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0 0 0 = 6
Using basic arithmetic operations ( + , − , × , ÷ ) , parentheses, and any other operations on the left side of the equal sign, can you make this equality hold true?
Note : Some of the other operations you might try are the trigonometric functions ( sin ( ⋅ ) , cos ( ⋅ ) , and tan ( ⋅ ) ) , factorial ! , floor ⌊ ⋅ ⌋ , and ceiling ⌈ ⋅ ⌉ . You can put in as many operations as you like, but you can't put in any additional numbers or digits!
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58 solutions
What a great solution.It made me smile!!
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Thanks :) Check out the other solutions, there are so many ways to solve this
I used this solution as well, but truth be told, I have NO idea why 0! = 1. Anyone know why?
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Since we have P ( n , r ) = ( n − r ) ! n ! P ( n , n ) = ( n − n ) ! n ! c c c c c c c let n = r n ! = 0 ! n ! 0 ! = n ! n ! = 1
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None of that makes any sense to me. All you are doing is demonstrating that you know it works and not explaining why it works !
You should definitely see this
How is 0!=1??? Can anyone explain???
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If you have 3 distinct items, there are 6 distinct ways to arrange them: ⟨ 1 , 2 , 3 ⟩ ⟨ 1 , 3 , 2 ⟩ ⟨ 2 , 1 , 3 ⟩ ⟨ 2 , 3 , 1 ⟩ ⟨ 3 , 1 , 2 ⟩ ⟨ 3 , 2 , 1 ⟩ If you have 2 distinct items, there are 2 distinct ways to arrange them: ⟨ 1 , 2 ⟩ ⟨ 2 , 1 ⟩ If you have 1 distinct item, there is only 1 distinct way to arrange it: ⟨ 1 ⟩ Likewise, there is 1 way to arrange 0 items: ⟨ ⟩
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But I thought factorial meant to multiply down from the number itself i.e. 5! = 5x4x3x2x1 therefore 0! = 0. But factorial is just the number of ways to arrange a set of n things?
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@Clover Ross – You should definitely see this
@Clover Ross – 5 ! = 5 × 4 × 3 × 2 × 1 4 ! = 4 × 3 × 2 × 1 3 ! = 3 × 2 × 1 2 ! = 2 × 1 1 ! = 1 0 ! = ( e m p t y p r o d u c t ) If you multiply together no numbers, you don't get 0. If anything then you get 1, since it's the multiplicative identity, just as the sum of no numbers is 0, the additive identity.
Why 0!=1? Well that's the definition of factorial.
Solved it the same way you did.
This is so confusing. How can you get three 0's to equal 6. It's the same as saying 0+0=2. It doesn't make sense. Please explain so I can understand because my brain can't wrap around that concept.
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look carefully at the solution.he proved that 0 ! = 1
now, if we do, 0 ! + 0 ! + 0 ! = 1 ! + 1 ! + 1 ! = 3 ! = 1 × 2 × 3 = 6
hence,proved.
That's inaccurate. N!/N = 1!/1 but this does not mean n=1, in this case, n=0 and by that logic, n!/n is dividing by 0
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n = 1 because you have that n ! = n ( n − 1 ) ! , substituting we have:
1 ! = 1 × 0 ! 1 1 ! = 0 ! 1 1 = 0 ! 1 ! = 1 1 = 0 !
0! Has to be 0 because you multiply 0 by every number smaller than it x*0=0
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You are correct that 0! = 1 for reasons that are similar to why x^0 = 1. Both are defined that way. But there are reasons for these definitions; they are not arbitrary.
You cannot reason that x^0 = 1 by thinking of the meaning of powers as
"repeated multiplications" because you cannot multiply x zero times.
Similarly, you cannot reason out 0! just in terms of the meaning of
factorial because you cannot multiply all the numbers from zero down
to 1 to get 1.
Mathematicians define x^0 = 1 in order to make the laws of exponents work even when the exponents can no longer be thought of as repeated multiplication. For example, (x^3)(x^5) = x^8 because you can add exponents. In the same way (x^0)(x^2) should be equal to x^2 by adding exponents. But that means that x^0 must be 1 because when you multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense here.
