Solving a popular puzzle. How many boxes are there?

[@Qermaq]

For those confused: original upload had the title "olving a popular puzzle". It was fixed later.

[@TonboIV]

THANK you!

[@stoatystoat174]

Thanks, i assumed letter S or an S shape was part of the solution and got real confused

[@MindYourDecisions]

I am thankful to the early notification quad for alerting me of that typo

[@Qermaq]

@@MindYourDecisions You are blessed with so many comments, that alone cannot serve as an alert. :D

[@lailoutherand]

​guys i think he did it again

[@hello_hi1]

The S got trapped in a box

[@Becky_Cooling]

51 of them

[@MindYourDecisions]

I am thankful to the early notification quad

[@JustAnotherCommenter]

It wa obcured and wa only viible from the ide view

[@soundsoflife9549]

@@Becky_Cooling 17 rows by 3 =51 unless there is a hidden compartment under the boxes containing the S!

[@KKR-t5q]

From top Black is filled and white is blank. 1st layer... ⬛⬜⬜⬛⬛⬛⬛ ⬜⬛⬜⬜⬜⬜⬛ ⬜⬜⬛⬜⬛⬛⬛ 2nd layer... ⬜⬛⬜⬜⬜⬜ ⬛⬜⬛⬜⬛⬛ ⬜⬜⬜⬛⬜⬜ 3rd layer... ⬜⬜⬛⬜ ⬜⬜⬜⬛ ⬛⬛⬜⬜ Total no. Of black boxes is 21 here . This is lowest possible no. in this case. ​@@MindYourDecisions

[@Hokiebird428]

Lesson: Orthographic views are good for dimensions, but you need to include at least 1 isometric to give the full picture. And show your hidden lines!

[@TonboIV]

Engineering drawings always require making certain assumptions and following conventions to interpret them. That's in common with any system we have for understanding reality. Everything is a simplification involving assumptions and agreed conventions, unless you want to try to stack assemblies of fundamental particles using quantum theory (which won't work anyway because of the quantum gravity problem and Heisenberg and various other problems). Presented with just those three views, the common sense assumption would be 51 boxes. Otherwise you should do a drawing for each layer and eliminate all doubt.

[@Gideon_Judges6]

This is not realistic anyway. If these were really 3d perspective, even if orthographic, the layers further from you would look smaller. You would be able to tell.

[@xenontesla122]

@@Gideon_Judges6Orthographic projection ignores perspective

[@nevarmaor]

You can get the full picture without an isometric view if you have six orthographic views. Drawings omit the views that do not add further information.

[@nevarmaor]

@@Gideon_Judges6 Orthographic views are 2D.

[@brandonvillamizar1216]

Don't forget to olve the puzzle by yourself, and then, watch the olution.

[@caput_in_astris]

That’s a mart comment

[@jo_hauser]

That' *

[@MindYourDecisions]

I am thankful to the early notification quad for alerting me of that typo

[@DAMusic-qu2ec]

I ink here’s somhing wrong wih your s key

[@acarbonbasedlifeform70]

And don't forget to mind your deciion

[@stromboli183]

If the boxes are filled with helium, they don't fall and you can have an empty spot under a box. This allows for several more boxes to be removed. Alternatively, the boxes underneath can be slided to non-integer grid positions so two boxes can carry three.

[@BobfromSydney]

Good thinking, in order to not affect the lines showing the edges of the boxes we could only remove the ones in the middle row underneath the second row. So that should allow another 2 boxes to be removed.

[@neuralwarp]

Or ½ of them if they need to be supported.

[@thunkwolfman1561]

bro said olving

[@RobertMeredith-sv6dd]

At 6:12 you can actually remove 2 more boxes. You can remove them from the corner to get down to 33 boxes. It's still not the least but when you said "no more boxes can be removed from this configuration" I just saw that 2 more actually could. Nice video though. I love your content!

[@Archy_The-Wizard]

It's even possible to go as low as 21 boxes if you ignore gravity.

[@mzrhub]

Exactly what i thought. Glad i found someone who thinks like me

[@derToblin]

That was my first guess as well, but I couldn't find a way for it to work. Did you?

[@anyajenkins4611]

i can go to 24 without gravity, but 21?

[@Archy_The-Wizard]

​@@anyajenkins4611 @dertoblin So here's how you can do it: Bottom: XOOOXXX OXOOOOX OOXXXXX Middle: OXOOOOO OOXXXXO XOOOOOO Top: OOXXOOO XOOOOOO OXOOOOO

[@SirRebrl]

@@anyajenkins4611 Without gravity, there's a 4x3 section where 2 boxes can be lifted from the bottom layer to satisfy the side and back views. It's kind of a sudoku-ish problem, where you need three collections of unique arrangements occupying the same space on different axes. I built it using little cubes that can be stuck together, with one color representing boxes and another color to bridge empty space. It's pretty easy with a good tool like that.

