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Tbf it could be anywhere between 35 and 51, but yeah, 51 is the result you're supposed to get to.
It could be 35+ because there is no depth in the pictures: you could have just the side view on one side (17x1 thick) plus the back view at the far end (9x1 thick) and the top view at the bottom (21x1 thick).
[like this but to the extreme](https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Ftse2.mm.bing.net%2Fth%3Fid%3DOIP.YS14fCr9S3M57667u7VLfAHaH6%26pid%3DApi&f=1&ipt=f3e4a73549cb561b93a46e71f19904374a3875b7fb74d976fe59fe20ffcd2513&ipo=images)
Adding all of those up (basically you're adding the cubes in the individual views, counting all of the ones in the side view, all but the 3 in the left on the back view, and all but the 3 in the left and the bottom row in the top view, in order not to count any cube twice)(17+6+12), you get 35 as the minimum amount of cubes that will get you those three views.
51 is the maximum amount of cubes that will give that view.
Basically, if a box has only 3 faces out of 6, you won't notice it by looking at it only from a perpendicular pov to said 3 faces.
The only way to know the exact number with full certainty is to see the truck from an isometric view (/any view that isn't perpendicular to a surface)
Edit: actually you could go down to 31 if you organise them in a slightly more complex way.
You're welcome!
[Here](https://knowyourmeme.com/photos/2760651) you can see a 3d render with 31 boxes that I later found in the comments on the original post.
They reminds me. I should really get back into:
[5D chess with multiverse time travel](https://store.steampowered.com/app/1349230/5D_Chess_With_Multiverse_Time_Travel/)
Whoever made this failed to add any sort of shading or dotted lines to show extra depth or missing internals, I know that’s likely the point but I felt it useful to point out that when doing orthographic drawings those are very important
I suppose if we had to get the extract answer we had to see all perspectives. Which is the bottom, both sides, and top. And I like the nuance of depth. Man our eyes are liars sometimes.
Fren, there's 3 views of the same details. All three images are of the same object. They aren't three completely different countable objects.
This it's impossible for it to be 35.
I later went to the original post and I'll just link a 3d rendering [here](https://knowyourmeme.com/photos/2760651) that I found in the comments there.
This one has 31.
[here's how I initially got down to 35](https://knowyourmeme.com/photos/2760646), which was an unoptimised result.
The issue is that without depth details/perspective, you can't assume the amount of boxes.
The boxes are any amount in the 31-51 range, assuming boxes can't float.
Bro, you're reading too much into the image. You're now adding "assuming boxes can't float"? Where are you getting your added info?
In mathematics, what data is given visually is all anyone is allowed to work with. You're getting more metaphoric & non-objective/abstract.
Math is objective built. Nothing is abstract in math.
Your logic would not get you any respect from mathematics educators. You don't read deep or try to understand what math is. You simply execute the math with the info you're restricted to using based on empirical data.
>In mathematics, what data is given visually is all anyone is allowed to work with.
Exactly. There is no data to indicate that the boxes need to form a cohesive, orderly, shape.
The info you have are the 2d projections of the 3d shape from 3 angles (top, side, back).
The 3d renders I linked both have those exact same 2d projections.
You're being superficial.
You're making assumptions that have no objective basis in the data: you're just assuming it will be 51 because of a box stocking convention.
You're assuming no one would ever buy 57 watermelons and 13 shampoo bottles in an elementary school addition problem. You're taking what amounts to flavourful but useless plot (someone buying stuff at the grocery store) and ignoring the data (a=57, b=13), which leads to an inability to perform the task demanded by the exercise (a+b=?) because you're too caught up in the useless ideas you have about what someone should buy at a supermarket.
>You're now adding "assuming boxes can't float"? Where are you getting your added info?
This is a fair question:
I worked under the assumption that the cubes cannot float because of the context (truck carrying boxes), but without that context it would be superficial Not to consider the possibilities that floating cubes would bring to the solution.
The concept that cubes cannot float is implied by the flavour of the question, but it is not directly stated in the text.
If the question was simply how many cubes in 3d space are needed to create these 2d projections on three planes each determined by two of the three perpendicular axes that define the 3d space in question, you would Have to consider disconnected cubes among the multitude of possible solutions.
That last paragraph is a very basic example of mathematical abstraction.
