[@fahad5593]

Cheesy

[@aravinthsrisivasritharan4098]

Cheeky

[@Coolkid245]

Cheesy

[@hellohumans9721]

Cheeky

[@babitasaxena2369]

What does it mean Cheeky ckeesy whatever

[@moetocafe]

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!

[@OliviaAndreoli]

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

[@yehor_ivanov]

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

[@rs5256]

@@yehor_ivanov 24/8 = 3 not 6. You'd have to say: square(8+8) = 4 -> 4! = 24 -> 24/8 = 3 -> 3! = 6.

[@cythism8106]

@@yehor_ivanov say sqrt(x) not square(x). It almost makes it look like you're saying (x)^2.

[@324_Sli]

Just this, 8+8=16 aight? 16 root=4 4 root=2 8-2=6

[@los-lobos]

You used 4 10’s

[@Kriegter]

Damn the maths developers really need to nerf this new "factorial" boost

[@gjproductions9337]

They said they will in update 3.14

[@CrazyAsians123]

@@gjproductions9337 finally

[@slaughterhouse5585]

@@gjproductions9337 Mmmm! Pumpkin 3.1416.

[@benr77]

noooo! now I can magically make almost any equation just work

[@gorgonix2264]

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.

[@imonsanyal]

If you can't use "³√", using "√" should also be against the rules because it is just the short form of "²√".

[@Stubbari]

And 4 is a short form of 1+1+1+1. With your locig the whole problem is impossible.

[@imonsanyal]

@@Stubbari you're not making any sense.

[@Stubbari]

@@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.

[@imonsanyal]

@@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.

[@jhndvdvdd]

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?

[@robertp9297]

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.

[@mjorozco3786]

this is a high quality good viewer that we need to protect at all costs

[@Kumra_Podash]

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!!!

[@kumarsaurav8885]

(0!+0!+0!)!=6 For all the rest: (sgn(x)+sgn(x)+sgn(x))!=6

[@asgarrahmani939]

It is not true

[@dorondaniel318]

@@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)

[@trabadix]

@@dorondaniel318 interesante!

[@bruhbruh1580]

Kumar Saurav or for 6 6 +6 =12 12-6=6

[@mikel4879]

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.

[@kay710]

*888 is hardest* Me : 6 CIRCLES!!!

[@dhanajon5528]

😂😂😂

[@zirconescalus]

😂

[@Lililili0170]

Keerthi Mahesh genius👍👍👍

[@walterfischer456]

Ehm.... Yes?

[@Broody90]

888 was easy 1. third root of 8 is 2 2. add them up 3. .... 4. profit!

[@kaderen8461]

challenge: do the ones with square roots without them, as square roots technically involve other numbers like cube roots do

[@Jamesdavey358]

@@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

[@H0uxdubxston]

@@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

[@falling_banana]

@kaderen8461 👍 finally, someone's thinking what i'm thinking

[@falling_banana]

@@H0uxdubxston exactly

[@projectmoonsleeperagent]

@@JossWainwright So I can write a square root, but say it’s implied as a cube root and then cube root now counts?

[@MarcusPereiraRJ]

This challenge is a classic one in Brazil we learn as younglings. Nice to see it here!

[@marvelstark3797]

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√

[@Stubbari]

@Ro Bert Yeah because CR introduces a new digit "3" while SR doesn't.

[@videopoetic7101]

@@Stubbari You don't have to write it but in SR is number 2 so it's a problem in rules.

[@Stubbari]

@@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: " "?

[@loganxavier]

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.

[@bachlamtung5131]

point is, it’s literally the same, the 2’s implied thus you dont have to actually write a number

[@gregatherton4688]

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 √ :)

[@andreaq6529]

Bro 🤯👍

[@static3479]

You’re the type of guy that does his homework on time

[@sajamily2404]

😳😳

[@YSFmemories]

can someone tell me if this guy is legit or trolling?

[@isjosh8064]

Who tf even are you?

[@user-uj5gh3xy7n]

For 10, i did [Log( 10 ) + Log ( 10 ) + Log ( 10 ) ] ! = 6. Usually log is base 10, so no new digits

[@Able89535]

Nice, or we can use use ln or even sqrt directly if we introduce ceiling or flooring

[@user-uj5gh3xy7n]

@Able89535 yeah but I don't like to use floor or ceiling because they annot be described with regular math.

