Can You Solve The Three 3s Challenge?
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- Опубликовано: 20 окт 2024
- Thanks to Yale from Hong Kong for suggesting this problem! This problem went viral after Carl Ho's video (in Chinese), and there are many fun ways to solve it. My challenge is: find two answers to 3 3 3 = 10 using only the symbols + ! ( ). There is a way you can do this!
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Carl Ho's video (Chinese) • 考倒大學生的數學難題! 3個3如何運算才得1...
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Can You Solve The 6s Challenge?
• How To Solve The 6s Ch...
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Glad to see my question in this channel :D
and yes, the "subfactorial" should be the better way to solve =10 question.
Thanks! I encourage everyone to watch the video on Carl Ho's channel, you'll definitely enjoy it:
ruclips.net/video/6acBMe0Ugy8/видео.html
And here's another clever solution to the bonus, which was emailed to me: !(!3 + !3) + !(!3) = 10. This solution breaks down as !(!3 + !3) + !(!3) = !(2 + 2) + !(2) = !4 + !2 = 9 + 1 = 10.
(But please don't email me any more as I get too many emails!)
i found two simple ways:
3.3(repeating) *3 = 10
3*3 + Floor function of sqrt(3) = 10
(3+3+3)++ = 10
hi 卡爾
3square + (3/3). (3 cube +3)/3.
Arjun(3+3+3) = 10
Arjun is a function created by me which adds 1 to the result
I feel like that’s cheating. That function doesn’t even exist, and makes no sense whatsoever.
@@SlipperyYayas Its a joke.
@@thepanthar It doesn't feel like a joke honestly
@@SlipperyYayas lmao its obviously a joke
function Arjun(value) {
return value+1;
} ... now it exists
Easy!
(3+3+3)++ = 10
As notation used by programmers seems to be acceptable.
++(3+3+3) or else it will become 10 but still give you 9
He hav to use pre increment .
programmers: 333 != 10
@@Mn-Fe-N Webprogrammers, maybe
The U.S. 5
I definitely think using a decimal point as an operator is definitely cheating. .3 is not an operator acting upon 3, it is simply a different number. You can make a similar case for 3% as well, it is simply a different number.
And rounding just stinks.
Exactly.. I lost interest due to this twisting..
Smart ^^
Though I will say that using a decimal point is the least egregious of various ploys that I've seen. E.g., to do 11 = 33 / 3, you're concatenating two 3's together to form 33. Is that allowed? You could also do 11 = 3 x 3 + Γ(3).
3! is 6 so it’s a different number it doesn’t worj
When you were showing alternate solutions for 0-9, I saw a decimal point being used for 7 and I realized an easy way to do 10 is:
(3! - 3) / .3 = 10
I had this same. Until then, I assumed that the use of the dot itself is not allowed, because I would have to add 0 in front.
√3*√3/.3=10
3!/(.3+.3)=10
Yeah I guess they were being pretty slack with rules for this one. The arctan solution is a stretch too, it needs to be equal to 10 not 10 degrees. Then they started using floor functions, then I rage quit. I mean I may as well make up an arbitrary function and use it. eg f(x) = 10, so f(3+3+3) =10. and done.
Also I didn't think bit shifting was a mathematical operator...
(3! - 3) /3 = 1+0
3.3 repeating (use a bar on top of the second three) multiplied by 3.
3.3....×3=10
This is more satisfying than any of the proposed solutions tbh
Yes, you genius hahaha
But you only can use + ! and (), not .....
@@hafizhperdana Well that was for the "Super hard bonus" rules. I'm offering a simpler solution.
It doesn't equal 10. It comes as close as possible to 10 without being 10. 33/3 does equal 10.
the only hard thing about the bonus is knowing that subfactorial exists, lol.
Ive never heard of it either. Does it have any practical use anyway?
Same is here
I have never even heard of FACTORIAL
I have a Masters in Physics, so I know what factorial is, but I have never ever heard about subfactorials.....
@@eriknystrom5839 you dont really need a master in physics to know about factorials though
I was expecting double factorial, so I found this:
(3!)!! = 6!! = 6*4*2 = 48
((3!)!!)/3 = 16
((3!)!!)/3 - 3! = 16 - 6 = 10
genius
You found a solution following the rules, congrats
This is the best way to solve the problem, congratulations!!