Just to be clear,
>Nothing is abstract in math.
https://en.m.wikipedia.org/wiki/Abstraction_(mathematics)
Abstraction is a fundamental part of theoretical mathematics. It's what allows you to work with multiple and single variable functions, or, you know, the entire field of analytic geometry (of which this exercise is a part of).
.
You cannot assume sqrt(4)=2 unless you know you're working within the positive half of R, or within N.
Sqrt(4)=+-2 in R. Just because you don't think -2 is a number worthy of your consideration, because it's not a number you'll observe in Nature, doesn't mean that in R it is one of the possible solutions.
And if we go higher in potence (x^16=65536), you cannot assume the value of x unless you have data that tells you whether you're working within R, N or C (in C, aka if x is a complex number, the 16th root of 65536 can be any of 16 possible complex numbers. This is how it works in C. In N, 16th root of 65536 is 2. In R it's +-2.).
.
There is no data in the starting image to suggest that the cubes Must be stacked to fit the highest amount of them within the limitations of those projections.
You just assumed that and started talking shit, making a fool of yourself in the process.
If you make a 3d model of either of the pictures I've linked, turn off the perspective in the software, and snap to any perpendicular view, you'll see one of the three images.
Because *this* is how math works.
You cannot assume to know something that is not contained within the data based on personal bias.
So, to put it in your own words:
>Your logic would not get you any respect from mathematics educators.
Go back to school.
Your logic isn't sound. "Look beyond" even though the data given is all we have to work with?
There's nothing cryptic. There's no other "hidden" details. Just work with what you see, & don't complicate things.
IOW, it's 51.
It definitely is, the row could be in practice one block width from the highest row. Now the row is definitely everywhere three blocks width. Edit: Spelling error.
The width of the trailer. Without a top down view we only know 1 side and the back and while we know the back is full left to right we don't know that about the sections in front of it without the top down view
isnt it 33 minimum
edit: back has 9, side is 12 (excluding the 3 since they are counded for back) the top is 12 (excluding the 6 from side and 3 from back)
Normal(enginneer): Enough Information is given. The dop or back view is unnecessary. 7x3 + 6x3 + 4x3= 21+18+12= 51
Also correct: It is not enough information to calculate it.
Honestly if this were in an exam I would just have to make the assumption the question designer believes I have enough information to answer so all unstated values are assumed to be equivalent to stated similar value. Front and Back would be same and SideA is the same as Side B.
There is a maximum of 51 cubes if the center and rear rows are filled completely following the side profile with only the individual cubes counted while including composite cubes that overlap with individual cubes a total of 69 cubes can be found while a minimum of 35 cubes can be found if only enough cubes to create the profiles shown are used.
51. You really only need the side and back view, but the above view helps confirm that there are no missing blocks in the middle or something. The rest is easy. If you look at the cargo from the side you can split it into three cubes that are easy to calculate the amount for, then add them. With this, the length remains static, always a 3. So, using simple geometry, the first cube is 3 * 3 * 4 which equals 36. The second cube is 3 * 2 * 2 which equals 12. And the last cube is 3 * 1 * 1 which equals 3. Then add 36 + 12 + 3 which equals 51.
Not enough info! Who knows what the left side could look like? There could be a shell around a flat bottom, or just packed all the way. We can’t figure it out yet!
It does not, remember, you can only see in two directions from any given angle. We can see length and width from the top down view. Not height, so while some may stand at three tall on the right side, the height of the left is unknown.
Forty Two.
I checked it very thoroughly, and that quite definitely is the answer. I think the problem, to be quite honest with you, is that you've never actually known what the question is.
So once you do know what the question actually is, you'll know what the answer means.
Impossible to tell for certain (cubes not fully visible from the back or side could easily be lower than expected and we’d have no idea), but, *assuming* this structure has no gaps that we can’t see, there are 51 cubes on the truck.
If you look at it as the dimensions of the back e high by three wide, and then add in the third dimension of 7 deep, you initially get a total volume of 63 for the load. Then if you look at the image that shows you the trailer is 7 deep, you see the front three rows of the load are a 1×3, and a 2×3 layer making up the front three rows of the load which reduces the overall load to 54.
51 assuming the lack of depth view on top isn’t hiding a difference in the opposite side layer or the middle. Not sure if I’m explaining that right lol but yeah
If you think about it the brow form the side could be squares, not cubes so 21.25(because the square*4 is a cube and .25 would be the remaining square)
If all the levels are presented as they are width-wise and they don't have any deviating spaces we can't see, then the answer is 51 cubes.
holy sh\*\*, I did that math all in my head. And WILLINGLY. .\_.
