Viral logic test from Cambridge

This is a fun logic puzzle and one of the easier questions in Cambridge's entrance exam (source of puzzle: told by email).
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[@jamesparochetti5279]

I got it! Took the approach of assuming that one grid was the correct grid, and compared to the others. I was able to get E by elimination of the other grids. I guess I should apply for Cambridge.

[@Bibibosh]

Don't assume nothing!

[@hypercubemaster2729]

@@Bibibosh That statement is a double negative.

[@kaikyc.5072]

@@Bibibosh Assume everything!

[@applmango]

@@kaikyc.5072 Not exactly... Don't assume nothing means you shouldn't assume that there is nothing This is different from assume everything In languages, a double negative is not a positive

[@archturusdeydas3989]

@@applmango lol I feel like he meant to say don’t assume anything

[@jimmyc3238]

Hey, I got it!

[@nurdauletturar9596]

and I did in less than 15 seconds!

[@无聊的一生]

Me too

[@AalbertTorsius]

Congratulations! You're in Cambridge!

[@Shreysoldier]

Yea i got it within a minute

[@quax]

Yeah, that was pretty easy.

[@AWOOGO]

I went about it sightly different. I saw a,b,c and e had 4 squares the same. So safe to assume those 4 squares are correct. Then I looked at Cs 2 other squares. C is gurenteed to be wrong as its missing a square the other 4 have. One of them is not repeated anywhere which means this square is wrong as we know C cant have 2 wrong squares. The other is repeated in e so E must be the correct answer.

[@lmp4x]

did it ecaxtly the same way

[@katherineh9432]

Yessss thats what i did too

[@todanyeah]

Exactly the same way!

[@asimkoirala8074]

Same here 😊

[@MrPuzzleCodes]

Yep thats how i did it

[@kindreon]

I thought the problem meant 1 grid has a property correctly, and all other grids are 1 square away from that property. In this case, B is symmetric along the minor diagonal, but the others are 1 square away from being so. 🤦‍♂️

[@keyboardtalk]

I thought the same

[@dj_laundry_list]

I think that's a totally reasonable interpretation of the problem, and even highlights a problem with questions like these. Institutions are selecting for people who think in the same way they do instead of seeking to diversify

[@blablablubb1]

Similarly, you could say there is a rule that shaded neighbors are only allowed diagonally. Again, B would be the only one.

[@r-prime]

@@dj_laundry_list umm you say that as if this is how Cambridge actually selects their students... As someone with an offer to Cambridge for Maths, which I'm hoping to match, the questions I'll be asked for my entry will have much less to do with whatever this is and much more to do with actually challenging maths... For example I had a question concerning the evaluation of integral_0^b (sec(x)cos(b)+tan(x))^n dx, as one part of a 3 part question of 6 questions to choose of 12 given in a paper... Not exactly a spot the difference

[@dj_laundry_list]

@@r-prime Excuse me for believing the title of the video

[@alphabyteseeker]

Another approach. C is the only grid that does not have even one row or column that is completly blank. In grid A, the second column is blank. In grid B, the second column is blank. In grid D, third row is blank. In grid E, second column is blank. That leaves us only with grid C in which no row or column is blank.

[@thomasaskew1985]

This one blew my mind. I had no clue.

[@EvsEntps]

The 'solving' of this puzzle really hinges on 1 word: 'wrong'. If they had written 'different' instead of 'wrong' then the instructions would be very obvious and most would solve it quickly. The use of 'wrong' however means the solver has to intuit that 'wrongness of square' (in any pattern) is a quantity relative to the 'correct' pattern. It would be unfounded, however, to make this leap, since you could have a situation where 'wrongess of square' is a quantity relative only to a property unique to its own pattern. In fact, 'correct' and 'wrong' are typically meta-level terms pertaining to the objective of a question, in effect just artefacts of a questioner's arbitrary construction of a problem; it's not obvious at what level the terms in this problem are communicating specific mathematical relationships between squares and patterns. For example, you could construct a set of patterns where all of them are 1 square away from having symmetry about some axis except the 'correct' pattern which is already symmetrical (this was in fact the first thing I entertained in the problem: B has symmetry about the 45 degree diagonal, so I tried to see if all the others were 1 square away from their own symmetry) - here the squares in the 'correct' pattern do not inform the 'wrongness' of squares in 'wrong' patterns, since every pattern has its own (and potentially unique) possible axis of symmetry, with its own (and potentially unique) possible layout of squares about that axis of symmetry. In this example it's the property of symmetry of a given pattern that informs 'wrongness' of its squares, not the 'correct' pattern's layout - that is in fact irrelevant here, unlike in the original question. Hence, use of the words 'correct' and 'wrong' at both the pattern level and the square level in the original question makes the problem statement very ambiguous in its formulation. So while this question might test some sort of 'Lateral Thinking agility', I don't think it's particularly well-formed for testing Mathematical or Logical ability.

