Can You Solve The Three 3s Challenge?
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- čas přidán 2. 09. 2018
- 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...
Cut The Knot www.cut-the-knot.org/arithmeti...
Can You Solve The 6s Challenge?
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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:
czcams.com/video/6acBMe0Ugy8/video.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
@@mnfen9792 Webprogrammers, maybe
The U.S. 5
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
Use the zero power. Anything to the power of zero always equals one, so you can just do
3 x 3 + 3º and 3º equals 1
You can't use other digits than 3, and 0 is a digit
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
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.
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!
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. 😥
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
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
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?
@@thuglife1219 i did
The solution for 10 really baffled me as I didn't know about subfactorial, but a great new thing to have learnt! Thanks :)
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
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
The derivative solution for 10 was really cool
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
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
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
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
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
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.
One way that I had in mind for (3 3 3=10) is by using Sigma, or small sigma to be accurate (as I understood from math class, small sigma can be used to sum the number with all numbers before it up to one, like factorial but sum instead of multiply, for example: Σ3=3+2+1=6, so my answer is Σ(3/3+3), this leads to Σ4 which is equal to 4+3+2+1=10.
is this solution acceptable as I had a long disccusion with my friends that Σ is not known as I used.
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.
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?
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
300IQ
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
Haha, I came up with 3 x 3.3 (with an added dot/bar to show it's recurring) tried the bonus for a few minutes and... well now I see the answer I still like mine best ;)
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
:)
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
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)
I haven't ever known the !number before. I only knew number!, so this was soooo helpful. Thanks for this video!
Nice video MindYourDecisions!
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.
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.
You can use !! (double factorial or semifactorial) as well. Double factorial or semifactorial of a number n, denoted by n!!, is the product of all the integers from 1 up to n that have the same parity (odd or even) as n. 4!! = 2 * 4 = 8, thus, (3 + 3/3)!! = 8.
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.
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
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.
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?
I was led to your video by a short question 3x3+3=? And I wanted to know why it wasn't 4 but 10. I never got that far in math. Heck, I never heard of a pectorial or subpectorail. This is a video I would have to watch a hundred, or maybe a thousand times to get up to speed 😮 That's why I subscribed and glad I did.
Thank you for sharing
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
czcams.com/video/EZSiQv_G9HM/video.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)!!!?
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
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.
omg i love these i tried one like this that i found on this channel and it was a blast
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
4:57 WAIT. Arctan must return the value as radian...
that's what i thought, but firstly i think in ° than in radian
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)
Here's my solution for 3 3 3 = 10:
(3!)!!/3 - 3!
I personally doesn't like it that much because the double factorial is a rather uncommon operation.
(Btw it means that if the number is even then: n!! = 2 * 4* 6 * ... * n , so you multiply all even numbers until n and if it is odd then:
n!! = 1 * 3 * 5 * ... * n , so you multiply all odd numbers until n)
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
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 +
To do the 3 3 3, you can use factorials and sub factorials. To give a prior brief explanation, a sub factorial is written with an exclamation point before the term, rather than after. 3! represents the combinations with 3 values, called A, B, C. Those can be arranged in the following order:
ABC
ACB
BAC
BCA
CAB
CBA
Those are all the possible combinations with using each letter once, and is calculated by 3•2•1. Sub factorials represent derangements instead of arrangements.
DEF
def
dFE
EDf
EFD
FDE
FeD
Those are written out where lowercase letters represent times where the individual letter is in the same spot as the original position. For example, EDf from DEF means that the f is in the third spot both times, while D and E are in different places. By counting the number of full combinations with all uppercase terms, those are called derangements. There are exactly two derangements in !3, as shown above with EFD and FDE. Thus sub factorial 3 equals 2, written as !3.
By saying (!3) + (3!) + (!3) = 10, this means 2 + 6 + 2 = 10,
and last, 10 = 10. This can be solved in other ways, but this is just one example.
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 ]
Cos(d/dx 3) + 3 * 3. This was my initial thought after staring at that problem for about an hour.
Nice
initial?
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.
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
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
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
can i take the square root of 3 an infinite number of times, therefore 3^1/2.1/2.1/2... which would approach 0. So my solution would be 3^0+3x3. Would that be an acceptable answer?