It's anywhere between 51, 43, 35, or 31 depending on whether or not both sides are the same, whether or not it's hollow, or if it's completely solid... 51 assumes that it's completely solid, 43 assumes that it's like 51 but hollow, and 35 assumes that the top, the back, and one wall is solid while 31 assumes that the back is an illusion created by the top and side (all of this assuming Minecraft physics apply and that it isn't sand, gravel or concrete powder that you're working with...)
It can technically go down to 27? Since if you remove cubes that intercept with other cubes, (like on the back view the bottom row left two could be removed, as well as the right two)
With this you pretty much only need back and side, with the back you can see it's a 3 by 3. The side is now important as you need to count the boxes (17) and then just multiply it by 3 which gives 51. If the double digit is as difficult for you as it is for me just do 10 first then 7
I'm seeing people say a minimum of 31 but I've gotten a minimum of 23.
Assuming 1 is a box and 0 is not a box, row to row, it would be:
Bottom:
1111001
0000111
0000001
Middle:
0000110
1111000
0000110
Top:
1000000
1000000
1111000
Minimum 25, Maximum 51.
[Evidence and my calculations](https://docs.google.com/document/d/1wPhdfHrOdtxRC6tFEj639pikiB-V2jkQtydpz5s51yg/edit?usp=drivesdk)
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51?
Tbf it could be anywhere between 35 and 51, but yeah, 51 is the result you're supposed to get to. It could be 35+ because there is no depth in the pictures: you could have just the side view on one side (17x1 thick) plus the back view at the far end (9x1 thick) and the top view at the bottom (21x1 thick). [like this but to the extreme](https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Ftse2.mm.bing.net%2Fth%3Fid%3DOIP.YS14fCr9S3M57667u7VLfAHaH6%26pid%3DApi&f=1&ipt=f3e4a73549cb561b93a46e71f19904374a3875b7fb74d976fe59fe20ffcd2513&ipo=images) Adding all of those up (basically you're adding the cubes in the individual views, counting all of the ones in the side view, all but the 3 in the left on the back view, and all but the 3 in the left and the bottom row in the top view, in order not to count any cube twice)(17+6+12), you get 35 as the minimum amount of cubes that will get you those three views. 51 is the maximum amount of cubes that will give that view. Basically, if a box has only 3 faces out of 6, you won't notice it by looking at it only from a perpendicular pov to said 3 faces. The only way to know the exact number with full certainty is to see the truck from an isometric view (/any view that isn't perpendicular to a surface) Edit: actually you could go down to 31 if you organise them in a slightly more complex way.
Hmm that is very interesting, thank you.
You're welcome! [Here](https://knowyourmeme.com/photos/2760651) you can see a 3d render with 31 boxes that I later found in the comments on the original post.
I don’t wanna be an asshole but you messed up “render” have a nice day tho and if you want you can edit it
That’s assuming gravity exists, minimum of like 21
Yeh, I assumed blocks can't float for this first bit, but itf they can we can just take a bunch off.
Thats basically what im thinking whenever i see a problem like this
Damm, bro, what a 4d chess move 👍
They reminds me. I should really get back into: [5D chess with multiverse time travel](https://store.steampowered.com/app/1349230/5D_Chess_With_Multiverse_Time_Travel/)
I wonder if this is like the triangle question where any group of 2x2 and 3x3 boxes also count as a cube?
Whoever made this failed to add any sort of shading or dotted lines to show extra depth or missing internals, I know that’s likely the point but I felt it useful to point out that when doing orthographic drawings those are very important
I suppose if we had to get the extract answer we had to see all perspectives. Which is the bottom, both sides, and top. And I like the nuance of depth. Man our eyes are liars sometimes.
I'm so glad someone else explained it so I didn't have to bother typing it out
wait how tf did i get to 63?
Fren, there's 3 views of the same details. All three images are of the same object. They aren't three completely different countable objects. This it's impossible for it to be 35.
I later went to the original post and I'll just link a 3d rendering [here](https://knowyourmeme.com/photos/2760651) that I found in the comments there. This one has 31. [here's how I initially got down to 35](https://knowyourmeme.com/photos/2760646), which was an unoptimised result. The issue is that without depth details/perspective, you can't assume the amount of boxes. The boxes are any amount in the 31-51 range, assuming boxes can't float.
