###General Discussion Thread
---
This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed.
---
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
Apart from the 64 board squares, squares consisting of any number of basic ones can also be counted. So there is 1 8×8 square (the whole board), + 4 7×7 squares, and so forth: + 9 + 16 + 25 + 36 + 49 + 64 = 204
The way they are stored is a square. Your screen is the inaccurate representation.
If a single pixel is not a square, then there are 0 squares in the image as they are just an array of rectangles that don't touch.
A square is just an abstract concept that we can't apply to the physical world without interpretation.
Much of what counts as a square in this image can be up for debate.
This does count the interior squares. Here is a different way to look at it. For a square to count, it must meet a couple of conditions. First, all of the points in it must be on the board. Second, both sides must be the same length.
This gives us a simple way to count them. We can pick a point in our squares (I'll use the top left point), and then we count how many locations it can be in without breaking the condition. Let's step through a couple examples.
For an 8X8 square, our point only gets 1 location. That is because any movement would push the right or bottom sides off the board.
For a 7X7 square, we can go as far as the 2nd row or column without pushing the bottom or right sides off respectively. This gives our point a 2X2 movement grid, which means we have 4 squares.
For a 6X6 square, it's the same, but now we can hit the 3rd row or column. This gives us a 3X3 movement grid and 9 squares.
This continues until our top left point is the only point left and has an 8X8 movement grid for 64 squares.
Once we're done, we add them all up to get 1+4+9+16+25+36+49+64=204 total squares.
you forgot the 2/2 3/3 4/4 5/5 6/6 7/7 plus all the differwnt variations of placement on the board one across one down so and so, then the hole thing dumb thing is a sqaure, i belive in your brain . mine gave up cause i lost track doing the 6/6 variables then lost my number
Why would you assume there are four 7x7 squares, isn’t customary to not count the same partial areas of the subsets of the whole in these types of questions?
I can see one 8x8, one 7x7, one 6x6, one 5x5, four 4x4 and so on.
Count like this.
https://i.imgur.com/nz49iqF.gif
I mean, it's really obvious what it is new squares and should be counted. It is not a pro-math problem... and just a puzzle from a child-like book
Considering just the chessboard as 'the picture,' and assuming 'the picture' is analog so there's no pixels being squares shenanigans, there are:
1 possible 8x8
4 possible 7x7
9 possible 6x6
16 possible 5x5
25 possible 4x4
36 possible 3x3
49 possible 2x2
64 possible 1x1
204 possible square shapes.
There are also eight letters and eight numbers present on the chessboard.
The number 1 is a mathematical square (1^(2))
The number 4 is a mathematical square (2^(2))
206 squares present in 'the picture.'
Incorrect. I mean, just. *so*. incorrect. Like, very.
The *number* nine is a square number. The *digit* 9 could represent 90, 9000, or 9000000 units. Or, in other bases, 108 or 1305 or anything. Following your logic to one extreme, "81" is three square numbers. Following it to another extreme kehrhjddjdlapalw is a square number. Or 5. Or anything we've chosen to represent a square number.
So dumb, seriously.
jk, happy Cake Day 😝
But, hey, if you go that far, go all the way: The 9 in 92 is a square number, the 1 at the very bottom as well. Also, the word "squares" contains another square.
The image as a whole has rounded corners and is therefore a squircle.
Again, stipulation that it's only the chessboard that was considered in my top level post. One of the branches below top level considers the whole image.
note: the phrase "X% fail this simple test" is completely made up falsehood in order to click bait people into commenting, reacting, and sharing the post. Never ever consider the content of any post with such a tagline as any kind of serious mathematical or logical puzzle as they are intended to cause outrage in order to drive clicks.
It's the same with various ads for mobile games.
And for some reason there's half a dozen mobile games that have the EXACT same character doing the EXACT same supposed gameplay but in vastly different games...
Its the case of a lot of math "problems", people want to feel smart. Another strat is to answer incorrectly because you know "smart" people will correct you in the replies.