[@headcrab4]

Small nitpick, but in the 35 box solution you showed, could you not remove the top and middle boxes from the top left corner of the truck? Those removals shouldn't effect the final views, i dont think.

[@jesss.5864]

Correct, I agree. That would be a 33 box solution in that configuration.

[@ShawnF6FHellcat]

My first thought with these puzzles is "Assuming I take it at face value, then the answer is ____, but technically there are xyz number of answers"...

[@noc8076]

Lowest number is 1. If you are dealing with cargo thieves, there is no reason there wouldn't be one box with lines drawn on it.

[@Nathan_Avril]

Lowest number is 0. The orange thing you can see isn't a box at all. It's just a big chunk of orange rock with lines drawn on it.

[@u2bear377]

@@Nathan_Avril Why rock, just three carboard panels with square pattern.

[@ahmadmazbouh]

my favorite hobby in my free time is olving puzzles I love olving

[@ThomasNimmesgern]

Who doesn't?

[@erikziak1249]

After looking at the picture: Since it is drawn without perspective, we do not know if there are any "voids", like the middle row of lower boxes having lower height, because it is indistinguishable from the image. If I were to assume that the whole thing is packed "fully" as the views suggest, then the number of boxes is 51.

[@taflo1981]

Assuming we need cubic boxes (so using one large box instead of several smaller ones in a place we can't see is not allowed), each box is of uniform density and rigid enough so that it can support itself (and possibly another box on top of it) as long as its centre of gravity is placed on another box or two opposite sides are resting on other boxes, I can come up with a solution that uses 30 boxes in total. For notation, let's say that the side length of a box is 1 and the grid we see from above consists of the three rows A,B,C and the columns 1-7 (with 7 being the rightmost column that doesn't need any boxes in the middle or top layer). In the bottom layer, put all boxes in rows A and C (with A the topmost row and C the bottom row), plus boxes in squares B1, B2, and B7. In the rectangular area from B3 to B6, place three boxes with distance 1/4 between each other and the adjacent boxes in squares B2 and B7 (i.e. the leftmost box is occupying 3/4 of square B3 and 1/4 of B4, the next one uses half of each B4 and B5, and the last one is 1/4 on B5 and 3/4 on B6). In total, that's 20 boxes in the bottom layer, 6 in row B and 7 each in rows A and C. On that bottom layer, stack two boxes each on squares A1, C2, B3, and B4 (8 boxes). Finally, place one box each on squares B5 and B6. This makes 10 boxes stacked on top of the bottom layer, giving us 30 boxes in total. The boxes stacked on A1 and C2 rest safely on a box right below them, while the ones stacked on B3 and B6 rest on 3/4 of a box below them (centre of gravity safely above said box). Finally, the boxes stacked on B4 and B5 do not have their centres of gravity above a box (just above the edge of a box), but they rest on two boxes each that support their left and right parts, respectively. So this construction would physically be able to support itself. Finally, let's check that the three views are as described. (a) In the side view, we see all seven boxes of row C in the bottom layer, the boxes in the middle layer in squares A1, C2, and B3 to B6, and the boxes in the top layer in squares A1, C2, B3, B4. This is the side view we want. (b) From the back, we see the boxes on squares A1, B1, C1 in the bottom layer and the ones on squares A1, B3, C2 in the middle and the top layer. Again, this gives us the desired view. (c) From the top, we see the boxes in the top layer (A1, C2, B3, B4), two boxes in the middle layer (B5, B6), and all 15 boxes in the bottom layer that are in any of the squares apart from those six just mentioned. In particular, the three boxes that do not align with the grid are blocked from our view by the boxes stacked on top of them. I can't prove that this is the minimum (under the stated assumptions), but no approaches I tried for reducing the number of boxes any further have worked. Shifting the centre of gravity of a box extremely close to its boundary will enable solutions with fewer boxes.

[@ianmalcolm2552]

Another terrible phrasing of a question. The answer is that it’s not determinable from the information provided. How about, what is the minimum number of boxes on the flatbed?

[@eyozin]

Told my wife the same when she asked me when I got home last night. Didn't work...

[@GemingB2]

He doesn't state the answer is a single number tho

[@ianmalcolm2552]

@gemmingb2, proving my point. Multiple correct answers.