In the same way, when thinking about combinations we can derive a
formula for "the number of ways of choosing k things from a collection
of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a
group of 5 people shakes hands can be computed using n = 5 (five
people) and k = 2 (2 people per handshake) in this formula. (So the
answer is 5!/(2! 3!) = 10).
Now suppose that there are 2 people and "everybody shakes hands with everybody else." Obviously there is only one handshake. But what happens if we put n = 2 (2 people) and k = 2 (2 people per handshake) in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the value of 0!. The fraction reduces to 1/x, which must equal 1 since there is only 1 handshake. The only value of 0! that makes sense here is 0! = 1.
And so we define 0! = 1..........................[took help from dr.math forum]
if it helps,let me know please.
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But at the same time, 0 0 is undefined. Here, we have two conflicting rules, namely x 0 = 1 and 0 y = 0 . You could think of this as similar to how x x = 1 and x 0 = 0 , and 0 0 , which conflicts and is undefined.
Indeed, you could define x 0 = x x , and then the undefinedness at x = 0 matches up. (Indeed, I would imagine that wheel theorists, if they define exponentiation (at least by a nonnegative integer), would consider that 0 0 = 0 0 on this basis.)
Furthermore, you can liken 0 ! to x 0 on the intuitive level as well. x ! is the number of permutations of x distinct items; x y (for non-negative x and y ) is the number of possible lists of y items taken from a set of size x . If x = 0 , then we cannot form such a list as there are no items from which to pick, and so 0 y = 0 . If y = 0 , then we have just one possibility which is the empty list, so x 0 = 1 . But I realise this does suggest that 0 0 = 1 ....
Thanks a lot sir for the proof of 0!
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Thanks a lot for solving my problem :)
0! is also defined as 1, primarily for nCr purposes.
I have a very basic understanding of how to use mathematical tools, and this is the first time I've seen an exclamation point. Could you break this down for me? WTH does an ! mean in math? (It doesn't mean it's in Spanish apparently)
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Hi Kenneth, the exclamation point ( ! ) means the factorial function. You can see this wiki: Factorials
A factorial means you take a number 1 or above, and multiply it with every number below it.
Ex. 5! = 5 x 4 x 3 x 2 x 1 = 120
0! is 1 by definition not by proof.
The proof is wrong. The first line of the proof is simply the definition of n! for n > 1
If n = 1 by definition 1! is 1 (the product of all positive integers less than or equal to 1) not 1x (1-1) because 1-1 or 0 is not a positive integer.
Thought the same answer
(cos0 + cos0 + cos0)!
I forgot that 0! equals 1 !!
how does 0! = 1 ?
This is also possible for 1, 2, 3, 4, 5, 6, 7, 8 and 9,.
( cos 0 + cos 0 + cos 0 ) ! = 6
Nice solution...
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Thanks. :)
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Hey are you delete your report of this problem? I updated my solution. Please check...
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@Md Mehedi Hasan – Yes. I think I am wrong in that case.
I didn't saw this before... nice solution!
Wonderful!
this makes more sense than the top one
I also thought about that :D
What does the! Stand for
n factorial (n!) means you multiply all the numbers from 1 to n. In this case 3!=1x2x3=6
i also thought that buddy
I tought about that too
⌈ arccos ( 0 ) ⌉ + ⌈ arccos ( 0 ) ⌉ + ⌈ arccos ( 0 ) ⌉ = 6
Isn't this going to be 6π or 0?
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That would be 3pi/2
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Joy Jaoude he also used a 'ceiling function' for each arccos. These bring the number put into the function to the next highest integer.
I did exactly the same.
This was my first idea as well.
I’m only 13 can someone explain sin, can, and tos for me
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Its good to see young people interested in learning themselves rather than waiting to be taught. http://www.bbc.co.uk/bitesize/standard/maths_ii/trigonometry/graphs/revision/1/ This should give you a basic understanding of the Sin(x), Cos(x) and Tan(x) graphs.