[@DmitDmit1]

The puzzle is poorly worded. I was trying to find the "correct" property of a grid too. But there are many ways you can choose this property and every grid can be "correct".

[@Tiqerboy]

He said 'shaded' squares are wrong, so I took that to be the blue squares only, ignoring the white ones. Doing that you see the blue pattern is the same for 3 fixed squares on each grid and go from there.

[@jamessaliba1096]

@@Tiqerboy white is a shade too

[@buddythecow]

Yeah next they’ll be asking Cambridge applicants “I’m thinking of a number between 1 and 100…”

[@Tiqerboy]

@@jamessaliba1096 Uhh no. Shaded implies colored in. Blue is a color. White is nothing in this context. You don't color anything in. Just common sense here.

[@La4geas]

When you stacked them all on top of each other, I started questioning if the objective was to find the correct pattern, given that all the grids were off by one square.

[@sarankalra4406]

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[@zoraackerman1684]

When you stack them on top of each other the darker parts are the ones they have in common which is the correct one seeing as they all correspond to the correct one. So if you take out the pattern you get, you can see it's the same pattern as grid E. He also said the lighter squares are the different ones so he checked one by one to see which of the grids had those squares, and got A, B, C, and D. Leaving E as the correct one. But I like the other method better. To me, it's better.

[@ErikBongers]

Like many others, I immediately noticed that B is the only one with 2 touching blue squares. Given the wording of the question, I think B is a valid answer. If you interpret "exactly one square wrong" as "move exactly one of the touching blue squares away to solve the problem", then I'm right. I checked that for all the 4 other grids, one blue square can indeed be moved so it doesn't touch another blue square. The only problem I see with my interpretation is the word "exactly". If you interpret "exactly" as exactly one move can fix the problem, then again, I'm right, but if you interpret it as "in each grid there is exactly one square that can unambiguously be defined as 'the wrong square'" then I'm wrong, because you can take either one of the touching blue squares and move it away to solve the problem. The easiest way to get rid of the ambiguity is to change the first sentence. Replace "Only one grid is correct" with, "Only one grid has the shaded squares in the correct locations", or even "Only one grid has the correct layout." (nope..."correct layout" still allows for ambiguous interpretation). "Correct locations" is the phrase to use.

[@richardwhite2304]

I proposed a rule that each grid should display bilateral symmetry along a diagonal drawn from bottom left to top right. Grid B is the only grid to meet this condition and every other grid has one square error.

[@acarbonbasedlifeform70]

My solution was exactly the same

[@MeatBunFul]

That is how I looked at this problem

[@mrosskne]

why did you propose an arbitrary rule? this doesn't follow from anything in the problem

[@richardwhite2304]

@@mrosskne The problem doesn't in fact define 'correct'. Therefore you have to assume some rule for this.

[@acarbonbasedlifeform70]

@@mrosskne the problem is formulated in an ambiguous way. No criteria for what is "right" or "wrong" are given. So... why not?

[@pauldifolco5736]

D works if you allow rotating the grid

[@WahranRai]

One fastest way to solve is to find the set having *max intersection* with everyone ! Number the cells of rows from1:16 ---> A = (3,5,8,11,15) B = (3,5,8,13,15) C = (3,8,9,14,15) D = (2,3,5,8,15) E = (3,5,8,9,15) you will find that correct one is E Remark : You could also use matrix (lines and columns ) A,B,C,D,E and the value of M(i,j) is the intersection / commun cells and the line having max sum is the correct one

[@mrcoolguy1989]

I did this by the paper folding method. For grid E, Consider 16 squares in the grid name them in order like 1,2,3,4,... Then fold the paper in such a way that square 8 overlaps square 5. Next overlap square 3 with square 15. You will see that every square is overlapped and no square is left overlapped. Consider doing this for other grids also and u can even find the correct place the squares should be placed. I hope this helps.