[@neuralwarp]

A set or range is a valid solution.

[@ianmalcolm2552]

@neuralwarp, the question as phrased is “how many boxes are there?” not what is the range or set of possible solutions. And therefore, anything other than a single number would not be considered a valid answer.

[@nmrs4342]

I can olve this too

[@silviamayo2993]

6:08 Couldn’t you take two of the boxes from the layer at the back? The top right and middle right ones from the back perspective? Since the side layer should stop the back view from changing

[@silviamayo2993]

That doesn’t change the answer of 31-51 but still

[@TheTriomo]

Yep. My first solution too. But it's just 33, not minimum.

[@IljaMuromec114]

And cant you also take out most of the bottom layer from the side? Because that could allow you to remove 7 extra boxes from the 35 making it 28 boxes

[@TheTriomo]

@@IljaMuromec114 No. Bottom layer must be filled completely.

[@jasilos9109]

7:20 you could remove the boxes from the bottom layer on which are holding the upper boxes in middle and so the upper boxes do not fall you must place underneath them some kind of support. If you do that the number of boxes goes as low as 27 boxes and 4 supports that aren’t boxes

[@derToblin]

Disregarding gravity, the minimal number of boxes would be 25, because we can remove all base layer boxes with boxes on top of them from the 31 boxes solution. This solution can also be achieved using glue or strong magnets.

[@konstantin.v]

21 😎

[@derToblin]

@@konstantin.v That was my first guess as well, but I couldn't find a way for it to work. Did you?

[@konstantin.v]

@@derToblin , I believe so. It's purely in my head, though. Needs double-checking 🙃

[@pangronostaj7080]

@@derToblin I did (sorry that i am replaying after month, just watched this video). I don't know what is the best way to present you my answer here so i'll wrote down layer by layer arrangement of boxes (B) and empty spaces (X) from top to bottom: top layer: B X X X X B X X X X B B back view after this one: B B B X X X X X X side view after this one: B B B B X X X X X X X X X X X X X top view after this one: B X X X X X X X B X X X X X X X B B X X X mid layer: X B X X B B X X B B B X B X X X X B back view after this one: B B B B B B X X X side view after this one: B B B B B B B B B B X X X X X X X top view after this one: B B X X B B X X B B B B X X B X B B X B X bottom layer: X X B B X X B B X X X X B B X B X X B X B back view after this one: B B B B B B B B B side view after this one: B B B B B B B B B B B B B B B B B top view after this one: B B B B B B B B B B B B B B B B B B B B B So all views filled. It is one of the possible answers. There are other possible solutions.

[@neosmagus]

Logically I would say you have to, at minumum, fill the bottom layer. This is to satisfy the top view and will fix the issue with floating boxes, if we assume we cannot have floating boxes. This is 21 minumum. Then to satisfy the back view, you know you have to add 6 additional boxes, 3 per layer, in some orientation. 21 + 6 = 27. Finally you have to satisfy the side view. The top two rows have 10 boxes. However we can subtract 6 boxes that we placed to satisfy the back view, leaving 4 boxes to still be placed. 27 + 4 = 31, which gets to your minimum answer. However if we drop the assumption that we can't have floating boxes, you have to again start with 21 boxes to satisfy the top view, but these can be orientated in any layer, meaning that the side view and back view are subsets of this 21, therefore the minumum is 21 boxes.

[@DonaldRoy-nr9xe]

If gravity exists for this problem, then the range of boxes required to satisfy is between 31 and 51; but if it is not stipulated that gravity is present as a rule, then the range is between 21 and 51. The only 3 boxes that don't change in this differential is the "front" 3 (on the top left diagram labeled "side view") being the ones occupying the right side. The rest have the configuration of: X =X = X = X= (123) X = X=X = X = X =X X X= X = X=X =X X= X =XXX (7 * 3 = 21)

[@jarrodkopf6813]

I actually solved it by calculating the volume of the boxes on the flatbed since the boxes would have formed a rectangular prism. Then, I subtracted 12 from the total, since there are 12 boxes missing from the prism to get the answer. In other words: V=l*w*h=7*3*3=63 63-12=51

[@yurenchu]

Here is another fun puzzle somewhat related to this: The three primary orthographic projections ( _top view_ , _front view_ , and "side" view/section view) of a 3-dimensional shape each show a profile that is a _circle_ with diameter R . Furthermore, this shape is the convex 3D shape with the largest volume that satisfies these three projections. Question 1: What is the volume of this shape (in terms of R)? Question 2: What is the surface area of this shape (in terms of R)?