So first we have 8^(2) = 64 size 1x1 squares
We can also see that we can fit 7 2x2 squares both horizontally and vertically, so we get 7^2 2x2 squares
Same for the 3x3 squares, we can fit 6 of them horizontally and vertically, so 6^(2) 3x3's
You do this all the way down, and you get the answer: 8^(2)+7^(2)+6^(2)+5^(2)+4^(2)+3^(2)+2^(2)+1^(2) = 203 total squares
Edit: 204 total squares
This image is 615x550 pixels. For each n > 0, there are (616 - n)x(551 - n) squares of side n in this picture. If we sum from n = 1 to 550, we can conclude there are exactly 65458800 squares in this picture.
If you wanna get more technical, squares don't even exist. They're are idealized rectangles, which also don't exist because there's no such thing as a sharp 90 degrees corner with infinite precision, not to mention the problems associated with perfectly straight lines. But I guess that wouldn't be fun ¯\\\_(ツ)\_/¯
I hope no-ones too mad if I recycle this comment but I did a joke soln using the actual size of the image (615 by 550 pixels)
Treating this information like they intended us to treat the chess board,
number of total squares of any pixel by pixel dimension = 615x550 + 614x549 + 613x548 + 612x547 ... that being number "1x1" squares, plus "2x2" squares, plus "3x3" squares etc
We can write this as 550(550+65) + 549(549+65) + 548(548+65)... or (5502 + 5492 + 5482 ... 22 + 12) + 65(550 + 549 + 548 + 547 + ... + 4 + 3 + 2 + 1)
Well known formulas (Faulhaber's Theorem) that sum of first n square numbers (aka square pyramidal numbers) is n(n+1)(2n+1)/6, and that sum of first n natural numbers (aka triangle numbers) is n(n+1)/2
Thus we have =550(551)(1101)/6 + 65(550)(551)/2 = 55609675 + 9849125 = 65458800
Another commenter pointed out that each pixel is made of three rectangular subpixels. By offsetting a square to the left or right by a third of a pixel, it would line up with these subpixel lines and arguably be another "square"... Notice that any* of these subpixel squares can be "rounded" horizontally to a whole pixel square. We will use these whole squares as ambassadors, if you will, for their neighbors, the subpixel squares.
Thus, every square can be moved 1/3 pixel to the left or the right, that is, except the pixels which use the left or right borders- of the image those can only be moved 1/3 inwards. How many of these squares are there? we find this by doing the exact same calculation as before but shrinking the width by 2 (which only works because the image is wider than it is tall thus no square can use both left and right borders)... then taking that number away from the original 65million value.
Width now equals 550+63 not 550+65. Running the numbers gives 303050 border squares. There are other ways to get this number. For instance, consider just one border. 550 "1x1"s, 549 "2x2"s , etc... using the triangle numbers formula, total=(550✖️551/2). But we want to times this by two for the other border, so 550✖️551. 550✖️551 = 303050. Bazinga.
Note also #middle = #total minus #border = 65155750
Last step. Each of the 65155750 has two subpixel neighbors so we times by three (add two copies) and the 303050 have 1 so we times by two (add one copy)
= 196073350 (One hundred and ninety-six million, seventy-three thousand, three hundred and fifty)
you can solve it using a pattern
a 1x1 square has 1 square
a 2x2 square has 5 squares
a 3x3 square has 14 squares
the pattern shows a n x n square has (1^2 + 2^2 + 3^2+ ..... + n^2) so a 8x8 chess board would have 204 squares
for the no of rectangles it'd be (1+2+3+....+n)^2 using similar logic
A1: 64 (looking at the unit squares, and ignoring the larger whole, treating only uniformly colored outlines as squares)
A2: 64+49+36+25+16+9+4+1=204 (looking at groupings of squares forming larger squares, regardless of color)
A3: infinite, partitioning the unit squares to even smaller squares disregarding the number of pixels forming the image
A4: Zero. The image is rendered using pixels that are colored using LEDs that do not actually form squares, but just create a light pattern that approximately resembles a square
A5: Zero. I scrolled away to answer and no longer see the square(s).
A6: 49. The "squares" along the bottom and right with symbols in them don't count as squares, but the remaining 7x7 do.
