Make 0 0 0 0 = 8

Question

Can you find a way to make:

$0\ 0 \ 0 \ 0 = 8$

by adding any operations or symbols? You can use only these symbols:

$+,\ -,\ *,\ !,\ /,\ \hat\, ,\ ()$.

It is limited to this list, and concatenation is also allowed. You cannot add other numbers to the equation.

@YoutRied: standard loophole: neither funny nor creative. It's usually the first answer on any math puzzle. — Thomas Weller, Sep 07 '18 at 18:40
Are you allowed to concatenate operators? (e.g !!, **, --, and ++) — Ole Tange, Sep 09 '18 at 01:09
@OleTange The accepted answer does concatenate operators. Also, the question allows it. — bb216b3acfd8f72cbc8f899d4d6963, Sep 09 '18 at 13:59

score 119 · Answer 1 · edited Sep 07 '18 at 14:01

119

A lateral thinking answer:

0! 0 0 0, because the binary equivalent of 8 is 1000 :)

edited Sep 07 '18 at 14:01

El-Guest

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answered Sep 07 '18 at 13:44

let_the_coding_begin

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I like this! very direct and minimal. – Ruadhan2300 Sep 07 '18 at 14:25
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My favourite one! I did wonder if someone would go binary. – m-smith Sep 07 '18 at 15:07
I was just checking existing answers to see if anyone else had thought of ! In the sense of negation, and binary. So "yes" :) – Stilez Sep 10 '18 at 08:24
I have reached my daily voting limit (DVL), but Imma vote this when I can, get it to $99$, and then hope that another user actually puts another upvote :P – Mr Pie Oct 03 '18 at 07:11

score 114 · Accepted Answer · edited Sep 07 '18 at 06:42

114

I think that

$\left( 0! + 0! + 0! + 0! \right)!! = 8$.

This is because

$0! = 1$ and $4!! = 8$. Note that $\left( 0! + 0! + 0! + 0! \right)!! = \left( 1+1+1+1 \right)!! = \left (4 \right)!! = 8$.

This works and is valid because

The question says I’m allowed to use any of the following symbols in my answer, I am not restricted to using $!$ as an operation only.

edited Sep 07 '18 at 06:42

Laurel

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answered Sep 07 '18 at 00:40

El-Guest

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Wow, this was fast! – SteamCode Sep 07 '18 at 00:43
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for those who are confused, !! is a mathematical operator that gives the product of all positive integers upto the argument which have the same parity (odd/even) as the argument, its called a semifactorial. (thus 4!! is 2*4 = 8) – Nobody Sep 07 '18 at 07:16
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@casualcoder most people assume $n!!$ for an arbitrary non-negative integer $n$ is the same as $(n!)!$, but it's not. That brings up some confusion, so I usually write $n!_k$ to denote $n$ with $k$ factorials, but not on this site. – Mr Pie Sep 07 '18 at 11:49
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Interestingly, use of ! works just fine if you're a coder too. since it means "not". and in binary boolean, !0 is 1 because !false = true. But it wouldn't scale up as well because !!0 is just 0 again. – Ruadhan2300 Sep 07 '18 at 14:24
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@casualcoder Google disagrees with Wolfram on this. – user1717828 Sep 07 '18 at 14:46
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Google disagrees with Wolfram on a lot of things. When it comes to math, chances are Wolfram is the correct one. – Alex Jones Sep 07 '18 at 23:03
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@user1717828, user477343’s comment above explains why Google’s interpretation is not correct in this context. – El-Guest Sep 07 '18 at 23:09
@AlexanderJ93 unfortunately though, Wolfram (and also Google) is wrong if you do !!!. It interprets it as (!!)! rather than as triple factorial (multiple of positive integers i≤n where i mod 3 == n mod 3). – Jim Cullen Sep 10 '18 at 11:47
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Dang, @user477343 I’m so close to 100 on this one haha! – El-Guest Apr 09 '19 at 02:42
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*SOMEONE UPVOTE @El-Guest's ANSWER!!! GET IT TO 100+ VOTES!!!* – Mr Pie Apr 09 '19 at 03:04
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@user477343 OH MY GOODNESS you are incredible, my friend!! Thank you thank you thank you!! – El-Guest Apr 09 '19 at 11:40
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@El-Guest no problemo! Glad to help :P – Mr Pie Apr 10 '19 at 00:56

score 75 · Answer 3 · answered Sep 07 '18 at 08:43

75

$0 + 0 + 0 + 0 ~~!\!=~ 8$

because

$ !\!= $ is an alternative way of writing $ \ne $.