[@ChristopherWanha]

At pause the video, my first impression is that this is a voxel and there is not enough info to give an exact answer. But you can give a range that the answer has to be assuming boxes are not floating, or have scaffolding/shelves under them.

[@Silas_MN]

I got 33-51 inclusive as my answer before watching the whole video, reaching 33 by doing the 35 solution and then removing the top two boxes from the back corner. if I'd thought it through a bit more, I would've reached the same conclusion you did

[@57thorns]

Minimum? 21 for the bottom layer as shown by the top-down view. +6 for row 2 and 3 (row 1 is the bottom) as seen from behind. +8 that are needed to fill up the side view 35 Maximum? Three time the side view, si 3x17 = 51

[@57thorns]

I forgot to stagger the two layers in the rear. That would remove another four boxes for 31.

[@gaijininja]

I can get down to 27, but ONLY if the blocks lock together. ie. Surrounding boxes hold the ones above. In the 31 box answer, take out the four boxes in the middle row of the bottom layer, (1x1 and 3x1) allowing the blocks around to support the ones above those removed. Legitimate mechanical answer, but may not be permitted mathematically as points and lines are infinitely small, so vertices can not technically support adjacent vertices.

[@BCS-IshtiyakAhmadKhan]

We can do it in 17 boxes also ,the bottom layer doesn't need to contain 21 boxes we can reduce bottom layer to contain only 7 boxes

[@gaijininja]

@@BCS-IshtiyakAhmadKhan Bottom layer has to have at least 16 blocks. All those not covered by blocks above them.

[@BCS-IshtiyakAhmadKhan]

@@gaijininja consider this arrangement [] [] [][][][][] The above arrangement only for bottom layer(Top View).Now of you see it from your mobiles left side it will show you 3 boxes and from your keyboard side it will show you 7 boxes.Now above these boxes you can place your rest of 2 layers

[@olive-robin]

@@BCS-IshtiyakAhmadKhan the top view has 21 boxes though. If you go below that then how can you get the top view correct?

[@gaijininja]

@@BCS-IshtiyakAhmadKhan But in the arrange you show would leave a lot of empty spaces when viewed from the top, even if blocks can stack with only vertices touching. (ie. No support directly underneath.) The bottom layer must have blocks in spaces not covered in the top view by the layers above. In only end and side view, you solution works though.

[@tungpham9219]

Minimum 30 boxes; in the first layer we need only 20 boxes. Two boxes (instead of 3 boxes) in this layer can support 3 boxes in the middle row in the second layer @ 7:38 in the picture on the upper left corner. In the first layer you will then see 3 gaps in the middle row, but not visible from outside.

[@dbumjr]

Am I the only one who's tired of tupid worded questions that allow the olution to be any answer between 31 and 51. I can't wait to see that Simi truck driving down the road with that awesome 31 box tack job😉

[@jonmccormick6805]

And a lot of glue holding everything together.

[@Barreloffish]

They would be fired.

[@Hypericus2]

You can get solutions with less than 31 by concealing voids underneath the central boxes. In the top left diagram of the solution, swap the second and third slices from the rear to give a horizontal block of four, and space three boxes underneath them to give a robust 30, or 29 if you rely on gravity to hold the middle two up when the four are balanced on just two. Of course if the boxes can levitate (or are held inside a clear block of ice or glass) then there are solutions as low as 21.

[@mecadragoon]

thats not how you should load a trailer i hate this

[@kenmore01]

Just hot melt glue them in place!

[@jawstrock2215]

between 35 and 51 :) max is 3*7+3*6+3*4 = 51 Min is 3*7(top view)+6(side row)+4(side row) + 4(to finish the back 9) = 35 After seeing his answer, yeah I admit I forgot you could have column put in zigzag to fill the back view requirement

[@michelpi5675]

7:42 In fact, in combinatorial optimization, it often happened to have a lower bound on the problem we want to minimize and at the end, discover that we can reach it and this is a mathematical proof that the solution is optimal

[@ProactiveYellow]

Observing that the "maximum load" of the trailer is a cuboid of dimensions 3×3×7 we get a full total of 63 possible boxes. Because the side view shows 4 missing rows of 3 boxes each, we get a maximum of 63-12=51 boxes on the trailer. However, we lack any sense of depth from these images, so we may be missing some boxes on the interior. Counting carefully, we know that, by the top view, the bottom layer has a total of 21 boxes, and that, excluding the bottom layer, the back has a total of 6 more boxes. Now excluding those already counted boxes, we see the side showing 8 more boxes for a minimum total of 21+6+8=35 boxes. Because we cannot see any depth in these images, we must conclude that the number of boxes N fits the bounded inequality of 35≤N≤51 for some natural number N, and any natural value within that range is possible.