[@Functional_Somehow]

I fact-checked this and this is absolutely correct. Good job! 👍

[@AlbertWang1]

Square root is controversial for all these kind of quest, as it's square 2.

[@ojojojojojoje]

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

[@thebanditterra2917]

Agrees I too thought it was out of bounds and despite having the solutions using it still believe thusly

[@diablo6250]

@@JossWainwright bruh, do you scan this comment section for this one question? if so- *i shall follow you kid*

[@JLvatron]

The rule should clarify you can't introduce other numbers, but numberless symbols are allowed. Like √ .

[@JLvatron]

@@JossWainwright Good point.

[@friedfries8432]

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.

[@clavio3082]

haaaaaaaaa, this is the equivalent to the infinite money glitch hahhahah

[@anshsharma2652]

Bravo!

[@user-ec6wu4hv7l]

невероятно, я восхищен

[@darkdesmond6706]

Big Brain Time

[@ronaldanderson4995]

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.

[@Foxy_Fan1200]

I learned what a factorial is because of this video!I love learning new things!

[@robertorodriguez5226]

What a fantastic way of teaching operations. It is inspired.

[@mrchoon2010]

He said the word "factorial" and I knew I was in over my head, haha

[@is1hair]

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.

[@mrchoon2010]

@@is1hair Yeh, I went and googled it, thanks

[@ayushgangwaropz1088]

@@is1hair thank u bro .. it help me a lot.. and prevent to Google it?!!!

[@NERONRR]

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"

[@mrchoon2010]

@@NERONRR Can you solve them without factorials?

[@maxbrandt1324]

3:29 (9 + 9)/ sqr(9)

[@AwesomeCreatorBen]

I did that too. Sorry im late to the comment section

[@nicolasrozenberg5209]

You cannot use sqr(9)

[@faradaykhaleesi877]

@@nicolasrozenberg5209 do you even know what sqr() is?

[@30IYouTube]

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.

[@nicolasrozenberg5209]

@@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

[@CT-pi2gl]

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?

[@dablitter5719]

this is one of the few puzzles i could actually do and it was actually pretty fun

[@kiancroxall2099]

Literally the only two I’m smart enough for: 2+2+2=6 And 6+6-6=6 😂🤣😂

[@crazybox9424]

Same

[@pikachu-dr1vs]

wow same lmao

[@rajakanijes]

5/5+5=6

[@melkor991]

kingmaker (5^2 + 5)/5

[@stefanmichaelsenzegarra5688]

Me too lmao

[@sandrasweeney798]

Wow! This is your best yet. I love your videos.

[@gtaserisiturkiye]

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

[@Jan_Heckmann]

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.

[@M3lodicDeathmetal]

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.

[@germanandosov2586]

One more solution with zeros (it's mine before watching this video): (cos(0) + cos(0) + cos(0))! = 6

[@benoitgrs2786]

@@nikizhu78 it's factoriel

[@minh9545]

@@nikizhu78 Factorial.

[@nicholasjarrett4480]

(log(10 * 10 * 10))! = 6

[@ogandreyka]

@@nikizhu78 не позорь русских

[@vladserg9829]

@@nicholasjarrett4480 not log but lg

[@horsesh8e]

The whole is about solving to 3 and taking its factorial.

[@swarnavabiswas425]

@@leinfare Wow!

[@asmitoghosh2193]

@@leinfare Yup

[@arnabchoudhury2651]

Great solution... your way of thinking is just awesome.

[@Courruptedmantealpink]

Thanks for creating this!

[@Tasarran]

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!

[@Numerixx]

As soon as I saw this, I thought 0 0 0 would be (0!+0!+0!)! bruh

[@minafawzy5086]

To use 10 10 10 there is another solution (log(10*10*10))!

[@Sid37612]

Awesome dude! That's an innovative solution!

[@ExploitRX]

@20 Subs Before Tomorrow? I think, he mean lg (log with base 10)

[@andsalomoni]

@20 Subs Before Tomorrow? A piece of trunk.