Nice search man
Better than Yale, congrats!
its old but maby somehow you end up reading it. you can do !(3+3+!3)=10
No rules specifically mentioned about adding operations to the right side of the equal sign:
(( 3 - 3 )/ 3 )! = log(10)
Also allows for other neat stuff such as:
( 3 + 3 )!/ 3 ! = 5!
I don’t think casually changing 3 to 0.3 in one of your solutions is acceptable.
I agree. I would go even further and say applying arbitrary functions like arctan is cheating as well. What stops me from defining a function such that it transforms 3 to whatever I want, and then applying that function. 3 should remain as 3 and then use a given set of operators to relate those 3's to produce a given number.
By changing 3 → .3
3! /(.3+.3) =10
Lol😆😆😆
@@PavelSTL It's certainly because I'm not native speaker but your sentence "What stops me from defining a function such that it transforms 3 to whatever I want, and then applying that function." doesn't make any sense.
@@PavelSTL Oh ok thank you I understand now
Yea... if it was, I would do
sqrt(3×3)/.3 = 10
who else didn't know the term subfactorial exist until this day?
I knew
I do not understand!
I didn't and I still don't
I didn't knew about ! in the first place, but it's actually very simple
ME! Didn't even know 'factorial' until today... got 3! + 3! - !3 = 10 after the 'sub' revelation. Loving these videos
The other eleven relationships are:
☆ncr(3!-sgn(3) ,3)=10
3×3+sgn(3)=10
3!+3+[Ln3]=10
3×3+[arcsec3]=10
[Ln(3+3)^3!]=10
((The symbol [ ] is the correct bracket or component function.))
Also, the following six relations are based on the number of divisors of the numbers:
3!+d(3^3)=10
d(3^(3×3))=10
3^d(3)+sgn(3)=10
(3!-sgn(3))×d(3)=10
ncr(3!,3)÷d(3)=10
npr(3!,d(3))÷3=10
In the above relations, d(n) is the number of divisors for natural number, n.
We begin by writing the number as a product of prime factors: n = p^a× q^b× r^c...
then the number of divisors, d(n) = (a+1)(b+1)(c+1)...
To prove this, we first consider numbers of the form, n = p^a. The divisors are 1, p, p^2, ..., pa; that is, d(p^a)=a+1.
According to the law, we must act in such a way that we only use the number 3, 3 times, and therefore we can not use decimal, non-3 and radical power, which is a kind of power.
you forgot this -> (3! - 3) / .3 = 10
and these
-> 3 x 3. ͞3 = 10
-> (3/3)^3 = (1+0)
-> Finally only using : + ( )
.
.
.
.
.
.
3 = (1)(0)(3)+3)
Thank me later.
Who asked?
The set of allowed operators really should've been defined before asking for solutions. The subfactorial thing for 10 makes it seems like it was a trick question. Everyone knows ! as the factorial, and hardly anyone knows about the subfactorial, so those who do might think you only mean to use that factorial. Hell, without a defined set of operations, I could define my own operation ? that just brings everything to 10: 3?+3-3=10. Done!
I mean, because of comments like yours I can learn some new set of operations, so I think it's valid to not define it beforehand
I was a math major in college, and I never once saw (!n) used for subfactorial n as defined in this video.
You just defined the Termial operation en.wikipedia.org/wiki/Termial
And you also can use it as well!
Damn man, you killed me with that 3?
For "0", you could just do (3 - 3) * 3.
or (3-3)/3 too
-3+3x3
@@user-hf6vy8xc4i No cuz that is -3+9=6
@@PlanesAndGames732 ok (-3+3)x3?
@@user-hf6vy8xc4i makes sense
I see in a lot of the solutions for 3 3 3 = 10 that the square root was taken. This is the same as raising to the power of 1/2 which leads me to assume that I can raise a 3 to any power I want...like 0. this means I can say (3 x 3) + 3^0 = 10
that's right
Я решил так же
You cannot do this, because it uses a new 0, but you could take the square root an infinite number of times maybe, which amounts to ^0 as well.
This is a fantastic interpretation of the problem and I actually thought something similar that sq root means ^1/2 which albeit an operator, who's to say you can't do 0th power of 3?