[@chinareds54]

Most: 17×3=51. Least: 17 in one of the three left-right columns + 14 in the bottom of the other 2 columns + 4 in levels 2 and 3 in one row of the other columns = 35.

[@olive-robin]

The assumption you make at 7:20 is quite interesting. That there needs to be a base, otherwise the boxes will fall. Which would include gravity and that there cannot be any other mechanism for vertical gaps. Such as hidden metal supports or adhesives. I didn't make this assumption. Even if I assume they experience gravity, they could meet the three view points while falling. It just wouldn't be a permanent state. You've also presumed the shape of the truck bed which isn't visible. Which is the same presumption that gives you the maximum number of 51 when applied to the orange boxes.

[@ejreob]

If the boxes are allowed to be non-cubes you can replace the middle seven boxes from the bottom layer in the 31 box solution with one 1x1x7 box and it should look the same from the top.

[@johnburgess2084]

Here's another "hypothetical" answer. The bottom layer need not be completely filled with 21 boxes. You can remove some middle boxes leaving a hollow space the size of some boxes. If some of the upper boxes are massless, they can sit above the hollow spaces and the air in the hollow areas will support boxes above them (like a piston in a cylinder). I didn't actually count the minimum number of boxes in this configuration, since it's sort of cheating. But, in so many puzzles, we assume "ideal" conditions in which there is no mass, no friction, etc. So, in the ideal situation, with some massless boxes sitting on a cushion of air, you could end up with fewer than 31 boxes.

[@masterkj0912]

or it could have stuff that simply isnt a box propping it up

[@MrValdesbg]

Any 3D modeler would know they have to look for more pictures online or in real world. Building anything from 2D blueprints alone leave a lot of room for speculation. I knew 51 was the most number of boxes but I knew there could be fewer than that.

[@kolega881]

technically we could get a smaller number than 31 as a minimum, we could remove the boxes from underneath the boxes of the middle layer in the 31 solution, meaning that technically they would be suspended in the air, as nowhere does it state that they cannot be in the air (if it does then i'm blind), and we can suspend them by rods or something, as a concept it would be like this a-air b-box 1st layer: a-b-b-b-b-b-b b-a-b-a-a-a-b b-b-b-b-b-b-b 2nd layer: b-a-a-a-a-a-a a-b-a-b-b-b-a a-a-b-a-a-a-a 3rd layer: b-a-a-a-a-a-a a-b-a-b-a-a-a a-a-b-a-a-a-a this makes it 26 boxes, it doesn't work in reality (putting it on a truck), but as just a conceptual answer it does seem plausible. if we look at it purely graphically, it makes sense, but not otherwise. as in all of those views, it will look identical, so this is my answer, 26 boxes somehow packed on the truck, is the smallest amount possible to get. (i may be wrong) (if someone made the same comment or concept then i didn't see it)

[@yurenchu]

With your approach, it could even be done with just 21 boxes. (The 21 boxes are all seen in the "top view".) For example, the following solution: 1st (= bottom) layer: a-b-a-a-a-a-b b-a-a-b-b-a-b a-a-b-a-a-b-b 2nd layer: b-a-a-b-b-b-a a-a-b-a-a-b-a a-b-a-a-b-a-a 3rd (= top) layer: a-a-b-a-a-a-a a-b-a-a-a-a-a b-a-a-b-a-a-a

[@pierreardouin6441]

[ 31 ; 51 ] For 31: the whole bottom floor has to be there frop the top view, for gravity reasons, then you put the remaining boxes to match both side views, first guess would be filming a L-shape only, lile this: ####### # # but you can be even mote efficient using a diagonal for the upper part: ##### # #

[@rikschaaf]

I initially thought it was 35-51 before hearing the answer, but once you mentioned being able to go lower, I realized that some boxes could do double duty. If you remove the assumption that a box on the 2nd layer needs one below it (because they attach to each other at the corners or something), you can actually go even lower still.

[@Fedra3443]

I remember a guy said that it had to be 51 boxes (no other answer was valid according to him) and his "proof" was that he worked drawing the different views for objects. Many people told him that he sucked as his job

[@felipevaldes7679]

I'll explain why zero: the truck could have the bottom part of the truck conceivably have been painted to look like boxes, and the back of the truck could also have been painted, and also, consider that the truck could conceivably be next to a WALL (or another truck) which has a painting the boxes with the pattern described, therefore, the min is zero and the max is whatever you want because we don't know if the boxes themselves have boxes inside of them, so this 31 or 51 is nonsense, the answer is we don't know how many tiny teensy boxes are inside the larger boxes, and we don't know what the truck was painted like. I would also like to point out that we don't know how many dimensions the truck is made of, consider a 4d truck...