[@manuelmontana2827]

What is log? Baby don't hurt me~

[@daapdary]

For 10 10 10 = 6, I did: ( log(10) + log(10) + log(10) ) ! = 6

[@KyleDoveGaming]

In the beginning you had a rule of not introducing cubed roots because that counts as adding a number. However, you use square roots in the majority of your answers. I'm sure you are aware that a square root can be written as X^(1/2), which if we go by that notation, it makes the use of square root against the rules that you put forth in the beginning.

[@katarsiz2]

Then you just don't go by that notation and simply use √ symbol. Multiplication N*K can, too, be written as N+N+...N+N with K terms, which obviously involves writing more digits, which is forbidden. But we take as granted that multiplication is allowed, therefore we can conclude that simple "existence of different notation that does include more digits" is not enough to forbid an operation by these rules.

[@sumanjangid1250]

According to the rule 1. at the start of the video, we can not also use square root as it involves the number 2 . But considering the problem, we are bound to use that otherwise a solution won't be possible . By the way it was a cool problem. Thanks for uploading.❤

[@samkhan-lu9ww]

you don't need the two for square root even though you can put it there

[@danielattema8596]

That's what I thinking, so I checked the notes to see if anyone else saw the inconstancy. The square root symbol is short hand without the 2. So, like the cubed root of x can be written as x^-3, the square root of x can be written as x^-2.

[@derekchase5462]

Reading that first equation very excitedly “Zero! Plus zero! Plus zero! (!!!)”

[@TheIvasyl]

@@user-SG717 what's 230 - 220 x 0.5? You probably wouldn't believe me, but the answer is 5!

[@TheIvasyl]

@@user-SG717 just because you're right doesn't mean I'm wrong

[@TheIvasyl]

@@user-SG717 it's 5!

[@TheIvasyl]

@@user-SG717 5! = 120

[@SteveSharps]

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.

[@ellanvanninalde]

And logarithms, since (log10+log10+log10)!=6

[@MajaxPlop]

Steven Song take ln0 so I know for e ;)

[@ArcticFoxWaffles]

Most people wouldn't know about factorials

[@xzZZZ799]

so what do we call a root symbol alone and how does it work

[@user-ob5hj5vn8c]

Yah Root 2 is basically X ^ (1/2)

[@mariusvr]

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

[@alexmark6580]

Was able to do all of them except 0, I forgot that 0! = 1. This is a fun puzzle exercise.

[@hamraj7231]

for 999 just flip the nines upside down

[@conanchan9033]

the 6 on the right also get up side down

[@dystopia-user181]

Nein nein nein, that won't work.

[@ogjrap6928]

Enin Enin Enin Nothing worked

[@liorem4003]

(9+9)/√ 9

[@myguy4w164]

Whoaaaaaaaaa there brother maybe we can relax

[@1ups_15]

each time you showed the solution I was like "how didn't I think of that?!!" haha

[@NikhilTheGreatest]

Same with me😂😂

[@aubreyundi]

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

[@huyxiun2085]

@@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.

[@TheRenegade...]

@@aubreyundi Technically multiplication is adding new digits because it's repeated addition

[@memebaltan]

@@aubreyundi a*b=a*a*a... b times ah yes, maeth

[@cliffordmorris6091]

You are doing what teachers have been doing for years getting children to think about different ways of arriving at answers. Eg Using partition to find all the ways we can make a number like 24. Maths does not have to involve roots and factorial so can be used at a simpler level. It is interesting to challenge people who believe maths is about right answers by giving the answer and requiring all the ways to get to it instead.Indeed most real life problems are open to innovative solutions and there is not an answer but a set of possible solutions and the requirement to provide often cost effective ones using an iterative approach towards the best one.

[@ali94hn]

For 8, I did √(8+(8)/(8)))! (the square root symbol is supposed to be over everything EXCEPT the factorial symbol) which gives us a result of 6 8+8/8 is 9, then I take the square root of that, which is 3, and then I put it into factorial EDIT: I just realized that's your second solution

[@JianJiaHe]

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.

[@RGP_Maths]

That's brilliant and deserves more attention than it's got so far!