ONLY +, !, and () - u didn't solve
The derivative solution for 10 was really cool
3*3+sgn(3) = 10
Simple and elegant. Since we can use arctan, using signum fucntion should also be allowed.
nicely found!
@@lucasmartiniano6915 Thank you.
Old memes: on the Pentium, 3+3+3 = 10 for very large values of 3
The good old days, when you were the envy of the engineering class when you had a brand new Pentium.
If we're going the CS method (and he did set the precedent with the
That was quite tricky!
I made 10 using the !! (double factorial), which is defined as the factorial but counting two by two, for example:
2!! = 2
3!! = 3x1
4!! = 4x2
5!! = 5x3x1
6!! = 6x4x2
and generally n!! = n(n-2)(n-4)... until you reach 2 if n is even or 1 if it's odd.
With this in mind, combining ! with !! we can do
(3!)!! / 3 - 3! = 10.
I wasn't able to use only + ! ( ), though :-D
Nice! I have been searching to put the double factorial in one of my videos too (perhaps I'll do it for my next calculation puzzle ;). I came across the double factorial in the derivation of the the Wallis product formula for pi, but that video was nearly 12 minutes so I didn't include that topic there. For those curious, here's an amazing formula for pi!
Wallis product formula for pi
ruclips.net/video/EZSiQv_G9HM/видео.html
@@MindYourDecisions you should do more videos like that one. I really like them :)
That’s how I did it too😂😂😂
I didn't know that, I was using double factorial like Google interprets it 3!! = (3!)! = 720 😂
Hmm, I don't know if it's valid but what about (3!-3/3)!!!?
If we follow the convention used for hexadecimal number representation but use base twenty seven then 3 ( 3 ( 3 ) ) = 10 or 3 * 3 * 3 = 10
In base nine, 3 + 3 + 3 = 10
If we are already shifting bits in previously displayed examples, we can also choose to not limit ourselves to 10(base two) numbering systems.
10 has a disguise for every day of the week.
I mean, you don't *need* to shift bits for any of 0 thru 10
3! - 3 - 3 = 0
3! / (3 + 3) = 1
3 - 3 / 3 = 2
3 - 3 + 3 = 3
3 + 3 / 3 = 4
3! - 3 / 3 = 5
3 - 3 + 3! = 6
3 / 3 + 3! = 7
(3! / 3) ^ 3 = 8
3 + 3 + 3 = 9
!(!3) + 3(3) = 10
I don't think that change of base is fair because you need to (rather, ought to) write in your bases as a subscript
At the beginning of the video, the instructions say: “Make each equation true using mathematical operations”. So I just did:
3! / (3+3) = 1+0
Hahahahahaha
nice
3 3 3 != 10
Ez
Of course that's not allowed, but you gave me an idea if we are allowed to add symbols on the otber side.
3+3-3 = [[√10]]
because √10 is close to three
Cheating :/
EDIT: But Smart
A Better Way is (3 X 3) + !(!(3)) = (9) + (!(2)) = 9 + (1) = 10
(3 X 3) + !(!(3)) = 10
My immediate thought was (3 x 3) + (3^0) = 10.
I wasn't sure if using "to the power of 0" was a legal move, so I watched the video.
i think one of the rules of the challenge was that we cant introduce a new number (0)
@@daneshjunior3387 Oh, I know my "solution" doesn't actually work. I just thought my fellow math-minded viewers would appreciate my thought process.
@@pibb2474 understandable, have a great day
@@daneshjunior3387 however the square root function technically introduces the 1/2 number
@@suriya8857 well you don't write it out though. Maybe taking the square root an infinite number of times could introduce ^0 though similarly.
Why do none of the 10 solution seem satisfying
Yeah they seem like cheating
Try 3²+3/3
Here is one satisfiyng
3(at power 2)+3÷3=10
@@carlospereyra6144 you cant just add the number 2
try ceiling(33 x .3)
The solution for 10 really baffled me as I didn't know about subfactorial, but a great new thing to have learnt! Thanks :)
3.(3 repeating)*3=10
(3 with the line on top to make it 3.33333333...)
EDIT:
3.3... * 3 = 10
LOL
one of the best solns i've found
I also thought of this one and actually jump right into the video to check if it was there.