answered Sep 07 '18 at 08:43

Teemu Piippo

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This is the answer! – Sep 08 '18 at 02:20
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@TheodoreWeld nope, this question is tagged [tag:mathematics], and you have to use a slash similar to Yout Ried instead of this – phuclv Sep 08 '18 at 04:24

score 64 · Answer 4 · answered Sep 07 '18 at 06:25

64

Lateral thinking!

$$0+0+\substack{0\\0}=0+0+8=8$$

answered Sep 07 '18 at 06:25

ə̷̶̸͇̘̜́̍͗̂̄︣͟

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First I thought "how does picking 0 out of 0 give you 8 combinations", then I got it. – Fabian Röling Sep 07 '18 at 12:01
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Looks more like vertical thinking to me. – Evargalo Sep 07 '18 at 12:02

malioboro · Answer 5 · 2018-09-08T15:29:56.237

39

let me try:

$0! \Vert 0 - 0!-0! =8$
$10-1-1=8$

$\Vert$ is a concatenation operation

edited Sep 08 '18 at 15:29

answered Sep 07 '18 at 00:43

malioboro

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good solution but in this case you cant use '[' or ']' – Nobody Sep 07 '18 at 07:13
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@casualcoder It was explicitly allowed. If this wasn't meant to be a valid solution, why even allow it, as 00 = 0+0 = 0? – Zizy Archer Sep 07 '18 at 07:51
@ZizyArcher your right, I missed a line below the allowed symbols which said concatenation was allowed. – Nobody Sep 07 '18 at 07:55
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I've never used $[\ldots ]$ as an operator before, let alone for what you use it as in this answer. I usually use it as a set notation; i.e., $[n]={1,2,\ldots n}$ :P – Mr Pie Sep 07 '18 at 11:53
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I think "concatenation allowed" in these puzzles usually means that, for example, 2 8 7 can be 28 ÷ 7 = 4, but not (2+8)7 = 107 – Chronocidal Sep 07 '18 at 14:22
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A commonly used symbol for concatenation is || – kasperd Sep 08 '18 at 12:15
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@Chronocidal What you are saying is probably true about most of these puzzles. But in this particular puzzle where all the digits are 0, that interpretation does not make a lot of sense. To me this answer feels more legitimate than the accepted answer. That's because concatenation is explicitly allowed, but the !! is not explicitly allowed (though it may be implied). – kasperd Sep 08 '18 at 12:20
@kasperd wow thanks for telling me, I'll edit it soon – malioboro Sep 08 '18 at 15:27

score 33 · Answer 6 · answered Sep 07 '18 at 14:46

33

It's different:

$\,++$
$0\;\;\;0$
$\,++$
$0\;\;\;0$
$\,++$

An ASCII art $8$ using only four $0$'s and $+$'s.

answered Sep 07 '18 at 14:46

JMP

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score 30 · Answer 7 · answered Sep 07 '18 at 14:04

30

$((0!+0!)^{(0!+0!)})!!$

Evaluation:

$((0!+0!)^{(0!+0!)})!!$
$\rightarrow ((1+1)^{(1+1)})!!$
$\rightarrow (2^2)!!$
$\rightarrow 4!! = 8$

answered Sep 07 '18 at 14:04

u-ndefined

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How did I not think of that?? DVL16 :\ – Mr Pie Oct 03 '18 at 07:12

score 18 · Answer 8 · answered Sep 07 '18 at 22:48

18

0 + 0 + 0 + 0 = !8

because

In C/C++, ! refers to the logical not operator, where all non-zero values become 0, and 0 becomes 1.

answered Sep 07 '18 at 22:48

l k

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I think this should be "the binary not operator". – Raimund Krämer Sep 11 '18 at 08:17
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@RaimundKrämer ITYM unary not. C99 calls it the "logical negation operator" and files it in the chapter "Unary operators". – Jens Sep 11 '18 at 15:56
But !8 can be a subfactorial of 8. – rus9384 Sep 12 '18 at 11:39
@Raimund Krämer, In programming, we have logical operators and bitwise operators. Logical operators (!, &&, || in C) operate on boolean values (true and false). Bitwise operators (~, &, |, ^ in C) operate on bits (0 and 1) of numbers. Calling either set binary operators would be confusing, as both set work on binary values. – ikegami Sep 13 '18 at 11:12
@Raimund Krämer Also, "binary operator" can mean any operator on 2 values (as opposed to a unary operator such as ! or ~) – l k Sep 18 '18 at 06:23
Thanks for clearing it up. I always thought binary operator means the same as bitwise, as opposed to logical operators which work on boolean values (independently from the technical representation of those boolean values). It makes sense that it is instead a counterpart to unary and ternary operators. – Raimund Krämer Sep 18 '18 at 07:00

rrauenza · Answer 9 · 2018-09-08T19:31:51.910

15

It's just a matter of perspective ...