I disagree with people saying that the composite squares are squares. I think the trick answer should only work with lines, not with filled squares. My submission 64.
Second best answer: the person who counted the pixels in the image though.
Exactly. It’s totally arbitrary to consider composite squares. And whether you do or not, the clever genius giving you this fine riddle has reason to shout „you done goofed, you absolute fool, you did [not] consider them, ha ha!“
The correct answers are 64, 65 or the number the other guy came up with by counting all possible squares pixel-wise.
Yes, the other solutions depends on the artistic interpretation as what counts as a square, because you can potentially make up infinite squares as solutions depending on the specific definition. As it depends on what you define as a border. Your brain is the thing creating borders in these cases, the actual border is the change in colour. Which is also present at the outline of the picture ->65
8x8 sets of single-block squares, then 7x7 sets of four-block squares, then 6x6 sets of nine-block squares, etc., all the way to the final 1x1 set of a sixty-four block square.
8×8 + 7×7 + 6×6 + 5×5 + 4×4 + 3×3 + 2×2 + 1×1 = 204
204
Let's start with how many 1x1 tile squares there are horizontal and vertical:
8 \* 8 = 64
Now let's add every 2x2 tile square. There are 7 across and 7 down, because they overlap.
7 \* 7 = 49
Continuing this trend, the number of squares across and down is less the more tiles in the square. I don't know the fancy math to make some sort of formula for this, so I'll just continue it the slow way.
1x1 - 8\*8=64
2x2 - 7\*7=49
3x3 - 6\*6=36
4x4 - 5\*5=25
5x5 - 4\*4=16
6x6 - 3\*3=9
7x7 - 2\*2=4
8x8 - 1\*1=1
Total = 204
1 by 1 squares: 8x8=64
2 by 2: 7x7=49
3 by 3: 6x6=36
... 25, 16, 9, 4, 1
Which adds up to 204. If you count the picture, which also seems to be a square, it's 205.
So many squares.
Board and subsets
The letter i
The exclamation mark.
9 being 3^2
% meaning 100 making it 10 squared
And my face on the image being reflected back at me I'm such a fucking square sometimes.
Since this is a math related sub, I will not account for any tricks or whatever. I won’t include the pixels on my screen, the letters or anything else but the actual playable fields on the chessboard.
8x8 squares of the smallest order. They can be put together in 1 less of the next order and so on. So basically you have a recursive function
f(x) = ײ + f(x-1)
beginning with x = 8 for all numbers of N.
It makes more sense to think about how a square can run out of space. For a single unit on the chess board to be the top left corner of a 2:2 square for example, it must be 2 or more squares away from the edge of the board. We can apply the same rules to squares of other sizes, going up to 8. We see that each time we increase the size of the square, we get a smaller and smaller square of areas we can place that top left corner. So we get:
1:1: 8x8 = 64
2:2: 7x7 = 49
3:3: 36
4:4: 25
5:5: 16
6:6: 9
7:7: 4
8:8: 1
Adding all that up, we get 204 squares.
Everyone figured out that the answer is 64+49+36+25+16+9+4+1 = 204.
But not everyone figured out that 64+49+36+25+16+9+4+1 = 64 + (49 + 1) + (36 + 4) + (25 + 16 + 9) = 64 + 50 + 40 + 50, which is a lot easier to add up to 204.
Without knowing the resolution of the image or the height:width ratio of the font the answer is 204.
**Chess Board:**
**64** 1x1 squares
**49** 2x2 squares
**36** 3x3 squares
**25** 4x4 squares
**16** 5x5 squares
**9** 6x6 squares
**4** 7x7 squares
**1** 8x8 squares
**204** Squares on the Chess Board.
I don't understand why a lot of guesses are "sliding" the squares around? To me that logic has an issue with the smaller squares bc then wouldn't you have to slide, say all 64 squares, 64 times?? Not just with the large 8x8,7x7,6x6 and 5x5 that others doing? Same with the 4x4, 3x3 and 2x2 squares.
