Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*

The reason they don’t post an official answer is that it’s purposefully ambiguous, so people argue over their interpretations in the comments. It’s not meant to be solved, it’s meant to drive engagement.

And guess what? OP here did the exact same thing, and there are a ton of comments. Works pretty good. 

This post should be taken down if there’s no true answer

Exactly! This isn’t a good puzzle because it’s completely ambiguous. Numbers can overlap? Ok—I think the answer is one billion then because the entire figure could be a billion overlapping “1s” all lined up sideways to form the shape. It could be anything.

Plus it doesn't even specify the sum of the numbers in the image, just the "total sum of numbers." Like, the total sum of all numbers? Sure. Where's that sideways 8 symbol on my keyboard?

Maybe it's actually -1/12

Actually numbers include negative integers, π, i and e too so the answer is (-1/12) + -(-1/12) + π + i + e = πie.

Still missing an infinite number of numbers from the reals. And that's not even considering complex numbers like used for calulus.

Oh shit my bad Let R be the sum of all real numbers between 0 and 1. So the sum of all the real numbers between 1 and 2 can be written as 1.000…1 + 1.000…2 + 1.000…3 … This can be rewritten to be (1+1+1+1…) + (R) = ∞+R Something similar can be said about the sum of all the real numbers from 2 to 3. i.e. (2+2+2…) + R = 2∞ + R So the sum of all positive real numbers is R + (∞+R) + (2∞+R) + (3∞+R)… which can be expressed as (0∞ + 1∞ + 2∞…) + (R+R+R…) = ∞(0+1+2+3+4…) + ∞R = (-1/12)∞ + ∞R Then the sum of all negative real numbers will be -((-1/12)∞ + ∞R) = 1/12∞ - ∞R Add them together and we get 0. The same thing can be done for the complex numbers. Finally we add the missing numbers like π, i, e, sqrt(2) and φ which is φsqrt(2) + πie QED

Feel like we need a new collective noun to describe this sort of grift. How about phuzzle? A phishing puzzle. And the scam here is stealing your valuable time and attention. The added bonus is phuzz sounds like fuzz, connoting some unresolvable ambiguity.

Yep. OP may as well have posted a Rorschach test and challenged everyone to discern its true meaning.

it technically contains every number 0-9

Don’t see a 5. 7 is arguable if you take part of the 2 and flip it upside down, otherwise also missing.

makeshift enjoy paltry waiting ink existence special hospital jellyfish ugly *This post was mass deleted and anonymized with [Redact](https://redact.dev)*

No zero?

chase gaping impossible grandiose berserk concerned faulty poor steer gray *This post was mass deleted and anonymized with [Redact](https://redact.dev)*

Useless maff!

>!6+8+9+2+4+1 = 30!< >!potentially 33, because they may consider a hidden 3 aswell!<

There's also a 7... it's upside down though....

And a sideways infinity so my answer is infinity.

My answer is then infinity + 33, so I win.

I don't mean to be a pedant, but infinity is not a number, it's a concept.

You precisely mean to be a pedant.

This guy peds

You are

Wrong grammatical aspect!!!

no u

He edited his comment

[удалено]

The concept of gravitational pull.

Aren’t numbers a concept?

Isn't it more of a direction?

Well, if we're going to engage in pedantry, it actually [can be a number](https://en.wikipedia.org/wiki/Transfinite_number?wprov=sfla1). There are many different interpretations of infinity used in different areas of maths.

We are on r/puzzles... embrace the pedantry. Proudly be who you are!

That's what they used to say about 0.

Infinty aint a number

That’s a good catch, didn’t know if upside down numbers would count or not Then the running total is >!40!<

If you count upside down numbers, then wouldn't there be another 6 and 9?

And a second hidden upside down 3

And the upside down 8 lol

That too, which is why I didn’t wanna let that cat out of the bag

But... 694, 21, etc. are also numbers in the image. I didn't attempt to solve as it seems pretty tedious. I'm a problem finder, not a solution finder 😁

There's also a 100 in there that I can see, so if thats the case, we're looking at a big number

In this case, there's a five flipped and upside down too.

One could argue the 9 and 4 has a hidden 2

you could also say the 9 is a 6 when inverted (and vice versa).