[@WowOafus]

@5:04 you can still remove the corner box on the top row and still have the correct layout.

[@Dysonnnn135]

5:56 you can still remove the corner and the middle box from from the corner

[@urbanbroccoli]

30 or even 29 is possible under some assumptions. Notation: top down view, 7 * 3 matrix of box counts. Columns 1..7, from left. Consider the following layout with 31 boxes: 3 1 1 1 1 1 1 1 1 3 3 2 2 1 1 3 1 1 1 1 1 Note the run [3, 3, 2, 2] in the middle row. The middle layer of 4 boxes in this run could be resting stably upon 3 boxes "spread out" in the bottom layer, for a total of 30 boxes. The gaps are concealed from the side view by the outer rows. Assuming a sufficient degree of rigidity and friction in the boxes, the middle layer of 4 boxes could even be resting on **2** boxes, spread such that the center of gravity of the column 3 and 6 boxes are above them, and the column 4 and 5 boxes are "perched" with centre of gravity above the gap, but resting together and unable to fall. This solution satisfies the visual requirements with 29 boxes.

[@ironcito1101]

You could get to fewer boxes if you use something else to fill spaces and stack boxes on top, so that boxes on top layers are not on top of other boxes, but on top of some other support. Not realistic, but possible.

[@lolishocks8097]

My answer would have been: You cannot determine the exact number of boxes, unless you assume the puzzle is honest. But what would an honest puzzle be? Even if you were to replicate this puzzle in real life, you still could not determine the exact number of boxes without making assumptions. There will always be a range of plausible solutions.

[@jesss.5864]

I love these logic puzzles! I first said “Well they want me to say 51, but the real answer is 35” and when you said “31” I was pleasantly surprised! I had to pause the video and figure it out again!

[@kokilabendamor7615]

Thank you for olving this popular puzzle! I was able to solve it, but i didn't know how to olve it. Thank you!

[@brianloch]

I don't think it said all boxes need to be the same size. You could reduce the box count by 4 (total of 27) by using four boxes that are equivalent to two boxes in height down the center row.

[@tannerarmstrong1496]

You said at the 35 box solution you couldn't remove any more boxes from that configuration without changing the views. That is not correct. You could have removed the two corner boxes in the back without altering the views to get to 33 boxes. From there, it might be intuitive that you could push the middle back stack forward 1 spot to allow you to remove another 2 boxes from the new back corner to get to 31.

[@tannerarmstrong1496]

Your reasoning was that the remaining boxes were all facing either the side back or top, the better metric would be to look for any boxes that are not supporting boxes above them, and who's x, y, and z coordinate are all occupied by at least one other box.

[@gavindeane3670]

​@@tannerarmstrong1496That's kind of what I thought. I got stuck on the minimum of 35. My tired brain couldn't see a better solution, but I had a niggling feeling that I still had quite a few boxes that were, at least from some directions, obscured by other boxes. It felt a bit inefficient because of that.

[@kmp8563]

Wouldn't there be boxes magically floating in the air for that solution?

[@gavindeane3670]

@@kmp8563 No. With the 35 box solution, go to the back-right corner of the trailer. In that corner is a stack of three boxes. Take the top box and middle box from that stack and the views remain unchanged.

[@baivulcho]

You can actually remove 4 boxes from the back and bottom in a way that will allow you to remove even more boxes. Consider the back view, and assign the bottom left box coordinates of (0,0). You can remove boxes (0,0),(1,0) and the two boxes behind them. Then push boxes (0,1) and (1,1) one row back, so that the top voew doesn't get changed. Now let's consider the side view and assign coordinates 0,0 to the bottom left box of the side view. You can also remove the top-left box(0,2) from the side view, for a total of 30 boxes. You can remove (1,1) as well, for a total of 29. Now if you push boxes (2,1), (3,1), (4,1), and (5,1) one box back, this will allow you to remove the boxes below them, and so hollowing out the mid-bottom section, for a total of 25 boxes. Finally, you can push boxes (1,2), (2,2) and (2,3) two boxes to the back, this will place them to the top-left on the back view and will allow you to remove 3 boxes below them, for a total of 22 boxes. I can't think of any more removal, and yes, this will leave a lot of boxes floating in the air.