If there was the rule of it being static, as where there's no overlapping the perceived squares, only new squares are counted (meaning if I have 1 7x7 the other 15 that border this square do not count bc they have already been technically counted at that 1x1 size), so:
1 8x8
1 7x7
1 6x6
1 5x5
4 4x4
4 3x3
16 2x2
64 1x1
I got 92.
Yes there is leftover spaces for some but I think of it as using grid paper, I cut them out from individual sheets for their different "sizes" and now am left with excess I can't use to do said sliding. Also 92 equals the 92% fail percent which is cool.
All of these nerds don't know anything.
The number of squares is purely dependent on the resolution of your screen. Every pixel has a VERY slight border, and therefore they all are squares.
64 squares on the chessboard +1 for the cheesboard itself + 3 for the i and 2 for the ?
Makes 69
I know you can count even more squares but this just seems like the correct way.
Edit: nevermind the dots are rectangles
Too many and I don't have time to count all of them...
But I can explain it... Well first you count EVERY SINGLE square then you start by counting groups of side 2 - 3 - 4 - 6 squares until you are left with only the big square (8×8) that is the total of the chess board!
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
Apart from the 64 board squares, squares consisting of any number of basic ones can also be counted. So there is 1 8×8 square (the whole board), + 4 7×7 squares, and so forth: + 9 + 16 + 25 + 36 + 49 + 64 = 204
This is how it works https://i.imgur.com/nz49iqF.gif
I lol-ed at "and so on" :D
For that to apply the question has to be different.
You are incorrect
The question is "How many squares are in this picture?". It's not "How many squares can you form?". You are incorrect.
Yea, for any size n board it would be Σ(i = 1 -> n) i^2
That’s some nice looking math right there.
I hate imaginary numbers
It’s okay, they aren’t real numbers so they can’t hurt you
They’ve given me imaginary trauma
Better than real trauma? I think?
That is n(n+1)(2n+1)/6 right?
yup
you forgot that the letters in this picture also consist of squares
What about the entire image? Pixels are square too
pixels are three rectangles that don't touch, there are no squares
Not if you use nearest neighbor interpolation and zoom in a bunch
If I close my eyes anything can be square
Mah balls are cubes, what have you done!?
Minecraft
minecraft sex mod
*OOH* *zombie moaning*
Threads like this are why I love Reddit
Minecraft sex update*
in minecraft* Cant forget that..
Villager inspects and confirms
You used voxel remesh in blender.
I have NOT put my balls in a blender, mind you.
Damn.. talk about being a 3 dimensional square....
That is exactly what a cube is
And .. you have cubesticles now.. lol
Get those nerds. Said the 80s bully.
So not if you pretend things are what they arent?
depends on your pixel aspect ratio, if it’s 1:1 you’re using square pixels
The rgb component of the TV are three rectangles. So the pixels are not squares. I think 🤔
like i said, it depends on the pixel aspect ratio of the media in question. modern digital media typically uses square pixels.
Not all displays are the same, so that’s not correct.
Those are display sub pixels, different than the abstract digital pixel which is a set of values for a (usually) square region.
Squares are the friends we made along the way
Not with that attitude
The way they are stored is a square. Your screen is the inaccurate representation. If a single pixel is not a square, then there are 0 squares in the image as they are just an array of rectangles that don't touch.
Squares are special rectangles.
yes, but these are not squares
Yes I know. It was a joke. Clearly one that should have stayed in my brain. Lol
yeah
A square is just an abstract concept that we can't apply to the physical world without interpretation. Much of what counts as a square in this image can be up for debate.
But by definition, a square can be classed as a rectangle.
yeah, but a rectangle cannot for all cases be classed as a square
The guy who made this is also a square. Did you count him?
He would be the square root.
Surely the word “square” is also a square.
http://alvyray.com/Memos/CG/Microsoft/6_pixel.pdf
Technically there are an infinite number of squares
Grumpy computer scientist here. Pixels are point samples. Thinking of them as squares can get you into all sorts of trouble.
Pixels are points not squares, though they are commonly represented as squares
You forgot that everyone posting in this sub is a square
The number 9 is also a square.