[@NestorAbad]

What an awesome and elegant solution!

[@donaldasayers]

I love it when someone just kills a puzzle.

[@boggless2771]

I hope this counts!

[@pentaxian7455]

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!

[@gamejunk2707]

3:30 NEIN NEIN NEIN

[@einfachben0399]

Xd

[@gamejunk2707]

Michael Foley kein Problem

[@tirthrajmahajan2508]

😂 Guter Witz!

[@andererletsplay8139]

geil

[@zwischendurundmoll3968]

GameJunk geiler typ 😬😁

[@theunknownspeedrunner276]

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.

[@AstroGalaxyTCG]

"!" Is just broken in mathematics 💀 bro almost solo'd the secret quest

[@Santhosh22NA]

You always have solution in mathematics when u stuck. "Let's assume 0 0 0 = 6"

[@wholesomehoorpari1971]

Hahah xd

[@arnavverma7308]

😂😂😂

[@vivek.r4832]

You are Savage until I am thug bro

[@ItsJustEthan1]

Ayyy best solution for 0 ( cos(0) + cos(0) + cos(0) )! Edit: Somebody already did this solution :(

[@MaxMathGames]

👍👍👍awesome solution dude, perfect .

[@ck3908]

0 to the 0 power added three times then factorial.... = 6

[@snirpleinad8592]

Your incorrect. cos (0) = 1, so ( cos (0) + cos (0) + cos (0) ) = 3

[@lAlexLunl]

@@snirpleinad8592 he wrote "!" in the end. So its (1 + 1 + 1)! = 3*2*1 = 6.

[@rohankorale6381]

@@ck3908 ahh, not exactly as 0^0 is indeterminate form 🤷‍♂️

[@upsidedownumop3psdn225]

I got 8-logbase(sqrt(8),8) for 8s and 10-logbase(sqrt(sqrt(10)),10) for 10s. Great video!

[@Rinneganpein389295]

|{x,x,x}|! = 6 for all integers x between 0 and 10. Fight me.

[@Anastasia___.]

Actually x can be any number :D Smart! :)

[@timduffy8935]

Underrated comment. This is the best solution.

[@restablex]

So... Is " |{....}| " The math notation for counting an array?... Please correct me if I'm wrong...

[@timduffy8935]

@@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.

[@restablex]

@@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.

[@rogeronslow1498]

You never explained in detail what was allowed or gave an example. I had no idea what I could do.

[@SkullDraker]

funny enough you can't use cubic root of a number but can use the square? under what logic? lul

[@prenomenomine9355]

@@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.

[@danieljared2307]

@@prenomenomine9355 but square root is power to 1/2.. basically new number is there.

[@SkullDraker]

@@prenomenomine9355 yes cause it's implicite, you don't NEED to write cause it is known, like that the sun is hot...

[@prenomenomine9355]

@@SkullDraker any operation can be expressed with more digits. +3 is just like +1+1+1

[@birdlegscass]

3 is truly wielding that exclamation point like a mighty blade today

[@peterhuroiye408]

So happy to say I found them all without cheating ! The 8 one was the hardest

[@chaincat33]

"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

[@Simplifier123]

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..

[@3possumsinatrenchcoat]

Exactly my thoughts, and seems I'm far from the only one.

[@jaakezzz_G]

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.

[@Simplifier123]

@@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.

[@archeosm8606]

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

[@danzirulez]

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 :)

[@XariksBeatbox]

Actually this argument wins, since sq root of any number is that number raised to the power 1/2

[@manawer720]

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

[@Dark_Voice]

@@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.)

[@Dark_Voice]

@@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.

[@Dark_Voice]

@@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.

[@evanfishsticks8010]

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.

[@CorghVosc]

square rooting introduces a 2 into the equation. sqrt shouldn't be allowed

[@lepassant478]

True

[@wyn2118]

But when you write the symbol of a square root you dont really write the digit 2 down, so technically it still works

[@MataMaticas]

Well, it is a little tricky because I can not see any "two" in the square root symbol. Quite clever, indeed.