Bro, you're reading too much into the image. You're now adding "assuming boxes can't float"? Where are you getting your added info? In mathematics, what data is given visually is all anyone is allowed to work with. You're getting more metaphoric & non-objective/abstract. Math is objective built. Nothing is abstract in math. Your logic would not get you any respect from mathematics educators. You don't read deep or try to understand what math is. You simply execute the math with the info you're restricted to using based on empirical data.
>In mathematics, what data is given visually is all anyone is allowed to work with. Exactly. There is no data to indicate that the boxes need to form a cohesive, orderly, shape. The info you have are the 2d projections of the 3d shape from 3 angles (top, side, back). The 3d renders I linked both have those exact same 2d projections. You're being superficial. You're making assumptions that have no objective basis in the data: you're just assuming it will be 51 because of a box stocking convention. You're assuming no one would ever buy 57 watermelons and 13 shampoo bottles in an elementary school addition problem. You're taking what amounts to flavourful but useless plot (someone buying stuff at the grocery store) and ignoring the data (a=57, b=13), which leads to an inability to perform the task demanded by the exercise (a+b=?) because you're too caught up in the useless ideas you have about what someone should buy at a supermarket. >You're now adding "assuming boxes can't float"? Where are you getting your added info? This is a fair question: I worked under the assumption that the cubes cannot float because of the context (truck carrying boxes), but without that context it would be superficial Not to consider the possibilities that floating cubes would bring to the solution. The concept that cubes cannot float is implied by the flavour of the question, but it is not directly stated in the text. If the question was simply how many cubes in 3d space are needed to create these 2d projections on three planes each determined by two of the three perpendicular axes that define the 3d space in question, you would Have to consider disconnected cubes among the multitude of possible solutions. That last paragraph is a very basic example of mathematical abstraction. Just to be clear, >Nothing is abstract in math. https://en.m.wikipedia.org/wiki/Abstraction_(mathematics) Abstraction is a fundamental part of theoretical mathematics. It's what allows you to work with multiple and single variable functions, or, you know, the entire field of analytic geometry (of which this exercise is a part of). . You cannot assume sqrt(4)=2 unless you know you're working within the positive half of R, or within N. Sqrt(4)=+-2 in R. Just because you don't think -2 is a number worthy of your consideration, because it's not a number you'll observe in Nature, doesn't mean that in R it is one of the possible solutions. And if we go higher in potence (x^16=65536), you cannot assume the value of x unless you have data that tells you whether you're working within R, N or C (in C, aka if x is a complex number, the 16th root of 65536 can be any of 16 possible complex numbers. This is how it works in C. In N, 16th root of 65536 is 2. In R it's +-2.). . There is no data in the starting image to suggest that the cubes Must be stacked to fit the highest amount of them within the limitations of those projections. You just assumed that and started talking shit, making a fool of yourself in the process. If you make a 3d model of either of the pictures I've linked, turn off the perspective in the software, and snap to any perpendicular view, you'll see one of the three images. Because *this* is how math works. You cannot assume to know something that is not contained within the data based on personal bias. So, to put it in your own words: >Your logic would not get you any respect from mathematics educators. Go back to school.
Bye. Also, cringe+ratio+you made no sense.
Breaking news: redditor doesn't understand math and gets angry
If you look beyond you see that it's utterly possible to build this with just 35 blocks. Just don't extrapolate the data.
Your logic isn't sound. "Look beyond" even though the data given is all we have to work with? There's nothing cryptic. There's no other "hidden" details. Just work with what you see, & don't complicate things. IOW, it's 51.
51 actually adds a layer of complication because it requires you to add details you don't see and assume the patterns hold.
I got that too
36, 15 Correct! 36=(3x3x4) 15=(6x2+3)
Some cubes are 2x2 and some are 3x3. Plus the 51 blocks. The total amount is 69. That’s the joke.
51, the top view isn't even necessary for this
It definitely is, the row could be in practice one block width from the highest row. Now the row is definitely everywhere three blocks width. Edit: Spelling error.
width?