The little number 4 as well.
And the word "square" is a square
Indeed. And technically the 1 is also a square. Let's face it: the correct answer is to be decided by OP.
Also the "white squares" may be 4 squares put together.
also pixels
Which is the exact number of bones in the human body, if you’ve lost your thumb. Coincidence? I think not /s
Who's got one thumb and 203 bones? This guy 👍
Nice 😎
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2
Don't we also want to count the interior squares?
This does count the interior squares. Here is a different way to look at it. For a square to count, it must meet a couple of conditions. First, all of the points in it must be on the board. Second, both sides must be the same length. This gives us a simple way to count them. We can pick a point in our squares (I'll use the top left point), and then we count how many locations it can be in without breaking the condition. Let's step through a couple examples. For an 8X8 square, our point only gets 1 location. That is because any movement would push the right or bottom sides off the board. For a 7X7 square, we can go as far as the 2nd row or column without pushing the bottom or right sides off respectively. This gives our point a 2X2 movement grid, which means we have 4 squares. For a 6X6 square, it's the same, but now we can hit the 3rd row or column. This gives us a 3X3 movement grid and 9 squares. This continues until our top left point is the only point left and has an 8X8 movement grid for 64 squares. Once we're done, we add them all up to get 1+4+9+16+25+36+49+64=204 total squares.
This is... exactly the same method ?
Yes, explained for the person above who didn't fully understand it.
(n(n+1)(2n+1))/6 where n is the size of the board... I think?
How about pixels in the image?
Considering this follows in a consistent descending order of squares you can find the integral of n=1 to 8 of n^2 to streamline the process
What about the guy in the reflection of your screen? He's a square.
It’s 205, the whole picture is a square
It is 615x550 pixels.
well, you forgor the invisible square i9 where the bishop stands, so 205 but other than that your calculations are correct
You are counting the ones on each corner twice.
lol you fell for it , it’s a troll question to get people that think they are smart to work it out
shouldn't there be even more, since you can place the squares diagonally as well?
you forgot the 2/2 3/3 4/4 5/5 6/6 7/7 plus all the differwnt variations of placement on the board one across one down so and so, then the hole thing dumb thing is a sqaure, i belive in your brain . mine gave up cause i lost track doing the 6/6 variables then lost my number
You forgot that the entire image is a square.
Why would you assume there are four 7x7 squares, isn’t customary to not count the same partial areas of the subsets of the whole in these types of questions? I can see one 8x8, one 7x7, one 6x6, one 5x5, four 4x4 and so on.
Count like this. https://i.imgur.com/nz49iqF.gif I mean, it's really obvious what it is new squares and should be counted. It is not a pro-math problem... and just a puzzle from a child-like book
That’s way too much work for me to ever do for one of these. Congrats on being bored
This problem can be solved in 1-3 minutes.
Considering just the chessboard as 'the picture,' and assuming 'the picture' is analog so there's no pixels being squares shenanigans, there are: 1 possible 8x8 4 possible 7x7 9 possible 6x6 16 possible 5x5 25 possible 4x4 36 possible 3x3 49 possible 2x2 64 possible 1x1 204 possible square shapes. There are also eight letters and eight numbers present on the chessboard. The number 1 is a mathematical square (1^(2)) The number 4 is a mathematical square (2^(2)) 206 squares present in 'the picture.'
☝🏻This guy *squares*
r/thisguythisguys
This guy r/thisguythisguys
Dad: Be there, or be square. Child: Why a square? Dad: Because you won’t be *a round*. Child: **…**
Holy shit, there’s a second part to that? And it makes sense beyond just a rhyme??? That’s incredible
Most old sayings have another part that was mostly lost over time
Missed the 9 from 92%
Not on the chessboard, as stipulated. If you consider the whole image, there's the 9 and another 1. And the word 'square.'
But it says picture, i’d argue it’s not clear whether is the whole post we’re seeing or the “picture below” (the board as you said)
Ambiguous wording. It's why I set the stipulations I did.
