Can't figure this one out.. What is the missing box? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Difficult IQ test question: What is the box suggesting?Mensa IQ test questionMissing number in figureSeemingly difficult culture fair question5008 out of the boxFigure out the missing cardFind the missing boxFind the missing brickFind the missing brick, againShapes and letters. What is the missing box?
3 doors, three guards, one stone
How to add zeros to reach same number of decimal places in tables?
Interesting examples of non-locally compact topological groups
Fishing simulator
How many spell slots should a Fighter 11/Ranger 9 have?
Single author papers against my advisor's will?
What computer would be fastest for Mathematica Home Edition?
How do you clear the ApexPages.getMessages() collection in a test?
How should I respond to a player wanting to catch a sword between their hands?
Slither Like a Snake
How are presidential pardons supposed to be used?
What is the largest species of polychaete?
Aligning matrix of nodes with grid
Are my PIs rude or am I just being too sensitive?
Replacing HDD with SSD; what about non-APFS/APFS?
What do you call a plan that's an alternative plan in case your initial plan fails?
When communicating altitude with a '9' in it, should it be pronounced "nine hundred" or "niner hundred"?
If A makes B more likely then B makes A more likely"
Autumning in love
I'm thinking of a number
Why is "Captain Marvel" translated as male in Portugal?
Is it possible to ask for a hotel room without minibar/extra services?
What are the performance impacts of 'functional' Rust?
How is simplicity better than precision and clarity in prose?
Can't figure this one out.. What is the missing box?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Difficult IQ test question: What is the box suggesting?Mensa IQ test questionMissing number in figureSeemingly difficult culture fair question5008 out of the boxFigure out the missing cardFind the missing boxFind the missing brickFind the missing brick, againShapes and letters. What is the missing box?
$begingroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
add a comment |
$begingroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago
add a comment |
$begingroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
logical-deduction pattern visual geometry progressive-matrix
logical-deduction pattern visual geometry progressive-matrix
New contributor
New contributor
edited 6 hours ago
user477343
3,05511061
3,05511061
New contributor
asked 8 hours ago
Rrz0Rrz0
1285
1285
New contributor
New contributor
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago
add a comment |
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) in a >!clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of >!sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81760%2fcant-figure-this-one-out-what-is-the-missing-box%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
add a comment |
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
add a comment |
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
edited 7 hours ago
answered 8 hours ago
user477343user477343
3,05511061
3,05511061
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
add a comment |
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
1
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
6 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
3 hours ago
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) in a >!clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of >!sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) in a >!clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of >!sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) in a >!clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of >!sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) in a >!clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of >!sides (+1) that the newly shaded shapes have.
New contributor
New contributor
answered 2 hours ago
Rrz0Rrz0
1285
1285
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
edited 1 hour ago
New contributor
answered 2 hours ago
SteveSteve
1011
1011
New contributor
New contributor
add a comment |
add a comment |
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81760%2fcant-figure-this-one-out-what-is-the-missing-box%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
2 hours ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
2 hours ago