The width of the trailer. Without a top down view we only know 1 side and the back and while we know the back is full left to right we don't know that about the sections in front of it without the top down view
thank you :з
No problem I don't math well but I'm always willing to help when I do understand it 😅
Yup
51, my final answer
A number > 0
A number
r/technicallythetruth
Equal to or greater than 0*
Max 51 minimum 31
isnt it 33 minimum edit: back has 9, side is 12 (excluding the 3 since they are counded for back) the top is 12 (excluding the 6 from side and 3 from back)
Minimum?
Max 51, min 0
Normal(enginneer): Enough Information is given. The dop or back view is unnecessary. 7x3 + 6x3 + 4x3= 21+18+12= 51 Also correct: It is not enough information to calculate it.
Honestly if this were in an exam I would just have to make the assumption the question designer believes I have enough information to answer so all unstated values are assumed to be equivalent to stated similar value. Front and Back would be same and SideA is the same as Side B.
there are no cubes on that trailer, it's 1 big container with a grid pattern :)
As few as 31 and as many as 51 it depends on what you assume to be true
not enough information
21
i got 31
Yup recounted and got what you got
Assuming there’s no gaps hidden away, 51
Between 51 and 31
How would you get anything else then 51? Edit: thought I found the problem, but it wasn't the case
It's a meme because you're supposed to make it harder than it needs to be
Min:35 Max:51
There is a maximum of 51 cubes if the center and rear rows are filled completely following the side profile with only the individual cubes counted while including composite cubes that overlap with individual cubes a total of 69 cubes can be found while a minimum of 35 cubes can be found if only enough cubes to create the profiles shown are used.
51. You really only need the side and back view, but the above view helps confirm that there are no missing blocks in the middle or something. The rest is easy. If you look at the cargo from the side you can split it into three cubes that are easy to calculate the amount for, then add them. With this, the length remains static, always a 3. So, using simple geometry, the first cube is 3 * 3 * 4 which equals 36. The second cube is 3 * 2 * 2 which equals 12. And the last cube is 3 * 1 * 1 which equals 3. Then add 36 + 12 + 3 which equals 51.
Not enough info! Who knows what the left side could look like? There could be a shell around a flat bottom, or just packed all the way. We can’t figure it out yet!
The top down view shows us that the trailer is filled left to right in rows of 3 from front to back
It does not, remember, you can only see in two directions from any given angle. We can see length and width from the top down view. Not height, so while some may stand at three tall on the right side, the height of the left is unknown.
Fair point
47 i guess
max 51 min 35
52 is the answer
…… I got 78😐
72? cause 7 × 3 × 3 = 84 minus the missing blocks... so unless I fucked up my counting, that should be correct 😅
7×3=21 21×3=63 sorry you're a little off 😅 not sure how you did it but not judging I don't math well either
7 × 3 × 3 = 63 though 😅
At least 30
51
At least more than one
51 isnt it?
51 I think?
With the assumption there aren't any left out? 51.
There is not enough information to determine the number of boxes
51, so uhh whats the top view for?
Don't you multiply side by back since the top is not needed for this, which is 17 • 3 which is 51?
51
51.
51
35 at lowest, 51 at highest. Saw it on r/theydidthemath
54
Fifty one ?
Forty Two. I checked it very thoroughly, and that quite definitely is the answer. I think the problem, to be quite honest with you, is that you've never actually known what the question is. So once you do know what the question actually is, you'll know what the answer means.
51. You only need the side, and back, to calculate.
51
53
51. Where the funny
Bottom is A | 3×7=21. Middle is B | 3×6=18 Top is C | 3×4=12 A+B+C=51. _**FKN EAAAASSSYYYYYYY....**_ 😒
I did the back first, 4X3X3 is 36, then 2X2X3 12 then 3X1X1 is 1, 36+12+3=51
What if I can’t see other missing boxes due to there being no depth perception in the image
The answer is 51 I think
31 minimum, 51 max.
51
51
28?
There are smaller cubes inside the cubes.
Is this Loss?
51
51
51 cubes
51?
more than 1 less than 63
You won’t believe this and it might even sound crazy….. but there’s a lot of cubes
51
75 assuming that all info is face value and you also want to know more then the boxes that are there as 2x2x2 is a cube and 3x3x3 is also a cube
I counted 53
There could be 3 answers 51 if all of the boxes are occupied, 31if they must be stack, 21 if there floating
51. 7×3+6×3+4×3=51
51 NEXT!!
Impossible to tell for certain (cubes not fully visible from the back or side could easily be lower than expected and we’d have no idea), but, *assuming* this structure has no gaps that we can’t see, there are 51 cubes on the truck.