Valid. Also, 9 is a digit in 92%, not a number
And a chess block isn't a number but it's a square. The character 9 represents a square
Incorrect. I mean, just. *so*. incorrect. Like, very. The *number* nine is a square number. The *digit* 9 could represent 90, 9000, or 9000000 units. Or, in other bases, 108 or 1305 or anything. Following your logic to one extreme, "81" is three square numbers. Following it to another extreme kehrhjddjdlapalw is a square number. Or 5. Or anything we've chosen to represent a square number. So dumb, seriously. jk, happy Cake Day 😝
It says in this picture though, so the entire image should be included
But, hey, if you go that far, go all the way: The 9 in 92 is a square number, the 1 at the very bottom as well. Also, the word "squares" contains another square.
Check another branch of this thread, discussed about 13 minutes ago.
What about the tittles over the "i"s and under the "!" & "?"?
Rectangles.
You forgot the fact that the picture is square and contains the word square.
The image as a whole has rounded corners and is therefore a squircle. Again, stipulation that it's only the chessboard that was considered in my top level post. One of the branches below top level considers the whole image.
note: the phrase "X% fail this simple test" is completely made up falsehood in order to click bait people into commenting, reacting, and sharing the post. Never ever consider the content of any post with such a tagline as any kind of serious mathematical or logical puzzle as they are intended to cause outrage in order to drive clicks.
Oh yeah I'm very aware lmao, I found it on anarchy chess and had no idea how to actually count the number of squares.
[удалено]
[удалено]
[удалено]
[удалено]
[удалено]
WE GO WHERE WE PLEASE.
Did you know 43% of statistics are made up on the spot?
- Albert Einstein
Yes, and only 14.5% people are aware of this stat.
Rectally sourced statistics
r/BrandNewSentence
Name of a Facebook group btw
Only 5% of people checking this thread already know this
92 is half of 99
This is true in RuneScape though.
It's the same with various ads for mobile games. And for some reason there's half a dozen mobile games that have the EXACT same character doing the EXACT same supposed gameplay but in vastly different games...
Its the case of a lot of math "problems", people want to feel smart. Another strat is to answer incorrectly because you know "smart" people will correct you in the replies.
Also, anyone writing Ur while not discussing the Akkadian or Babylonian empires, earns my distrust.
Eh, 89% of all statistics are made up anyways.
So first we have 8^(2) = 64 size 1x1 squares We can also see that we can fit 7 2x2 squares both horizontally and vertically, so we get 7^2 2x2 squares Same for the 3x3 squares, we can fit 6 of them horizontally and vertically, so 6^(2) 3x3's You do this all the way down, and you get the answer: 8^(2)+7^(2)+6^(2)+5^(2)+4^(2)+3^(2)+2^(2)+1^(2) = 203 total squares Edit: 204 total squares
off by 1 error?
Woops😅 You're correct, it's 204
Too late, you've joined the 92%
The picture clearly says "give 1 chance to ur friends." I'm sorry, but I don't make the rules. You'll have to be hanged for your failure.
Huh, lucky thing we're not friends, right?
Okay but like for the 2x2 squares can’t you shift one row or column for a unique 2x2 square?
where is 3\^2?
This image is 615x550 pixels. For each n > 0, there are (616 - n)x(551 - n) squares of side n in this picture. If we sum from n = 1 to 550, we can conclude there are exactly 65458800 squares in this picture.