[@michaelmullin106]

I don’t think you could do it with numbers 8, 9, or 10 without the sqrt

[@rgazsy8366]

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

[@bro_vega_1412]

According to rule 1,you can use d/dx,and d/dx will make any constant 0,then you know what to do.

[@Nathan-tg4gu]

Favorite comment on this video

[@suraj5957]

Good one bro😝😝😝

[@the_username_is_taken]

_Rice_

[@hedgehogmind3186]

I don’t think calculus is allowed

[@phrog7193]

wow thats kinda boring.

[@fernandoalonsooliveira6090]

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.

[@u.s.4129]

A nice one, thanks Presh

[@jachpi1080]

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...

[@m_th_m_t_cs]

good idea!

[@preyunknown1820]

cos is not operation it is FUNCTION

[@jachpi1080]

@@preyunknown1820 oh sugar lumps yur right. However, it was a nice try.

[@jachpi1080]

I think it should count, but, welp, rules are... rules I guess? (I think it should count xd)

[@preyunknown1820]

@@jachpi1080 yeah nt tho

[@LasTCursE69]

Wait isn't common operations just "+" "-" "/" "x" ? What's with the square roots and factorials? xD

[@ThumbsTup]

Have you ever been to school?

[@kaladin6199]

i partially agree i think that square roots are ok but i don't think that factorials count as simple common operations

[@LasTCursE69]

@@ThumbsTup Yeah.. have you??

[@ThumbsTup]

@@LasTCursE69 sorry, I misunderstood the question for a sec, my mistake. And, yes, I have

[@LasTCursE69]

@@JossWainwright It isn't about the solution of the puzzle.. It's about how they phrase the rules and the question..

[@theraven749]

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 :)

[@Archy_The-Wizard]

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'

[@Arcessitor]

@@Archy_The-WizardFactorials literally turn 0 into 1. How is that not introducing a new number?

[@lebeccthecomputer6158]

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

[@rperm834]

So it is {√[8+(8/8)]}!

[@ari998]

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

[@-.a]

you can't use cube root, which means you shouldn't be able to use square root either

[@zaqcarson2875]

@@-.a No, dude. That would be the fourth root of x. You can only build (power of 2) roots out of square roots.

[@alexkelley8342]

just use (8/8 + 8/8 + 8/8)! = 6

[@1987Videolover]

@@alexkelley8342 cant... because it only allow to use 3 digit not 6 like yours

[@MarcoOS05]

8 is as easy as number 10, but in 8 you add instead of subtract (square root ( 8 + ( 8 / 8))! =6

[@ValKS-0]

your sol. for 10 10 10 = 6 was sqrt(10-10/10)! the same way you could just do 8 8 8 = 6 as sqrt(8+8/8)! = 6

[@RafaelVogel]

8 can be done following the 10 approach, making a 9 (8+8/8) than square root turning it into 3 than factorial it, giving 6: ((8+8/8)^(½))!

[@Perezafer8]

On 10 i did: (lg 10 + lg 10 + lg 10)! just another way to do it

[@RB-cl8tc]

@Govinda Solanki Vlogs how about (ceil(log(8))+ceil(log(8))+ceil(log(8)))! :O

[@marcoasturias8520]

If ypu exclude log, you should also exclude sqrt, both have a intrinsic number to the operation

[@faraonzeu9462]

@Bjjs you can use

[@maithreebogoda8824]

When u include lg you automatic include log base 10. You can't include extra digits

[@mariush.215]

​@@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)

[@1q5]

Anyone else get 2+2+2 and then feel really proud of themselves?

[@ElectroGaming5]

Seb most of the comments are nerds lol

[@JamesCPotter13]

I got 2 × 2 + 2 and overcomplicated things.

[@extremenugget3658]

Shut up, using bad language is prohibited. Please refrain from such use of language, as it may result in ban (SOORY UTUBE ME SMOLL)

[@ofcrgry]

2! ^ 2! + 2!

[@draganminic4928]

The first derivative of a constant is zero. So for any real number, X, we can solve X X X = 6 by writing a "prime" sign after each X - which means the first derivative of each X, which is zero. Thus we reduce each X X X = 6 problem, to 0 0 0 = 6.