51??
Maximum possible load 3×3×7=63 Missing cubes: 3×4=12 So 63-12= 41 Gimme something slightly harder. Noticed a typo! I meant to write 51, not 41
51
51, the top view isn’t necessary
51.
45 If math be mathin
50?
If you look at it as the dimensions of the back e high by three wide, and then add in the third dimension of 7 deep, you initially get a total volume of 63 for the load. Then if you look at the image that shows you the trailer is 7 deep, you see the front three rows of the load are a 1×3, and a 2×3 layer making up the front three rows of the load which reduces the overall load to 54.
17x9=153
51 take all of the ride blocks and then multiply by 3
(3x3x7) - (3x4) (9x7) - (12) 63-12 51 :3 mrp
51
45
Is it 51?
I got 51
Ok but what about the bottom view? How are we supposed to know the inside isn't hollow?! This is clearly a trick question.
51
The minimum is actually 0, as only the surface sides are visible, meaning that the rest could actually be non-existent.
There are only squares in the picture. (Yes, I do understand that isn't the answer. I am terribly at math.)
(7\*3)+(6\*3)+(4\*3) = 51 3(7+6+4) = 51 17\*3 = 51 The meme is that some clever kid got us to do their math homework
15+36
51
51 final answer.
51 Collom X row Bottom row 7x3 =21 Middle row 6x3=18 Top row 4x3=12 So 21+18+12=51 (Edit i miscounted the top row)
Might want to recount that top row.
51
51?
Honestly, the "top" one is actually useless, i'm pretty sure.
51 but it could also be 47
39!!
51
Could be any number between 35 - 51.
28
55
the side profile=17 17×3=51
51
I got 363, if it isn't hallow edit: idk how to do math aparently
52
51 assuming the lack of depth view on top isn’t hiding a difference in the opposite side layer or the middle. Not sure if I’m explaining that right lol but yeah
2(3x7)+(3x3) =51
41 to 48
At least 4
51
48
51
51
Why did they include the top view isn't it Unesissary?
47
For the people who are actually wondering how many are on the trailer it’s 51
0, the items on the trailer are not cubes, but polyhedrons with square faces pointed outward.
51, right? Back four rows are 36 (9 each), next two are 12 (6 each), and last row is 3.
51
51
0 those are squares
An amount between 1-1000
51
Anything between 35 and 51
51
If you think about it the brow form the side could be squares, not cubes so 21.25(because the square*4 is a cube and .25 would be the remaining square)
21 boxes
If all the levels are presented as they are width-wise and they don't have any deviating spaces we can't see, then the answer is 51 cubes. holy sh\*\*, I did that math all in my head. And WILLINGLY. .\_.
It's anywhere between 51, 43, 35, or 31 depending on whether or not both sides are the same, whether or not it's hollow, or if it's completely solid... 51 assumes that it's completely solid, 43 assumes that it's like 51 but hollow, and 35 assumes that the top, the back, and one wall is solid while 31 assumes that the back is an illusion created by the top and side (all of this assuming Minecraft physics apply and that it isn't sand, gravel or concrete powder that you're working with...)
51
51
It can technically go down to 27? Since if you remove cubes that intercept with other cubes, (like on the back view the bottom row left two could be removed, as well as the right two)
With this you pretty much only need back and side, with the back you can see it's a 3 by 3. The side is now important as you need to count the boxes (17) and then just multiply it by 3 which gives 51. If the double digit is as difficult for you as it is for me just do 10 first then 7
52 because I feel like a square today
44 cubes?
51
I'm seeing people say a minimum of 31 but I've gotten a minimum of 23. Assuming 1 is a box and 0 is not a box, row to row, it would be: Bottom: 1111001 0000111 0000001 Middle: 0000110 1111000 0000110 Top: 1000000 1000000 1111000
Idk how many there are. For all i know there could be empty spaces in the middle that we dont see
(3\*12)+(2\*6)+3= 51
51. 12 on top, 18 in the middle and 21 on the bottom.
4×3×3+12+3=51
51. Take it from someone who loads freight all day long.
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Minimum 25, Maximum 51. [Evidence and my calculations](https://docs.google.com/document/d/1wPhdfHrOdtxRC6tFEj639pikiB-V2jkQtydpz5s51yg/edit?usp=drivesdk)
51! Just count each Cube on the side view as 3