@@0Clewi0 the same ;)
Just add round: round (3.(3)*3) = 10. I saw that he agreed with IT rules.
Let's point out that you can also do:
floor(ln(3))+(3*3) = 10 (plenty of similar solution with logs in different bases)
and for the bonus, since you displayed the bit shift, I guess I'm allowed the same kinda cheat:
3+3+3 != 10
Hey, I have an amazing answer using Gamma function.
We know , gamma(n) =(n-1)!
So, gamma(3)+gamma(3)+3!=10
Also used gamma but the other way:
3! + 3! - Γ(3) = 10
Hey what is gamma?
Diptojit Dhar Gamma is a function where n x gamma is equal to n-1!.
So if 5 was n, then 5-1 is 4, and 4! is 24. So 5Γ (5 x gamma) is equal to 24.
*@[**04:35**]:* For malicious compliers...
• round(exp(3)) ÷ (3!) × 3
• superfactorial(3) - 3! / 3
• 3! + round(exp(3!) / hyperfactorial(3))
• √(hyperfactorial(3) - (round(√3))³)
Another one: round(exp(-ln(ln(3) - 3 ÷ 3)))
Bonus question:
3+3+3+()! = 10
Parenthesis build a zero :)
That seems to be valid, however, implied 0(zero) would still be considered an extra number that may make it invalid.
GENEIOUZZZ
But in that case you could literally do any number so pretty boring
!3×3!-!3=10
Madhusudan Singh I guess square roots and logarithms are all of the table as well then as they also imply numbers
are we allowed to use infinite roots?
3x3 + sqrt(sqrt(sqrt(.....sqrt(3).....))) = 10
Hans Berkvens cheatER
If you can express infinite series of roots without number then I guess yeah.
Use GREATEST INTEGER FUNCTION outside that .
He showed a solution using the floor function, which just means "round down." Using the ceiling function ("round up") is equally valid, so you don't even need infinities. You just need to get close and add one of those two in. :/
Very clever, IMO
By the definition of Sloane and Plouffe of Superfactorial, you can make the following operation:
3$ - (3!/3) = 10
Because 3$ = 3!2!1! = 12
But $ and - r not allowed in the bonus ques.
He has just found out another way of doing it.. The answer need not be formed only from the given three symbols.
+,(),! Only these dude
Using double factorial, another solution can be attained for 3,3,3 = 10:
(3!)!! / 3 - 3!
! is obviously the product down towards 1 (ex, 4! = 4*3*2*1)
! is the product of the integers from n to 1 with the same parity (ex, 5!! = 5 * 3 * 1 and 6!! = 6 * 4 * 2)
Therefore,
3! = 6, and 6!! = 6 * 4 * 2 = 48
So 48 / 3 - 6 = 10
:)
With subfactorial you can generate any integer. Just take log base !3 of log base sqrt(sqrt(3)) of 3. By adding a square root you can increase the result by 1.
The problem with this is that the rules are ambiguous, there's no way to verify or to know in advance if you're allowed to transform one of the 3's into an exponent, or use arctan etc...
none of the problems NEED arctan or exponents and are just there for extra solutions
I have another answer for the problem how about
!(!3)+3!+3=10
or 3x3+(!(!3))=10
But what is 6 ÷ !(!3)? Should I divide the 6 by the ! first?
Also (!(!3)+3)! -!3 = 10
or (3!)! - (3/3)!
Shady Kardz That's subtraction. Not addition.
At that point, you may as well do:
(3!)! / (3!)!!!! / 3! = 10.
(An N-th factorial does n(n-N)(n-2N)... until it becomes less than 1)
5:00
It really should be specified in the question that you can do this.
Even had I considered trig, I would've used radians because that is the formal unit for angle.
Can you use gradians too?
Now it seems you could just define any unit for angle you want and go "huhuh arctan(3+3+3)=23874 of this unit I just invented."
and when they started using floor function....
Nice,
If you allow to play with the units, you could also use a conversion so you just have to get 1 cm, then you convert to 10 mm
Glad to learn that “!” Can be used another way
Have you heard of double factorial? Yet another way to use the ! symbol, double factorial is the product of every other number below it. So 6!! is 6x4x2 = 48. This creates the solution:
10=(3!)!!/3 - 3!