[@hopelessutopian]

Dang. I considered "log" a common math operation, so my solution for 8 8 8 = 6 was: 8 - log(8) / log(sqrt(8)) Which I didn't even think would work out until I just tried it, but it essentially captures the invisible 2 from the sqrt operation. I was proud of that one. And of course my 10 10 10 = 6 solution was (log(10) + log(10) + log(10))! = 6, since like sqrt, log is implicitly base 10 (with ln having base e). Probably should have listed out the acceptable operations explicitly in the video. Writing out all of your assumptions in tedious detail is the true Mathematicians Trick! :P

[@chitrakshsinha9007]

I have another way for 4 4 4 = 6. It is 4+4-√4.

[@7llininthedream]

Well done. BLOG IT

[@16anurag9]

Same

[@valdehuesagoddessphoebej.1082]

does that read 4 plus 4 negative square root of 4?

[@chitrakshsinha9007]

@@valdehuesagoddessphoebej.1082 Yes.

[@ba2138]

Sphynx it’s minus not negative

[@freewing3964]

for 10 10 10, i just took the log of 10 which =1, then added them up and took a factorial.

[@eriklagergren7124]

"Clever girl"

[@createyourownfuture5410]

But log 10 (10) introduces a new number

[@davisatdavis1]

@@createyourownfuture5410 not if you use natural logarithm.

[@freewing3964]

@@createyourownfuture5410 Log base 10 is implied, same as the two in a square root. If you can use one, you can use the other.

[@createyourownfuture5410]

@@freewing3964 I see. Log base ten can be written as lg in the same way as log base e can be written as ln.

[@MantasDabs]

i didnt even know what a factorial was so thank you

[@Sunset553]

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.

[@nathanisbored]

why is square root allowed? is it because the index isnt shown? would log or ln be allowed for simliar reasons?

[@SomeGuy712x]

I'm okay with allowing square root since that's the default for that symbol without a digit added to it. Also, I did end up using log for the 10 10 10 = 6 problem myself, coming up with log(10 x √(10^10)) = 6.

[@paulkennedy8701]

Yes. Log and ln would be allowed if it's the natural log, since they are written without using an extra digit. Logs to any other base, where a digit needs to be written, would not.

[@JohnDixon]

This problem is a test of our current mathematical symbols, so anything that doesn't explicitly use a digit in its symbol is fair game. The only non-numerical symbols that are usually excluded from problems like this are ones that would make the problems too easy or give some sort of "universal solution" that works for all numbers (e.g. the floor and ceiling functions).

[@connorhorman]

It has been noted that floor(x) and ceil(x) are fair game. Also hi nathanisbored

[@luisaguinaga9563]

Agree I did not use the squares. but I did not solve for 8, 9 or 10. I also tougth that the square root was not allowed

[@priyenswiss2002]

3:29 Adolf Hitler joined the room.

[@jaydani1996]

How?

[@user-bl1fe9mw7g]

Nain! Nain! Naaaain!!!

[@heorgegarrison5554]

N E I N N E I N N E I N

[@user-fo4ue9mo4z]

Heorge Garrison nine*

[@heorgegarrison5554]

ঊᴄᴏᴏʟ ᴘʀᴏ idk if you’re joking or not but 9 in English sounds like nein which is german for no, and Adolf Hitler spoke some form of german and who doesnt love a sprinkle of dark humour :)

[@thedudehimself69420]

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

[@isayahpesulima7297]

2×2×2=8

[@MrEscape314]

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.

[@thedudehimself69420]

@@MrEscape314 I only took 2 months. Try solving 0 0 0 = 4. Is is hard.

[@MrEscape314]

@@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!

[@thedudehimself69420]

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

[@jojotag5344]

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.

[@NestorAbad]

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

[@MindYourDecisions]

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

[@NestorAbad]

That's right! Or even easier: given a fixed N we can always find n big enough so that 1

[@Tehom1]

Interesting. One could do the same with the ceiling function and get everything into the form (2,2,2).

[@Mutlauch]

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 :)

[@Sam_on_YouTube]

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.