0!/0 + 0!/0 = ∞

My reasoning....

0/0 is undefined so we have to first change the 0's into 1's with 0!
(...and why did you write the infinity symbol sideways in your question?)

edited Sep 08 '18 at 19:31

answered Sep 07 '18 at 18:20

rrauenza

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x/0 isn't infinite, though... – Adam Smith Sep 08 '18 at 17:49
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Lateral thinking was yesterday. Vertical thinking is the new kid in town! – Jens Sep 11 '18 at 15:59
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x/0 == +infinity, per ieee 754 – j__m Sep 15 '18 at 12:40

score 11 · Answer 10 · answered Sep 07 '18 at 12:35

11

$$[+!0]+[0]-!0-!0$$

Works in JavaScript. Hit F12 and type this into the console (This equation editor uses "−" instead of "-" so copy and paste doesn't quite work). Otherwise, it works in the same way as @malioboro and @Arnaldur's answers.

In fact, you can make any JavaScript application run just by using a combination of 6 characters, which is what inspired me to make this. I substituted +[] for 0 when asking JSF**k to do 10-2.

answered Sep 07 '18 at 12:35

Namyts

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I like this abuse of javascript a lot. !0 for true, +!0 for 1, [1] for array containing 1, [1]+[0] for concatenate two arrays to get the string "10" ; subtract 1 from that twice to end up as 8. Love it. – Ross Presser Sep 08 '18 at 04:10
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but the question is tagged [tag:mathematics] and not [tag:programming] – phuclv Sep 08 '18 at 04:21
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Who says you can use square brackets.... – Sep 08 '18 at 06:57

score 11 · Answer 11 · answered Sep 08 '18 at 06:39

11

$0 + 0 + 0 + 0 \equiv 8$
Adding the symbol $-$ over the equals sign makes it a congruence sign. Considering the congruence relation, we must be working mod N, where N divides 8.

answered Sep 08 '18 at 06:39

Vaelus

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score 8 · Answer 12 · answered Sep 07 '18 at 05:38

8

$concat(0!,0) - 0! - 0! = 8$

becomes:

$concat(1,0) - 1 - 1 = 8$

and finally:

$10 - 2 = 8$

cool puzzle!

answered Sep 07 '18 at 05:38

Arnaldur

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Great answer! But it has already been answered (check out @malioboro 's answer). Nonetheless, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Since you are new to the Stack Exchange community, let alone this site, I strongly suggest that you visit the Help Center for more info; in particular, I suggest going here, then here for questions (not answers) :D – Mr Pie Sep 07 '18 at 07:44

David Browne - Microsoft · Answer 13 · 2018-09-07T23:53:53.453

4

Question limits the symbols, not the operations. So with the symbol + can make the operator ++.

(++(++(++(++(++(++(++0000))))))) = 8

edited Sep 07 '18 at 23:53

answered Sep 07 '18 at 21:23

David Browne - Microsoft

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I think you'd need the prefix form if you actually wanted that to work. – LegionMammal978 Sep 07 '18 at 23:48
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Doh. Of course you can't increment a literal in the first place. – David Browne - Microsoft Sep 07 '18 at 23:55
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but that symbol doesn't exist in mathematics – phuclv Sep 08 '18 at 04:20
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You need to pass an l-value to the ++ and -- operators, and neither their result or a literal is one. – Maya Sep 08 '18 at 14:35

score 3 · Answer 14 · answered Sep 07 '18 at 11:37

3

Just put the minus symbol over the first zero to give it the look of a figure 8 and use plus to add the zeros.