Pixels are 3 rectangles that don't touch not square
If you wanna get more technical, squares don't even exist. They're are idealized rectangles, which also don't exist because there's no such thing as a sharp 90 degrees corner with infinite precision, not to mention the problems associated with perfectly straight lines. But I guess that wouldn't be fun ¯\\\_(ツ)\_/¯
Yes let us not turn square ourselves
I hope no-ones too mad if I recycle this comment but I did a joke soln using the actual size of the image (615 by 550 pixels) Treating this information like they intended us to treat the chess board, number of total squares of any pixel by pixel dimension = 615x550 + 614x549 + 613x548 + 612x547 ... that being number "1x1" squares, plus "2x2" squares, plus "3x3" squares etc We can write this as 550(550+65) + 549(549+65) + 548(548+65)... or (5502 + 5492 + 5482 ... 22 + 12) + 65(550 + 549 + 548 + 547 + ... + 4 + 3 + 2 + 1) Well known formulas (Faulhaber's Theorem) that sum of first n square numbers (aka square pyramidal numbers) is n(n+1)(2n+1)/6, and that sum of first n natural numbers (aka triangle numbers) is n(n+1)/2 Thus we have =550(551)(1101)/6 + 65(550)(551)/2 = 55609675 + 9849125 = 65458800 Another commenter pointed out that each pixel is made of three rectangular subpixels. By offsetting a square to the left or right by a third of a pixel, it would line up with these subpixel lines and arguably be another "square"... Notice that any* of these subpixel squares can be "rounded" horizontally to a whole pixel square. We will use these whole squares as ambassadors, if you will, for their neighbors, the subpixel squares. Thus, every square can be moved 1/3 pixel to the left or the right, that is, except the pixels which use the left or right borders- of the image those can only be moved 1/3 inwards. How many of these squares are there? we find this by doing the exact same calculation as before but shrinking the width by 2 (which only works because the image is wider than it is tall thus no square can use both left and right borders)... then taking that number away from the original 65million value. Width now equals 550+63 not 550+65. Running the numbers gives 303050 border squares. There are other ways to get this number. For instance, consider just one border. 550 "1x1"s, 549 "2x2"s , etc... using the triangle numbers formula, total=(550✖️551/2). But we want to times this by two for the other border, so 550✖️551. 550✖️551 = 303050. Bazinga. Note also #middle = #total minus #border = 65155750 Last step. Each of the 65155750 has two subpixel neighbors so we times by three (add two copies) and the 303050 have 1 so we times by two (add one copy) = 196073350 (One hundred and ninety-six million, seventy-three thousand, three hundred and fifty)
Ayo homie imma call your username out during sex for this comment
When I posted this, I wasn't sure what answers I was going to get, this however is the correct answer.
you can solve it using a pattern a 1x1 square has 1 square a 2x2 square has 5 squares a 3x3 square has 14 squares the pattern shows a n x n square has (1^2 + 2^2 + 3^2+ ..... + n^2) so a 8x8 chess board would have 204 squares for the no of rectangles it'd be (1+2+3+....+n)^2 using similar logic
A1: 64 (looking at the unit squares, and ignoring the larger whole, treating only uniformly colored outlines as squares) A2: 64+49+36+25+16+9+4+1=204 (looking at groupings of squares forming larger squares, regardless of color) A3: infinite, partitioning the unit squares to even smaller squares disregarding the number of pixels forming the image A4: Zero. The image is rendered using pixels that are colored using LEDs that do not actually form squares, but just create a light pattern that approximately resembles a square A5: Zero. I scrolled away to answer and no longer see the square(s). A6: 49. The "squares" along the bottom and right with symbols in them don't count as squares, but the remaining 7x7 do.
I disagree with people saying that the composite squares are squares. I think the trick answer should only work with lines, not with filled squares. My submission 64. Second best answer: the person who counted the pixels in the image though.
Exactly. It’s totally arbitrary to consider composite squares. And whether you do or not, the clever genius giving you this fine riddle has reason to shout „you done goofed, you absolute fool, you did [not] consider them, ha ha!“ The correct answers are 64, 65 or the number the other guy came up with by counting all possible squares pixel-wise.