It isn't correct. arccos(0)=pi/2
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The symbol outside arccos(0) represents Ceiling function, you can have a look at Ceiling functions .Hope it clears your doubt.
This seems silly. What is meant by "any" operation?
let f(x)=x+6 -> f(0+0+0)=6. Done?
Or,
ProductOfDigits(Round(CelsiusToFarenheit(0+0+0)))=ProductOfDigits(32)=3*2=6. Done?
Of course, as indicated elsewhere, you can solve this just by using the functions listed in the question, so the answer is obviously "yes". But trying to figure out what is allowed by the rules of this question is...annoying.
What's silly about your solutions, exactly? The second one seems quite clever.
Yes, you can define "anything", but this is in the Basics as question 2, and not everyone here would know or think of that. (Especially given the current 62% correct rate.) Thinking of a creative solution is an interesting challenge in itself.
ceil(exp(sqrt(exp(cos(0))))) = 6
Works, if exp() is a function, --- as in python's math module or maybe in some scientific calculator. Does not work if e is needed explicitly as a constant. The scope of the word "anthing" is just annoyingly vague.
But as soon as we allow pythonic terms, even that would work:
floor(sqrt(ord(str(0)))) = 6
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The most ridiculous idea I've come up with so far is,
length(string(0/0))+length(string(0)) = length("ERROR") + length("0") = 5+1 = 6
:)
“Anything” is extremely vague and for that reason, this question should not be in the BEGINNERS section.
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The note is very specific with some suggestions for that reason (and there are certainly some answers that use the suggestions alone).
⌈ tan ( cos ( 0 ) ) ⌉ + ⌈ tan ( cos ( 0 ) ) ⌉ + ⌈ tan ( cos ( 0 ) ) ⌉ = 6
Nice solution, but you should use LaTeX.
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I went ahead and edited it for Paul. Note that not everyone knows LaTeX, especially newer users to the site! It's certainly worth getting the hang of though.
In degree mode, and really only needing 1 zero: ⌊ arctan ( cos ( 0 ) ) ⌋ + 0 + 0
In degree mode, and really only needing 2 zeroes: ⌊ arccos 0 ⌋ + ⌊ arccos 0 ⌋ + 0
Binairy -> decimal (0!)|(0!)|0=6
I define the operation s i x ( x , y , z ) = 6 for any x , y and z , and apply it to the 3 zeros. Done.
nice answer!
Awesome solution!
ceiling(exp(cos(0)))!+0+0
Wow! Creative use of a constant!
Great solution :)
An easy solution: (0! + 0! + 0!)! = (1 + 1 + 1)! = 3! = 6
( cos 0 + cos 0 + cos 0 ) ! = 6
(0!+0!+0!)! = 6
[(tan0)! + (tan0)! + (tan0)!]! = 6
[(sin0)! + (sin0)! + (sin0)!]! = 6
Hence the answer is y e s
Nice solutions
This is wrong.
( tan 0 + tan 0 + tan 0 ) ! = 0 ! = 1 = 6
( sin 0 + sin 0 + sin 0 ) ! = 0 ! = 1 = 6
Mine is discusting, but cos(0) is 1 tan(1) is 1.55 The ceiling function of 1.55 is 2. Do it for the three of them. 2+2+2 = 6
I did the same ;)
Define the binary operation * such that a*b = 6 for all real numbers a and b. Note that the operation is associative, among other things.
0 * 0 * 0=6.
We can use (0 ! + 0! + 0!)! =6
Another solution: ⌈ tan ( 0 ! ) ⌉ + ⌈ tan ( 0 ! ) ⌉ + ⌈ tan ( 0 ! ) ⌉
( ∣ { 0 } ∣ + ∣ { 0 } ∣ + ∣ { 0 } ∣ ) ! = 6
(cos(0) + cos(0) + cos(0))! = 6
( cos 0 + cos 0 + cos 0 ) ! = 6
The cosine of 0 is 1 therefore ((cos0) + (cos 0) + (cos0)! = (1+1+1)! = 3! = 6 - thus it can be done.