[@DasMonitor1]

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)

[@XYZGarfieldZYX]

But square was not allowed I guess

[@Joeljr110]

He isn't squaring 3 in his example he is sqrting 9 to get 3 which then he factorials to get 6.@@XYZGarfieldZYX

[@Joeljr110]

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

[@seanharris3447]

For 8s I found another way. Add two of the 8s to get 16 then square root to get 4. Then 4 factorial divided by the last 8 is 3 then use 3 factorial so ((square root(8+8))!/8)!

[@lenzworker]

For 10 10 10, I separated each number into its individual digits and took the factorials of the zeros and added them all up so: 1+0!+1+0!+1+0!= 6.

[@doczero1296]

This comment section is full of nerds. I've never felt more at home.

[@user-fo4ue9mo4z]

I only know about square roots but not factorials but factory is like saying 5! = 5x4x3x2x1 = which is 120

[@ArtisticScratch]

@@user-fo4ue9mo4z Factorials are easy to understand but this video made me think factorials were addition rather than multiplication: 3+2+1 = 6 = 3x2x1

[@AdithyaShankaran]

@@user-fo4ue9mo4z so

[@user-fo4ue9mo4z]

ADITHYA SHANKARAN just explaining how factorials world, because most people don’t know about it!

[@orngng]

ঊᴄᴏᴏʟ ᴘʀᴏ also written out as n! = n x (n-1) x (n-2) ... x 3 x 2 x 1 As n is a member of a natural number

[@samhecht1492]

In the future please try to be a bit more specific in ur directions because it was unclear which mathematical symbols were allowed

[@sondesobbaia1886]

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

[@thunderbolto7611]

​@@sondesobbaia1886 But why choose a base 10 logarithm? This is just a completely arbitrary base to choose.

[@Nuclearburrit0]

Thunderbolt O because it is the only base that doesn’t require a number to be written in order to represent

[@sanhakim1335]

@@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.

[@J7Handle]

@@sondesobbaia1886 log(10) = 2.302... using the natural logarithm.

[@user-eo2ed7lr2n]

10 10 10 - take decimal logarithm of 10. We know it equals to 1. Therefore lg10+lg10+lg10 = 3, but 3! = 6

[@digitalphoenix2787]

Another solution for 8 8 8, though a bit more complicated, is to add 8+8 for 16, take the square root to get 4, factorial 4 to get 24, divide by 8 to get 3 and then factorial again to get 6

[@tushermajumder7211]

In binary method 10 implies 2 Thr for, 10 +10+10 equal to 6

[@21nod]

In binary method there is no 6

[@lotsofd6739]

bachelors of trolling degree, 2013 there are 110,but it’s not 6 also 110₍₂₎=6₍₁₀₎ but it’s useing diferent number, “2” so we can’t do that

[@21nod]

@bachelors of trolling degree, 2013 I know my comment didn't encourage your creativity (which is something that has to be praised) but all I wanted to achieve is challenge you even harder.

[@supercool1312]

Tusher Majumder 6 is 110 in binary. 10*10+10 is 110. easy

[@ralphy1054]

No 10+10+10=110

[@yassinesafraoui4540]

for 444=6, there is an easier way to solve it: 4+4-√4=6

[@65EKS65]

Define easier? It's just different way to solve it. Also knowing that √4=2 we can solve it the same way than we solve 2 2 2 without need of inventing new method to solve it so I'd say √4+√4+√4 is more simple.

[@oenrn]

@@65EKS65 "Easier" as in requiring the minimum amount of steps.

[@65EKS65]

@@oenrn I'd see what you mean if the problem was long but in this it doesn't take any more time to calculate either way. The easy part is that 2 2 2 is already solved and everyone knows that √4=2 so I think it's more complicated to try calculate something different even if it requires less steps if the other answer is already on the table to be seen. But yeah idk. Not big difference either way.

[@threej4pope]

Given that square roots are technically illegal in a strict reading of the rules, I solved this one via (4 nPr 4) /4

[@InterestingStuffWithMe]

Personally I feel that subtraction is harder and more of a hassle than addition hence I don't think it's an easier way Anyone else thinks the same way?

[@macro144p]

For 8 I used: (round(sin(8) + round(sin(8) + round(sin(8))! I did a similar thing with ten: (round(-cos(8) + round(-cos(8) + round(-cos(8))!