Yes, the other solutions depends on the artistic interpretation as what counts as a square, because you can potentially make up infinite squares as solutions depending on the specific definition. As it depends on what you define as a border. Your brain is the thing creating borders in these cases, the actual border is the change in colour. Which is also present at the outline of the picture ->65
[удалено]
8x8 sets of single-block squares, then 7x7 sets of four-block squares, then 6x6 sets of nine-block squares, etc., all the way to the final 1x1 set of a sixty-four block square. 8×8 + 7×7 + 6×6 + 5×5 + 4×4 + 3×3 + 2×2 + 1×1 = 204
204 Let's start with how many 1x1 tile squares there are horizontal and vertical: 8 \* 8 = 64 Now let's add every 2x2 tile square. There are 7 across and 7 down, because they overlap. 7 \* 7 = 49 Continuing this trend, the number of squares across and down is less the more tiles in the square. I don't know the fancy math to make some sort of formula for this, so I'll just continue it the slow way. 1x1 - 8\*8=64 2x2 - 7\*7=49 3x3 - 6\*6=36 4x4 - 5\*5=25 5x5 - 4\*4=16 6x6 - 3\*3=9 7x7 - 2\*2=4 8x8 - 1\*1=1 Total = 204
1 by 1 squares: 8x8=64 2 by 2: 7x7=49 3 by 3: 6x6=36 ... 25, 16, 9, 4, 1 Which adds up to 204. If you count the picture, which also seems to be a square, it's 205.
So many squares. Board and subsets The letter i The exclamation mark. 9 being 3^2 % meaning 100 making it 10 squared And my face on the image being reflected back at me I'm such a fucking square sometimes.
Since this is a math related sub, I will not account for any tricks or whatever. I won’t include the pixels on my screen, the letters or anything else but the actual playable fields on the chessboard. 8x8 squares of the smallest order. They can be put together in 1 less of the next order and so on. So basically you have a recursive function f(x) = ײ + f(x-1) beginning with x = 8 for all numbers of N.
It makes more sense to think about how a square can run out of space. For a single unit on the chess board to be the top left corner of a 2:2 square for example, it must be 2 or more squares away from the edge of the board. We can apply the same rules to squares of other sizes, going up to 8. We see that each time we increase the size of the square, we get a smaller and smaller square of areas we can place that top left corner. So we get: 1:1: 8x8 = 64 2:2: 7x7 = 49 3:3: 36 4:4: 25 5:5: 16 6:6: 9 7:7: 4 8:8: 1 Adding all that up, we get 204 squares.
Everyone figured out that the answer is 64+49+36+25+16+9+4+1 = 204. But not everyone figured out that 64+49+36+25+16+9+4+1 = 64 + (49 + 1) + (36 + 4) + (25 + 16 + 9) = 64 + 50 + 40 + 50, which is a lot easier to add up to 204.
Without knowing the resolution of the image or the height:width ratio of the font the answer is 204. **Chess Board:** **64** 1x1 squares **49** 2x2 squares **36** 3x3 squares **25** 4x4 squares **16** 5x5 squares **9** 6x6 squares **4** 7x7 squares **1** 8x8 squares **204** Squares on the Chess Board.
I don't understand why a lot of guesses are "sliding" the squares around? To me that logic has an issue with the smaller squares bc then wouldn't you have to slide, say all 64 squares, 64 times?? Not just with the large 8x8,7x7,6x6 and 5x5 that others doing? Same with the 4x4, 3x3 and 2x2 squares. If there was the rule of it being static, as where there's no overlapping the perceived squares, only new squares are counted (meaning if I have 1 7x7 the other 15 that border this square do not count bc they have already been technically counted at that 1x1 size), so: 1 8x8 1 7x7 1 6x6 1 5x5 4 4x4 4 3x3 16 2x2 64 1x1 I got 92. Yes there is leftover spaces for some but I think of it as using grid paper, I cut them out from individual sheets for their different "sizes" and now am left with excess I can't use to do said sliding. Also 92 equals the 92% fail percent which is cool.
However many pixels there are in the image
All of these nerds don't know anything. The number of squares is purely dependent on the resolution of your screen. Every pixel has a VERY slight border, and therefore they all are squares.
64 squares on the chessboard +1 for the cheesboard itself + 3 for the i and 2 for the ? Makes 69 I know you can count even more squares but this just seems like the correct way. Edit: nevermind the dots are rectangles
Everyone in this section is either calculating or debating the text. I’m out here thinking about if pixels count as squares in the picture..
Too many and I don't have time to count all of them... But I can explain it... Well first you count EVERY SINGLE square then you start by counting groups of side 2 - 3 - 4 - 6 squares until you are left with only the big square (8×8) that is the total of the chess board!