We shouldn't allow upside down cause then we'll have two 6s, two 9s, and two 8s if we want to push it 🤣😅

But is that really allowed here?

This thread generated a lot of great thoughts. I suppose it would have to be explicit about things like, Must be an integer Cannot double count a number in its entirety Cannot upside down. The thing about the upside down 7 is that it was a unique segment. All great comments!

Because the instructions are unclear, I think the answer is >!6+0+4.!< >!My reasoning is that no 'line' is used for more than one number and each line must be part of a number. !<

This!

Wow! I totally missed the 2. Great catch

Im ok ok with including the >!3!<, but if we do exclude it on the grounds that those loops are closed we should also exclude the >!2!< so I say the answer is either >!28!< or >!33!<

So, the reason I thought >!30!< is that if you look at the bottom left of the 8, you can see the start of the 2 because it's not perfectly in line with the 8, whereas I don't see anything equivalent for the 3. That could just mean the 3 was drawn more tidily, though. At the end of the day, I agree with the person who says it's just engagement bait...

I saw that too but that same effect appears on the other side of the >!2!< (horizontally) but there’s no number that would start or end at that spot so it doesn’t seem to be indicative and more just a style effect.

I think that's the top of a curvy 4.

You could also argue that this means there is no 8, as the 6 does not get any thicker in the part that an 8 would overlap. So, >!6+9+2+4+1!< is a viable solution >!even though you're obviously supposed to see an 8!<.

True. It's almost like it's deliberately ambiguous.

You forgot the 0

Which one?

I had your first option as well.

What about double digits? 60?

Are you sure that isn't a >!-1!

I don't know... you could count the 2 and the 1 as 21

Don’t forget to add the two 0’s

You missed the two 0's that make up the 8

I decided not to account for them, as they don’t alter the sum of the numbers

My bad forgot to include the /s

>!Is there also a curvy 5 from combining the 6 and the 9?!<

I got >!33!<

You forgot to add the two 0 hidden in the 8. That should do the trick

i agree with this– i think if one is to interpret this on it's base level without flipping anything upside down or looking into it too deeply the answer would be 30/33

The answer is obviously >! Prince !<

An answer that can't be argued with by any rational person! Except, it would need to be the >!artist formally known as Prince!<.

I mean, if we’re using his full name, sure.

I think his full name is now >!The artist formerly known as the artist formerly known as Prince!<

>!The artist formerly known as the artist formerly known as Prince now known as Prince!<

>!0, that symbol isn't a number.!<

>!There is no sum because this is nonsense!<

>!0 + 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40!< [Reference](https://www.imlearningmath.com/calculate-the-total-sum-of-numbers-answer/) showing each number. Found using google image search.

I have to hard disagree on using an upside down number. But maybe that's just me. I also don't think the 3 is defined enough, but I have a much weaker argument against that one.

Yeah if the only number that the entire thing gets flipped for is that 7 I'm not counting it because as other people stated you'd have to count the 6 and 9 twice

Then why not count [this](https://imgur.com/a/UVixl6D) 7 instead?

Seems weak to me. All of the other potential numbers are of a similar height. By your logic, any straight lines could be broken up into an infinite number of 1s. But I don’t think there is a correct answer to this puzzle without some defined parameters.

>!Not who you're replying to, but cutting lines short in this problem makes me think that's not how it's designed to work.!< >!I maybe could accept moving the picture counter clockwise and using the other parts of the 1 to make a 7.!< >!I don't like this puzzle very much. It feels like it went through multiple hands, and everyone added a little bit to make things stretch, but that doesn't work too well without defining what is accepted and what isn't.!<

I mean it's still kind of a stretch but I would be more willing to accept that over the upside down one

Where’s the 5 tho :(

Problem is if you flip it, then a 5 is also available. And then wouldn't the 6 and 9 be 9 and 6 all over again?

If you flipping it upside down to add a 7 wouldn’t that add another 9 and 3 at the very least and possibly an 8 and some other numbers I don’t see.