=6!!/3 - 6
=48/3 - 6
=16-6
=10
English: Exclamation mark
Math: Factorial
@@Thebiggestgordon no, but good to know
@@shawnreinerdavid6848
Programming: logical operator
Same for me.
I really enjoyed this video, but I can't help but voice my concern that for many of its solutions, it _greatly_ hinges on the *lack* of rigorous definition for "mathematical operations."
In principle, I could define a new operation @(x,y,z)=N where "N" is any number I desire. Thus, what is meant by "mathematical operations" must be that which our world's mathematics community has come to accept as valid operations. This depends on history (culture at a given point in time and space), making it not only a mathematics question, but a matter of cultural trivia.
For example, one might call into question the validity of "." , "%" , "sqrt()". These operations can be said to correspond to "x/10", "x/100", "x^(1/2)". Unlike the elementary operations, these contain an implicit numerical component, not unlike my imaginary "@" function. It's only happenstance that I can conveniently use these function's properties to "extract" 10's and 2's.
Is bit-shift a valid operation? Does it matter what base I do the problem in then? Etc.
I liked the solutions by Yale. The 10° one feels like cheating to me because it's like adding a arbitrary constant, like I could say N=10/9 and do N(3+3+3)=10. Also I feel like using .3 shouldnt count because thats just an incorrect way of writing 0.3. That being said, an interesting way (imo at least) of getting 10 would be to do 3.3(with the recurring thing on top)*3
this is the simplest way to get to 10 since we already used .3 earlier.
(3!-3)/.3
True, but using the dot (decimal sign) felt like cheating to me. I don't think a dot can be considered a mathematical operation.
find a solution for 4
(3+3)-([sq.3])
then find 6
(3!)-3+3
(3!-([sq.3]))+3!
@@kingsley. what about this. (3! / 3)^2 + 3!
@@youleq3968 You can’t use 2
3:40 and 4:45: Hm. By taking the square root of 3, aren’t we slipping in a 2 that wasn’t in the original problem?
I agree if you can take the square root (1/2) then why not raise to power if 2? So SQRT is not acceptable.
@@n.gineer8102 why not make it 0= (3-3)÷(10×3)? Nodoby said stick to your side of the equation
This is also true for factorials, as you are technically adding a 2 and 1 (3! = 3*2*1)
@@dxtfx1912 I don't agree with that, as the factorial is an operation that does not require a base. In contrast, roots and powers do require a number for the base or exponent. In the case of the square-root, the base is 2. In this commonly used case, it is common practice not to write the 2 - but that doesn't mean it isn't there! But allowing that seems arbitrary as it is just a convention to eliminate writing it.
@@tiemen9095 I agree that operations that imply a quantity are mostly okay, like a natural logarithm. I’m only barely okay with square roots, because I think they’re a gateway to other fractional powers (cube roots would have my full support). I absolutely cannot, however, abide by the garbage that is dividing by 10 or 100 and pretending that adding the decimal or percent sign are “operations”. I almost stopped watching there - he is Fonzie and that decimal point is the shark.
Cos(d/dx 3) + 3 * 3. This was my initial thought after staring at that problem for about an hour.
Nice
initial?
if its not against the rules u can do: 3² + 3/3
3+3+3=10 base 9
Best answer
J G sadly, then you have an extra nine
Interesting. Persnickety, but interesting. I suppose you are correct, but I think it is a fair answer because Presh used several additional symbols and letters to generate additional answers. Presh ... your judgment?
Well done! 6 base 6 ... I missed that one.
No
4:57 WAIT. Arctan must return the value as radian...
that's what i thought, but firstly i think in ° than in radian
I've never heard of "subfaktorial" neither did any of my calculators
@@somebody700 I wouldn’t call it uncommon. If you follow mathematics, it’s used quite a lot
@@somebody700 Bruh my apologies. I read it as factorial, not subfactorial. Yeah this is the first time I’ve seen subfactorial
we are taught to use this in School [ 12th grade ]
One 3 equations:
0=!(!(!3))
1=!(!3)
2=!3
3=3
(Radians) 4=ceil(sqrt((3!)!°)
(°) 5=floor(sqrt(tan(acos(3%))))
6=3!
7=floor((3!)!%)
(°) 8=floor(sqrt(atan(3)))
(°) 9=ceil(sqrt(atan(3)))
I will think of more
And you could just use ++ to add one.