[@osmanbadroodin3215]

6:09 You an still remove 2 boxes in this configuration that doesnt affect the views The 2 blocks on on top of the bottom left block (the 2 on top of the first block that was placed)

[@cameronwebb7964]

My answer is 51 because of (7×3+6×3+4×3=51) following Bedmas.

[@golddddus]

Potemkin village solution. 1 box is cut into 31 parallel blocks 1/31*31*1/31. And then they line up to look like solid cubes. So the answer is 1 box, possibly 0 box. If you don't take a real box, but a cardboard that is colored like one side of the box.😎

[@brickworkorange2811]

Perhaps it would be possible to remove some boxes in the base layer, If the boxes in the second layer stand solid on half of a box. So the minimum solution could be 29 or 30, because you need only 2 or 3 boxes under 4 boxes in the second layer, second layer

[@aounelias]

could be done with 30 boxes where you remove 5 boxes from the middle row of the first level. then you stack on the sides 6 and 4 boxes for the side view and 2x2 boxes on the rear view. total 30 boxes 🙂

[@tychozzyx9439]

0:45 At least 35, possibly 43, guessing intended is 51

[@neuralwarp]

This is maths. We only guess (conjecture) if we can verify.

[@tychozzyx9439]

@@neuralwarp This is language and it's indistinct what people want. If this were an actual truck you would have a little paralax between ends of the truck and safe loading would restrict holes in the center. The full 51 would be the most real answer and what I would expect from someone I would give this question to

[@popvoid]

Without any depth perspective, I think you could create the same results with 35 boxes. 17 base boxes, a row of six, a row of four, and a square of four more boxes atop the base row at the back (or on any of the first four columns, really).

[@talhaaftab1135]

If the boxes could flote There could be less boxes in the bottom layer beneath the boxes on the second layres

[@dvorszkydavid9140]

Back in high school, my math teacher said that "There is not enough information to answer the question" was a possible solution, so that was my first answer. If I were forced to give one number as a solution, I would use Occam's razor (the simplest explanation is usually the one closest to the truth) and say 51. Otherwise, the solution is between 31 and 51.

[@InfDev_Yt]

I mostly just do it like if the first picture has 3 boxes and and the second one has 5 that means that 7 boxes will be the third one Which i assume if x=+2 How many boxes will be in the Seventh Picture?

[@acediamond5399]

Pausing to give my answer: anywhere from 31 to 51 (so long as there is no trick by which the boxes can be held up against gravity other than by being stacked on another box).

[@emptystudent]

Hmm I wonder how I hould olve thi one

[@DiamondTMZ]

hould*

[@clawag]

thi*

[@weibrot6683]

In the first solution, what would prevent us from removing the 2 upper boxes in the top left corner? It wouldn't affect anything

[@BCS-IshtiyakAhmadKhan]

We can do it in less than 31 boxes since we are only concerned with projection we can also remove boxes from bottom layer

[@michaelobrien1106]

And here I thought you were going for the different ways of visualizing/counting the boxes! This is a great extension to the classic version of these problems.

[@PB1997.]

i love puzzles when anyone else solves for me.

[@jewi71]

30 Boxes is also possible. In the second layer there are 3 boxes next to each other in the middle. It is possible that there are only two boxes under these 3 boxes ;-)

[@MisterVercetti]

Not enough information. All you know for certain are the following ("row" refer to the 3-box line from the front to the back of the truck): - The truck conit of a 3x7 grid of boxe, where each grid cell repreent a tack of 1-3 boxe. - The three tack in the front row can only have one box each. - The tack in row 2 and 3 can have, at mot, two boxe each. - The tack in row 4-7 can have anywhere from 1-3 boxe each. - At leat one tack in each column ha three boxe. Therefore, the anwer can range anywhere from 35-51 boxe, ince under the above rule there will alway be ten column whoe quantity of boxe cannot be definitively determined.

[@somanken]

I got 35 instantly and didn't realise it could go lower, nice when the simple ones still make you mind your decisions hehe

[@BrianFullerton]

Where did it say the boxes needed to be the same size? Can get lower by having long boxes strategically and partially surrounded/ bridging gaps/ empty positions in the first layer while achieving the same outward appearance.

[@Phonophobic]

If it doesn't stipulate that all the boxes are perfectly square, couldn't you have a 4x1 rectangle box on the bottom layer, centered in the shorter dimension and get down to 28? And possible another rectangular box in the center of the second vertical layer?

[@probablysatire886]

6:10 You could actually remove the 2 boxes in the corner of the wall without changing the view. Having a face exposed does not make a box unremovable.