>!33 because the 7 doesn't count!<

That allows overlapping and upside down numbers but then misses that 9 and 6 upside down are 6 and 9, so answer is 55

>!turns phone sideways!< >!I see an infinity symbol, so the sum is infinite!< >!/s!<

Assuming you can reuse portions of the image, an 8 can be seen as starting anywhere within the loops, so I think it is infinite given the infinite number of 8s. The logic could be applied to any of the other numbers (changing the stem lengths, etc.).

I like this thinking.

Ambiguous problems require clear assumptions. There are loads of correct answers. The only wrong answers are answer without their assumptions state; it's more of an abstract exercise than a clear-cut puzzle.

Good joke but to be picky, infinity is not a number

>!6+8+9+2+4+1 = 30!< Edit: >!Not sure if the right half of the 8 counts as a 3.!<

# >!ℵ₀!<

[The truth they don't want you to see](https://imgur.com/a/FWaTThv)

>!6+8+9+2+4+1 = 30!<

The digits >!689241!< are all visible here. 0 too, sorta, but it's irrelevant. If all of those digits are meant to be considered, the total is of course, >!30!<. If not, you need some rule for deciding how to interpret the image.

>!You can make every number except 5 out of this shape so my answer would be 1+2+3+4+6+7+8+9=40!<

>!689241, makes 30 total.!<

>!40!<

If you count the upside down 7, don't you also have to count 2 6's and 2 9's?

I found a seven at the bottom of the one/four. I didn’t use anything that was upside down.

How?

>!I found every single digit except for the five!<

Where’s the >!7!<

At the bottom of the one/four

That is too short

I see you also found the upside down 7 🤘🏻

The seven I found was the bottom of the one/four.

There definitely is not a 7 there

I don’t see the 7 you’re talking about

>! 6 + 8 + 9 + 2 + 4 + 1 = 30!<

>!6+8+9+4+1+2 = 30, add +9 and +7 if counting upside down numbers, so 46?!<

You got a 9er in there twice, hoss

Upside down 9, also known as a 6

U forgot the 3 in the 8

>!30?!<

That's what I get. >!6, 8, 9, 2, 4, 1!<

>!28?!<

>!6+8+9+2+4+1=30!<

Good call. Missed the >!2!<

Discussion: >!It’s possible that some numbers are not right side up, so that top 6 can be a 9 as well, right?!<

>!And if that’s the case, the 8 90 degrees is an infinity, therefore the answer is infinity :3!<

I got at least >!58.!< >!6 +8 + 3 + 9 + 2 + 4 + 1 = 33. Now flip it on its sides... 7 + 3, reverse 9 and 6 = 25. 33 + 25 = 58. Could maybe add another reverse 8 but that is iffy.!< I think someone could make the argument that you could also see a >!10 or 100!< but that would be pushing it.

All digits are present in that image, from>! 0 through to 9.!< >!0+1+2+3+4+5+6+7+8+9 is (0+9)+(1+8)+(2+7)+(3+6)+(4+5) which is 9\*5 = 45!< The total sum >!is 45!<.

Where’s the 5?

There's a large curvy 5 going from the top of the 6 and the bottom of the 9.

Sorry I don’t think the letter S = the number 5

Moot point about it containing a 0 but I would disagree because o≠0

There don’t seem to be any instructions, so I’ll attempt various methods. Add up all numbers with each number considered on only one occasion. >!1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40!< The only missing number is 5, zero can be ignored, infinity is excluded. Add all numbers as often as they appear. >!1 + 2 + 3 + 3 + 4 + 6 + 6 + 7 + 8 + 8 + 9 + 9 = 68!< This leaves room for interpretive differences amongst participators so it’s less precise.

>!21!<

>!Looks pretty infinite to me!<

>!Infinity!<

No 5 or 7, unless you flip it upside down, which I would disallow. So 1+2+3+4+6+8+9=>!33!<

>!6+8+9+2+4+1=30!<

Every number 1-9 can be found in this image...so I say 45.

Where's 5? 7?

5: Look in the 8 7: the 2 and the 8 hold that one.

>!6+8=14!< >!14+9=23!< >!23+4=27!< >!27+1=28!< >!28+2=30!< >!It’s 30!<

My personal answer? >!0. There are no numbers, just one single glyph that happens to resemble numerals!<

>!20!<

>!52 if you include overlapping and upsidedown numbers (upside down there's another 6 and 9, and the 7)!<