Also would doing -bit-wisenot(x)=x+1 or -infinity?
also, using subfactorial you can do
!(!3) + 3(3)
Okay but the 《 method was kinda bs
So was percent or moving the decimal point. Those aren't mathematical operations.
If you come across some programming language, you may convince yourself that there really is a
@@DerMarkus1982 no i believe that, it's just a bit weird that your use that for maths when it's meant to be for programming
The statement of the problem was ”make each equation true using mathematical operations”, not all mathematical functions. A mathematical operation is +, -, *, /, but arctan(), !, are functions. The correct statement has to be ”make each equation true using all mathematics you know”. So, come back with a correct statement of the problem.
U mad kid
Buttered side down(3+3+3) = -9
A function by me that does the opposite with the mathematical symbols like × now works like ÷
÷ works like × + works like -
- works like +
Your left shift operator,
Well, getting to 30 makes getting to 10 pretty easy, as well.
How do you get to 30?
@@gregoryfenn1462 left shifting one place is the same as multiplying by the base, right shifting is the same as dividing by the base. So base 10 left shifted is the same as multiplying by 10.
Ahh I see. I'm used to python3 logic whereby "10 > 1 == 4". That is, the operator always works in base-2 under the hood, even if you write in base-10 numerals.
I'll mention that WolframAlpha evaluates the expression as 6, as seen in this link:
www.wolframalpha.com/input/?i=3%3C%3C(3%2F3)
Another solution for the =10 equation :
(√3 × √3) / .3 = 10
Thats equal to 1 lol
Oh i didnt see the . Sry
Nope that's 10
Cool , good one
Simplest solution
5:32 you can do this without using %
(√3×√3)÷ .3
Nice
3! + 3! - 3! = 6
Edit: Another solution for the above is (3(3/3))! = 0
Another solution for 3 3 3 = 0 is 3(3-3) = 0
I got to zero as follows:
(3-3)×3=0
For one:
((3-3)x3)!=1
Another one for 8:
3+3+! 3=8
I got 0 like this:
!3-3-3
Thank you to let me know about subfactorial , it is my first time I hear about it.
Provided I’m understanding subfactorials right, you can also do something more convoluted like !(!3) + !(!3 + !3) = 10
!3 gives you 2, so it simplifies to !2 + !(2 + 2)
The subfactorial of !2 is 1, & the subfactorial of 4 is 9, so the math checks out.
Thanks for letting us know about subfactorials! I'm sure its gonna help me in the future!
Dude.. i'm 1 year from the future and it's still useless. 😥
Dude I'm 2 yrs from the future and it's still useless 😢
Am I going insane, or did you already do this?
Or was it with the 6's?
I was thinking the same but in the end it was all numbers from 0 to 10 that equals ( like 0 0 0 = 6 or 4 4 4 = 6)
It was with 6s. Or some other number, but this is the first with 3s 👍🏼
C
I think that it is easier to do it with 6 than it is with 3.
Which question should we answer first?
Since you once used a decimal point, I just came up with: (√3*√3)/.3 = 10
I came up with a solution for all of them. Make it 0=(3-3)×(10×3) easy. They didnt say stick to fhe left side
How about this answer:
3 = 3 < 3 =10
3 = 3 is TRUE = 1
3 < 3 is FALSE = 0
Would work for 3 = 3 > 3 = 10
I found 3x(3.3) with a repeating bar over the last 3. Another solution is any calculations gives a whole number greater than 1... in that base (ie., 3+3+3=10 base 9) although one could argue that this also involved a 9.
I could ruin your "solutions" by claiming that n! (factorial), !n (subfactorial, did you call it?) and square root are functions, not operations.
good point. it wouldn't ruin this though:
( 3 + .3 ̇ ) * 3 = 3.33333333333 * 3 = 10
I found the answers that used
If you accept ln and exp, then [3 + exp( -ln (3) )] x 3 = 10,
I find my solution more pretty than all the other solutions since it uses only usual functions and no strange stuff such as "degrees" or "percentage" or "derivates" or "subfactorials"(but I guess I'm biased)
log(3) root (3! - 3) = 10
Your solution uses functions that are defined by an additional numbers. Exp(x) is literally just e^x but as its own function instead of general power, which means that you kinda use additional number. Ln is log with base e, meaning that you kinda used additional number there as well.