answered Sep 07 '18 at 11:37

Sentinel

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Great answer, but unfortunately, this is a duplicate of @TheSimpliFire 's answer. – Mr Pie Sep 07 '18 at 11:51
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@user477343 No, not quite, since this answer puts a line through the 0 instead of putting two 0s on top of each other. In this answer, only one 0 becomes an 8. – Fabian Röling Sep 07 '18 at 12:02
@FabianRöling As I read it, it uses the minus symbol to be put over the first zero and make it look like an $8$. In my imagination, I assumed the zero is the bottom circle of $8$ since there is something over it, of which is another circle $\large\circ$. I did not fully understand what this answer was talking about in saying to use a subtraction symbol, but just assumed it was just like TheSimpliFire's answer due to their similarity; and since that is a great answer, I thought this was great, too. If you understand this answer, may you please explain to me in other words if you can? :\ – Mr Pie Sep 07 '18 at 12:09
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@user477343 It uses the horizontal line of the minus sign to cut through the middle of the $0$. It is "over" the zero in the sense of overwritten. – Jaap Scherphuis Sep 07 '18 at 12:55
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Yeah. 'over' in the sense of "in front of" on a dimension between you and your monitor. – Sentinel Sep 07 '18 at 14:44
@JaapScherphuis ahhh, thank you for that :D – Mr Pie Sep 07 '18 at 20:44
@Sentinel that is a very cool way to describe that. Thank you :D – Mr Pie Sep 07 '18 at 20:44

score 3 · Answer 15 · answered Sep 08 '18 at 00:20

3

Here is an answer that doesn't use the semi-factorial or any concatenation.

$$0 + 0 - 0! / 0 = (-8)!$$
The left side is $-1/0$ and the right side is $-\infty$.

Plugging the expression into Wolfram.

answered Sep 08 '18 at 00:20

user1717828

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Division by zero is undefined (the limit of 0!/x as x goes to 0 is plus or minus infinity) and so are negative integer factorials (they are not equivalent to negative infinity). Even if they both evaluated to +∞, you can prove nonsense assuming you can equate infinities and treat like a real number. E.g., |1/0| = +∞, adding 1 to positive infinity is still positive infinity, hence |1/0| + 1 = +∞, hence |1/0| = |1/0| +1 by transitivity of equality. Subtract |1/0| from both sides and prove 0 = 1. – dr jimbob Sep 08 '18 at 18:11

score 3 · Answer 16 · edited Sep 11 '18 at 01:59

3

concatenate(0!, 0) - concatenate(0! + 0!) = 8.

Note that 0! = 1

(0!, 0) = 10, and (0! + 0!) = 2, so 10 - 2 = 8

edited Sep 11 '18 at 01:59

Alto

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answered Sep 11 '18 at 00:57

Nick

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Unfortunately concatenate is not a valid operation. – boboquack Sep 11 '18 at 01:37
Congrats, concatenation is allowed now! Good answer, but people beat you to it. Welcome to Puzzling.SE! Here's a bonus puzzle: What's the word? Hilarious - extremely amusing – Alto Sep 11 '18 at 01:47
Also, concatenation is combination. Basically, (2, 4) = 24, you get it. – Alto Sep 11 '18 at 01:48

score 3 · Answer 17 · edited Sep 14 '18 at 02:45

3

Add a - above the equals to get $0000 \equiv 8$, which is true assuming we are working in the ring $\mathbb{Z}/\mathbb{2Z}$. (Note I'm trying to avoid writing $[0] = [8]$...)

edited Sep 14 '18 at 02:45

a stone arachnid

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answered Sep 11 '18 at 15:52

Gus314

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score 3 · Answer 18 · answered Sep 13 '18 at 09:10

3

This could work too:

(0!+0+0)/0 = ∞

Explanation

(0!+0+0)/0 = 1/0 which is infinity (8 but put horizontally)

answered Sep 13 '18 at 09:10

Michal B.

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That is not infinity — it is undefined :\ – Mr Pie Oct 03 '18 at 11:33
@MrPie - We can get into calculus if you want. It can still be considered infinity; though can also be considered as though it is approaching any number so we can even go directly to 8! IEEE 754 states that division by 0 leads to an infinity, so there's that, too! – Voldemort's Wrath Jul 25 '19 at 15:19

RothX · Answer 19 · 2018-09-07T13:30:07.777

0

$00^{00} = 0^0$ which is indeterminate. In some sense, an indeterminate form can be equal to any value, since in Calculus, a function that approaches "$0^0$" can approach any real value, including $8$. So in that sense, $0^0 = 8$.
If you don't like concatenating two zeroes as "$00$", then $0^0 + 0 + 0$ also works.