[@Tolbens]

3x3x7=63 boxes If the truck was full minus 9 (3 at the middle row plus 6 at the upper one) = 54

[@thespyfromteamfortress2656]

35-51. We don’t know if the shape is full, so the real number could be anywhere between those numbers.

[@neosmagus]

21 - 51, if we remove assumptions like boxes have to be supporting other boxes. 31 if they are supporting each other. You can still remove 4 from your solution while reasonably satisfying all the views.

[@yuryp6975]

At around 6:20 in the video, when the top 2 layers are L-shaped, you can remove 2 corner boxes andcetiol have correct views

[@garymucher4082]

If you are going to manipulate it, why not gut the entire center boxes so only the edge and end are left...

[@tonyennis1787]

Derp, good one. I got 35. For some reason it never occured to me to do the thing from the back as well as the side.

[@ybokors8524]

6:10 wrong, you can remove another box from that configuration, namely the top corner box.

[@Aderon]

I olved the upper bound pretty quickly, but the lower bound is much lower than I thought it would be.

[@mavairick]

From 0:46, The number of boxes could be between 35 and 51, If we assume that the truck is balanced and evenly loaded, and that from the side view, there are always 3 boxes, no holes, then it's 51. If the truck is possibly not balanced and evenly loaded, then we because the top view doesn't show height there is a possibility than from the side view, the front boxes are not hiding other boxes. in which case there are 16 boxes that we can't guarantee, therefore in worse case, that is 35 boxes. if my maths and logic were correct of course. edit : I indeed, missed the diagonal trick. nice one.

[@cd31000]

You could still achieve 31 and be symmetrical if you take the assumption that the boxes that are not immediately visible from above or the back are actually 3x1 boxes. This would mean that the entirety of the top layer is small square boxes while only the back two rows of the second layer, and the last row of the bottom layer are 1x1 boxes.

[@ghqebvful]

So for the final minimum count you get, I agree that a box requires a box below it to support it, no magic floating boxes here. However, why would the entire bottom need to be filled out? Any box on a higher tier would require a box below it but many of the boxes on the bottom tier don't support a higher tiered box and don't contribute to the side, top, or back views. I think we can slim the numbers down even further.

[@gavindeane3670]

Look at the top view again. Every box on the bottom layer, if it is not supporting higher boxes, is contributing to the top view.

[@ghqebvful]

@@gavindeane3670 Ah, youre right

[@1a1u0g9t4s2u]

A problem very similar to this was discussed in my high school drafting class. The lesson learned was the importance of the isometric view.

[@mouradbelkas598]

Thank you. The top view will look different if the behind boxes are removed. Therefore 51 boxes is the correct answer

[@JohnGisMe]

I'm going to put my answer here before continuing the video by saying it is between 41 and 57 boxes. However, if boxes could float, it could get as low as 34. This assumes there aren't boxes inside of boxes, different sized or non-cube boxes, or any boxes hidden in the cab or somewhere else nearby.

[@operez1990]

There are 2 answers because there is a limited amount of info provided by the visuals. The minimum amount of boxes possible with this information is 41, the maximum is 51.

[@Aqua2D]

the lowest is 25, u can remove the ones on the base that are covered by the protrusion

[@JaimeWarlock]

If the boxes can levitate, you can remove four boxes that are in the bottom middle and under another box. Gives a total of 27.

[@streq9199]

This is a really humbling channel. Fun puzzles and great visual explanations for the solutions. Subscribed!

[@alaster555]

Their are multiple ways to generate those 3 views. As such, I have deduced that the number of boxes is in the inclusive range of 35 to 51. That being said, the more boxes used will result in a more stable stacking method so the intended answer is likely closer to the maximum.

[@neuralwarp]

Remove ½ the boxes in bottom layer hidden by middle layer.

[@zhangbill1194]

35 min, 51 max. I dont think its possible to know the exact number Ahh, diagonally stacking boxes can get you even lower, cool puzzle I think the question should be how many boxes minimum does it take to make this shape, because that has a definite answer

[@jjcc2221]

but these are what if/depends scenarios. We can only work with what is given. The question would have include asking about a range, not a single answer. It just makes it frustrating. Or maybe I am just not good at these types of puzzles 😂

[@BobfromSydney]

Before I watch the video, based on the thumbnail there should be between 35 and 51 boxes, since there is no way to tell the "depth" of each stack based on the pictures.

[@jewi71]

That was also my result

[@ZuriekJT]

You got me. My first answer was a range from 35-51. You win this time...

[@simonvh7092]

5:03 you can remove two boxes from the corner still...

[@eds1942]

All I’m saying, is that the manifest and invoices better match with however many boxes you’re claiming.