So, while apparently the use of such methods is allowed, your solution is not prettier than for example using subfactorials.
f(3, 3, 3) = 10 where f is defined as the set of all functions that can use three 3’s and massage them according to Talwalker rules to get 10.
for '3', I came up with ³√(3³).
You can also get 0 Like this:
3x(3-3)
😁
For the regular 10 answer, I used logarithms in a weird way.
3!+(ln(3)/ln(sqrt(sqrt(3))))
Brandon Ignacio Wow, that's cool))
IMO, very good way 😁
Using square roots seems like a cheat to me, since there's implicitly a 2 in the symbol. If we can use square roots, then why not use any nth root?
Personally, I don't like square-roots and decimal points in problems like these because it's just a sneaky way of introducing 2's and 10's into the arithmetic.
Completely agreed.
10:2=5.
3×3.(3)=3×3.3333333..=3×10/3=10 I think this is the easiest 😁
@Nicholas Frieler yeah, you just show us how to put that bar over the 3 on a youtube comment and we'll admit you're right. Until then, all of the thumbs up for his comment knew exactly what he meant and that the bar over the three was implied.
Idk 'bout that guys i learn that way in school as I wrote down 🤔
Well, that depends on where you are, doesn't it? According to en.wikipedia.org/wiki/Repeating_decimal#Notation there isn't a universally accepted notation, and there are places where the standard practice is to use brackets.
@@NecRock Thank you for that, I didn't know this stuff 'til now.. and yeah I live in europe.. so here this is a legal and good solution I guess 😅
@Nicholas Frieler you would be surprised, but math notation is not universal across different countries. In most European countries a number in parentheses after a decimal point (or more often comma) is repeated infinitely.
For a non-super-hard bonus expression to produce 10, I came up with the following that uses the ceiling function instead of the floor function:
ceil(sqrt(33*3)) = ceil(sqrt(99)) = ceil(9.9498...) = 10
floor(33/3) = 10
Considering one of the 3s can be used as a power, (!3)^3 + !3 = 10 is also a solution. Btw i watched this video for the first time around the time it was released. I was about 9 at the time, and even though i would only understand about 80% of what you said, I watched every single video you released. Now im 13 and I just started rewatching you. I would like to thank you for being so consistent, posting regularly and teaching so well.
If I can take the square root, does that mean I can square as well?
Maybe not because sqrt can be written without 2 but square of a number needs the digit 2 in the notation
@@mysteryGuySaysHi the 2 is there, it's implied. the root is the operation, not the root base. if you're okay with implied numbers, then an easy solution would be log((3!-3)/3)=10 as a log without a base implies it's a 10.
@@johannesvahlkvist yeah but that's how they considered it
I came up with what I think is a fairly elegant solution to the 10 problem, though admittedly not as clean as the double factorial someone else mentioned.
3• 3.3... Eh? Eh? Is making 3 a repeating decimal allowed? If you can do a percent I feel like that’s fair.
3 x (3+1/3). If you can take square roots then you can take inverses.
Yes, but that adds an explicit "1" to the three "3"s
We can debate until the cows come home on whether there is a "2" in a square root symbol. My opinion is no. You are welcome to have a different one.
put a dot above the second 3. (3+3(with dot))*3 (or, in other words, 3 plus 0.333... times 3)=10. Problem solved.
3+3+3 != 10
!= is does not equal in various programming languages...
Request is specifically to make each quation true with equality - make it unequal is trivial simple (though a bit out of the box)
Great video! I really enjoyed it.
I came up with another one for 3 3 3 = 1, but it is not creative in any way. It's 3! ÷ (3 + 3) = 1. I just wanted to share it, in case anyone is interested.
Continue the wonderful work!
As with the six challenge, if you allow log, exp and sqrt it's possible to construct a function that takes a triplet of any real number and returns 10
If we can use decimals then (3.3 with a bar over 3 ) *3 = 10
3.3 with a bar means 3.33333333333333333333...........(till infinity )
That is as close as can be to 10 without being equal to 10.
I’d think the easiest way to do this is by drawing a single line.