edited Sep 07 '18 at 13:30

answered Sep 07 '18 at 13:22

RothX

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There are no widely accepted definitions under which your equation is considered to be true. – Tanner Swett Sep 07 '18 at 17:34
@TannerSwett Didn't think it was that unreasonable. My thinking is you can construct a function $f(x)^{g(x)}$ where $f$ and $g$ both limit to zero at some point $c$ such that $f(x)^{g(x)}$ limits to any real number at $c$, including $8$. Admittedly, writing $0^0 = 8$ is basically mathematically false. It is indeterminate for the reason I stated above. But this is Puzzling SE, not Mathematics SE, so I figured it was ok to think outside the box and not be formal with math. – RothX Sep 10 '18 at 13:23
Yeah, I dunno if I'd call it "that unreasonable". But if I saw a student say that the equation $0^0 = 8$ is true, I'd think that they probably misunderstand what an indeterminate form means. If I saw a math professor say the same thing, I'd ask them if they could please be a little more rigorous. – Tanner Swett Sep 10 '18 at 13:59
@TannerSwett Yeah, I completely agree. I'd never write $0^0 = 8$ in a mathematical setting, but since this is a puzzle, not a math problem, I thought maybe I could stretch a bit. – RothX Sep 11 '18 at 01:58

Eleftheria Chatziargyriou · Answer 20 · 2018-09-07T14:41:40.160

0

$ ((0! + 0!)\$)*(0! + 0!) = 2^2*2 = 8 $.

Further explanation:

The $\$$ operation denotes the superfactorial defined as : $ n\$ = (n!) \uparrow \uparrow (n!)$.

edited Sep 07 '18 at 14:41

answered Sep 07 '18 at 14:28

Eleftheria Chatziargyriou

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Good idea, but make sure to read the question fully. It gives you a list of what operations can be used and $ is not one of them. – Sensoray Sep 07 '18 at 15:26

score 0 · Answer 21 · answered Sep 08 '18 at 08:09

0

If you turn the problem around

answered Sep 08 '18 at 08:09

Florian F

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Similar to @rrauenza's answer – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 08 '18 at 08:11
Yes. A different way to express the same idea. – Florian F Sep 08 '18 at 09:13

score 0 · Answer 22 · answered Sep 08 '18 at 19:44

0

Similar to @Vaelus

$0+0+0+0 \leq 8$

Explanation

You can get the $\leq$ by adding a $-$ inclined on top of the $=$

answered Sep 08 '18 at 19:44

villasv

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The goal is to strictly make $0,0,0,0=8$ and not to "make the statement true" (therefore allowing this answer to be acceptable) :\ – Mr Pie Oct 03 '18 at 11:35
When you say "the goal is to make [math expression]", then you're under the influence of the interpretation of that mathematical expression. IMO, "make the statement true" is the one and only interpretation to be taken, while you could argue more strict rules about not manipulating already existing symbols. – villasv Oct 03 '18 at 13:18
I see, you have a point. I should've said that my comment is solely based on how I interpreted the question, possibly explaining why your answer was sadly downvoted. I am positive that there exists another answer that manipulates the equation in a similar way that you have. Sorry if I sounded mean. You can get a $(+1)$ but in at least $2$ hours, once my daily voting limit (DVL) is over :) – Mr Pie Oct 03 '18 at 21:00

score 0 · Answer 23 · answered Sep 10 '18 at 16:55

0

As per the list of allowed symbols we are clearly allowed to use "$,$" and "$.$"
This is doubly evident as otherwise how would we use the $\mathbb{concatenation}$ function without a comma to separate the arguments?

So the solution is:

$0! - \mathbb{concatenation}(., 0 + 0! + 0!!) = .8$
$1 - \mathbb{concatenation}(., 0 + 1 + 1) = .8$
$1 - \mathbb{concatenation}(., 2) = .8$
$1 - .2 = .8$

answered Sep 10 '18 at 16:55

SamYonnou

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You may simplify 0!! as 0!. – Cœur Sep 11 '18 at 15:28
I felt like $0+0!+0!!$ looked more elegant/purposeful as if it was part of a sequence like $a(n, m) = a(n-1,m)!; a(0,m)=m$ – SamYonnou Sep 11 '18 at 16:58

score -1 · Answer 24 · answered Apr 09 '19 at 23:59

-1

If the order can be changed:

(0 0! 8)0 = 0!

(0 0! 8) is a permutation.

answered Apr 09 '19 at 23:59

Artemmm

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$0 * 0 * 0 * 8 = 0$ would be much simpler, though no more a solution to this problem than yours. I don't think it's reasonable to think reordering is allowed. – Rubio Apr 10 '19 at 00:09
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nice try, but seems that this is not the kind of answer the OP was looking for. beware of downvotes, yet keep up your hard work and happy puzzling :) – Omega Krypton Apr 10 '19 at 01:49

Make 0 0 0 0 = 8

24 Answers24

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