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)
also, using subfactorial you can do
!(!3) + 3(3)
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.
*@[**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)))
i have a question about subfactorial, so after this video i searched up what are subfactorials and i learnt that is about how many deragements there are, but, why !3=2 and not 5? it adds the deragement of 231 and 312 but not 321 213 and 132, when i searched further i found that deragement are permutations and in the permutations it shows every single one, so why they changed it?
Because a derangement is a permutation where no element is in its original place. In the examples you gave (321), the 2 is in the middle, which is its original place in the set {1, 2, 3}. Same goes for 132 and 213, the 1 and 3 are in their original place.
@@jokernormangames6556 ohhhhh, thank you so much.
@@FlawlessM no problem!
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.
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.
I don't recall learning subfactorial in school. Thank you for that lesson.
If given a single 3, can you make any other number using only !, sqrt, and rounding down?
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)
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?
You can solve for both 8 and 10 using what I'm gonna call higher order factorials cause i don't know the name lol.
For anyone unfamiliar, normal factorials just multiply by every number below them until 1, so like 5! = 5x4x3x2x1
. But, when using a double factorial, "!!", you instead skip 1 every other value , so for example 5!!= 5x3x 1
, skipping 4 and 2
. So, you can use the solution for 4 where 3 + (3/3) = 4, and then just apply a double factorial to the whole thing
, which gives you (3+(3/3))!! = 4!! = 4x 2 = 8.
You can use the same concept for 10, but using a triple factorial on the solution for 5. A triple factorial skips 2 every other value, so for example, 7!!! = 7 x 4 x 1. Using whatever solution you got for 5 you can use the same to solve for 10. So, my 5 solution was 3! - 3/3 = 5. Now throw a triple factorial around that. (3! - 3/3)!!! = 5!!! = 5 x 2 = 10
I have no idea if this is legal but its what i was able to come up with. You people throwing in arctans and percentages are just cracked tho haha
((3+3)#)/3 = 10
Where n# is the primorial function, which is the product of all primes less than or equal to n. So 6# = 2*3*5.
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
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
Considering that square roots are equivalent to raising it to the 1/2 power, you're using a number that isn't 3. If we can arbitrarily raise to powers then this is easy. 3*3+3^0=10. I feel like there was some serious fudging with decimals, percents, and powers to get these answers.
Is pre and post incrementing (++x and x++) allowed? If so (3+3+3)++=10
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
Thank you to let me know about subfactorial , it is my first time I hear about it.
you can also use the gamma function Г the same way as the subfactorial, as Г(n) = (n-1)!
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.
5:32 you can do this without using %
(√3×√3)÷ .3
Nice
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
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!
The crazy hard bonus can be improved, since:
!3 + !3 + 3 = 7,
3 + 3 + 3 = 9, and
!3 + !3 + 3! = 10,
So we actually have 3 answers instead of 2!
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.
for '3', I came up with ³√(3³).
Can't you use the "is not equal to" symbol? (the = with a diagonal line through it) or is that not allowed?
You have to make the numbers (i.e. 3 3 3 = 10, the not equal sign denotes the challenge itself
you should see his previous vids as well since he said that there.
If u can use .3 like he said then you can use 30
(30+30)/3! Any solutions you got for “=1” might work to “=10” I u consider the 3 as a 30 (except for factorials)
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.
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
If making up new operators is acceptable then:
3 + 3 + ++3 = 10
Actually ++ is a unary C++ operator which can be used as prefix or postfix to apply pre-increment or post-increment to the operand.
What about 3^2 +3/3 ?
You can also get 0 Like this:
3x(3-3)
😁
I found the answers that used
Presh, Your notation for a list in your video, (e.g., {1, 2 ,3}) is identical to everybody else's notation for a set. Much better to say, for example, the list [1, 2, 3] has the derangements [2, 3, 1] and [3, 1, 2].
this video made me smile. thank you.
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 😁
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.
One can the sign(x) function. This is a much simpler function than acrtan and the reversed factorial. sign(3) = 1. 3*3 + sign(3) = 10.
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