3 3 3 ≠ 10
It said in the beginning "using the equation symbol"
and for 3 you can do 3=3=3=3
My solution, using the symbols of the C/C++ programming languages:
(3+3+3)++
The result is 10. Without any questionable "subfactorial"...
The ++ symbol is the symbol of incrementation in those programming languages. Valid, and frequently used symbol.
How about 𝑝₍₃₊₃₎ - 3 = 10. Because the 6th prime is 13, then subtract 3 makes 10.
Chad Faragher What is the "p" here? 😁
@@elinaevans5803 is a function that gives the nth prime number
Just a Random Name Ooh, I didn't know this)) Thank you 😄😘
Chad Faragher Then it's cool method, i think))
@@elinaevans5803 It could also be 3+3=10 base 3!, it's another solution
What about (3!-3)/.3 = 10
Which is 3/.3 = 10 ?
He used the ".3" for 7 so I guess that counts..
the first thing that popped into my head was putting a dot in front of the 3, and the second thing was putting a bar over top of it... so my 10 = 3.333... * 3
@@ifatreefalse lol! I don't really like the decimal point being used, but your idea is so creative I just had to say nice work
I've found an interesting solution for 10 using the sign function :
Note that sgn(3) = 1, since 3>0.
Thus, we obtain 3! + 3 + sgn(3) = 10
Or just 3x3 + sgn(3) = 10
3.3 × 3 = 10
10 ÷ 3 = 3.3
I forgot sgn exists.
@@arifyesehehehehhewahahahah3445 better so say "3.3... x 3"
Your unrivaled crowning achievement is your presentation platform. I would pay a large sum to know how you do it. Highlight, erase, disappear, reappear abracadabra
(3-3)/3=0 also works, as 0 divided by anything is 0.
3x3+3^0=10
@@factorizando8729 u used a 0
Sure, but that's not adding anything to what was already there. :D
(3-3)x3
Another solution for 3 3 3 =10
[d(3)/dx]! + 3 × 3 = 10
Oh ...
But this is in the video...
Another solution for 1 is
(3•(3-3))!
That would be 3
Rogue Ninja Why would this be three?
3 minus 3 equals 0 (inner brackets)
3 times 0 equals 0 (outer brackets)
0! is defined as 1, as far as I know
@@ThomasNimmesgern theres no factorial though
@@daytonjones05 then read again lol
No Punt Intended factorial is !
It can be solved without using sqr root and subfactorial; {log3}+3×3 where {} is a least integer function.. or [log3]!+3×3 where [ ] is greatest integer function (floor function)
First time I've heard of subfactoral.
There are Hyperfactorial, superfactorial and more as well :)
Good isn't it, we are learning many things for Presh
I wasn't even ready to use √3 since it's basically 3^(1/2) but using .3 and 3% is on another level. Hard to say if it's cheating or creativity 🤪
Square root for three and .3 aren't legal. 3%... Maybe.
I find this is the simplest solution to 3 3 3 = 10 outside the Crazy hard bonus (your challenge):
( 3! - 3 ) / .3 = 10
This is my favourite solution as it uses the least arbitrary and most natural operations.
This was the solution I came up with too. Can’t believe he didn’t include this in the video, I thought all the other solutions for 10 were kind of cheating. 60 degrees? Percentages? Rounding down?! This solution is definitely the most elegantly simple
5:40 I'm thinking returning degrees is impure because you'd have to divide by degree which is 180/pi and and percents are impure because they're kinda sorta technically zeros but to get 1 from only a 3, if you're allowed to use floor, I'd use -floor(cos(3)) or (floor(sin(3))! or even ceil(tanh(3)) or floor(ln(3)) etc etc.
These aren't really logic puzzles or even maths challenges - they're more "how much advanced mathematical techniques and standard answers do you know?".
You need both though, but advanced mathematic knowledge is quite an advantage.
What problem is not about the math techniques and standard answers you know?
[crumples paper and jumps out window]
[remembers window is at ground level when I get stepped on]
[pretends I meant to do that]
Zom Bee Nature 😂😂😂
Any time that I feel like I'm smart, I can just watch one of these videos and quickly get reminded that I very much am not.
You can solve 8 and 10 by squaring 3. A square of a number can be used as well because it does not serve as a number but a mathematical operation the same as the square root factorials and subfactorials.