(antilog(0) + antilog(0) + antilog(0))!= (1 + 1 +1)!= 3!= 6
Consider this, we can use basic trig functions parentheses and factorials. If you run through each trig function for 0 you'll see it equals 1 for cosine. So now you have 3 1's. So what next? Add them together in parentheses. You now have 3. What operation and make this 3 a 6? Factorial. You now end up with the equation (cos(0)+cos(0)+cos(0))!=6 (1+1+1)!=6
3!=6
3x2x1=6
6=6
( cos 0 + cos 0 + cos 0 ) ! = ( 1 + 1 + 1 ) ! = 3 ! = 6
(0! + 0! + 0!)! = 6 because 0! = 1
(cos(0)+cos(0)+cos(0))!=6
(1+1+1)!=6
3!=6
(cos(0) + cos(0) + cos(0))! = 6
(cos0 + cos 0 + cos 0)!=6
0! x ceil(exp(0!)) x floor(exp(0!)) = 6
0^0×6=6 seems to be a very simple answer, or am I missing something?
You can't add any other digit...
Or a more exotic solution: ceiling(|tan(tan(0!+0!)/tan(0!))|)
( a r c t a n ( c o s 0 ) a r c c o s 0 +cos0)!=6
(Cos(0)+Cos(0)+Cos(0))! = (1+1+1)! = 3! = 6
Wasn't this part of a scam school episode?
2[cos(0)+cos(0)+cos(0)] = 6
( cos 0 + cos 0 + cos 0 ) ! = 6
(Cos(0) + Cos(0)+Cos(0))! = 6 The answer is yes
⌈ c o s h ( c o s h ( c o s h ( c o s h ( 0 ) ) ) ) ⌉ + 0 + 0 = 6
(cos(0) + e^0 + 0!)! = 6 ( 1 + 1 + 1)! = 3! = 6
Nice thinking about inserting e .
Floor(|sec(sec(cos(0))) + sec(sec(cos(0)))| - cos(0)) = 6
I may even have found a solution for two zeros:
⌈ e x p ( ⌈ t a n ( 0 ! ) ⌉ ) ⌉ − ⌈ t a n ( 0 ! ) ⌉ = ⌈ e x p ( 2 ) ⌉ − 2 = 8 − 2 = 6
Now some of you might think that the exponential function adds digits to the term, but remember that sin() and cos() are actually also defined by exp(x).
I got (ceil(tan(cos(0))^(cos(0)+cos(0)))!
(Cos 0 + Cos 0 +Cos 0)! = (1+1+1)! = 3! = 3*2 = 6
Another solution, not as elegant as the others, using radians:
ceil(tan(cos(0)) + ceil(tan(cos(0))) + ceil(tan(cos(0)))
Did some cheating using Python to test. :-P
⌈ tan ( cos ( 0 ) ) ⌉ × ⌈ tan ( cos ( 0 ) ) ⌉ + ⌈ tan ( cos ( 0 ) ) ⌉ = 2 × 2 + 2 = 4 + 2 = 6
((cos(0)+cos(0)+cos(0))!=6
(ceiling(sec(cos(0!))))^3 - 0! - 0! = 6
as, 0! =1 ->
cos(1) = 0,99... ->
sec(0,99)=1,000149... ->
ceiling(1,000149) = 2 ->
2^3 = 8 ->
8-1 -1=6
so, Yes and in the first comment there is the prove why 0! =1 Nice puzzle !
∣ ⌊ Γ ( ln ( arctan ( cos ( 0 ) ) ) ⌋ ∣ − 0 − 0 = 6
A little excursion into negative numbers just for the fun of it...
More generally, to solve x x x = 6 the easiest (if set notation is allowed) is ∣ { x , x , x } ∣ ! = 6
(cos(0) +cos(0) + cos(0) )!= 6
ceiling(tan(cos(0)))+ceiling(tan(cos(0)))+ceiling(tan(cos(0)))=6
( 0 ! + 0 ! + 0 ! ) ! = 6 3 ! = 6 6 = 6
The answer is Yes.