[@l.w.paradis2108]

I've seen problems like this worded as: Find the most typical grid. "The most typical grid" is the one that differs from every other grid by exactly one shaded square.

[@Paolo_De_Leva]

That's how it should be worded. "One shaded square wrong" is very ambiguous. It does not necessarily mean "One shaded square in a different position relative to the *correct* grid".

[@l.w.paradis2108]

@@Paolo_De_Leva I still remember the Scientific American ran a puzzle like this when I was a kid. First hard puzzle I got.

[@mrosskne]

@@Paolo_De_Leva it's not ambiguous at all. that is exactly what "one square shaded wrong" means.

[@Paolo_De_Leva]

@@mrosskne That's only if you can't see the other possible meanings 🤷‍♂ Nothing in that sentence tells you that there is a "correct" grid among those shown, that you need to use as a reference. "Wrong" may refer to some sort of correct geometrical *configuration or pattern* (by rotational symmetry, for instance), that you need to discover, and in this case *all* the grids shown may be wrong. In other words, a "square shaded wrong" may be found in each of them ‼ In this case, you are supposed to reveal the hidden pattern, by removing the wrong squares *from all grids.* There's plenty of IQ tests where you need to find a correct pattern or configuration...

[@BestAnimeFreak]

@@Paolo_De_Leva "Nothing in that sentence tells you that there is a 'correct' grid among those shown" The first four words of the task are literaly "One grid is correct." How could you not see that as them literally telling you, that there is a correct grid in there?

[@nicolasgomez4052]

Ok, so possible spoilers but this is how I'll solve it: I'll assume A is right and then check if the other ones differ from it by exactly 1 square, if I find one that breaks the rule, I'll discard A, assume B is correct and so on until I find the winner. I think it's E, let's see.

[@nicolasgomez4052]

[Austin Powers transition music] *yeah baby yeah!*

[@emilyhaigh1244]

If you fold each grid on the top right to bottom left diagonal, B is only grid where the blue squares match up. All other grids are off by 1 square. Maybe this is a second answer haha

[@stevejeffryes5086]

If E is the correct grid, then each of the others differs in the shading of 2 squares, violating the setup of the puzzle. Labeling columns with letters and rows with numbers, grid A differs at a3 and c3, grid B at a3 and a4, grid C at a2 and b4, grid D at b1 and a3. If we assume that the grids may be correct or incorrect without comparison to one another, then each grid has 1 shaded square which voids symmetry except for grid B, which is symmetrical about a bend sinister diagonal.

[@DukeOfChevo]

There is another solution, where B is correct and others are wrong. B is the only square with symmetrical order of blue fields (alongside right diagonal). All other squares has one blue field out of symmetrical order...

[@Nikioko]

E is correct. If you move only one of the shaded squares of E, you can get any of the other grids A-D: A: Move the square in the third row two spaces to the right B: Move the square in the third row on space down. C: Move the left square in the second row to the second space in the fourth row D: Move the square in the third row to the second space in the top row.

[@MrPuzzleCodes]

My solution: (im gonna count the rows from 1 to 4,left to right and the collumns from A to D, up to down) Ive seen that 1C, 2D and 4C are shaded on all the 5 images so they must all be correct. No image beside C had 3A shaded, so C is automatically wrong, 3A being its wrongly shaded square. But we also see that C has the 4B square shaded, meaning that 4B is actually correct since we already found the wrong square in the C image. And the only other image beside C that has the 4B square shaded is E. So E is the correct answer. Sorry for the long explanation but i hope you understood what i mean.

[@acadelic841]

I drew out a blank grid, went through each square and wrote down the number of patterns from the set which occupied that square. Disregarded any square which had a 1 in it and was left with the answer!

[@easy_s3351]

Call the columns w, x, y and z (left to right) and the rows 1, 2, 3 and 4 (top to bottom) and look for similarities in the grids. All grids have y1, y4 and z2 shaded, so these have to be shaded correctly. Grids A, B, D and E also all have w2 shaded. Grid C and E both have w3 shaded. Grid A is the only one to have y3 shaded, grid B is the only one to have w4 shaded, grid C is the only one to have x4 shaded, grid D is the only one to have x1 shaded so these are all shaded incorrectly. So grid E must be the grid that is shaded correctly. In grid A w3 should be shaded instead of y3, in grid B w3 instead of w4, in grid C w3 instead of x4 and in grid D w3 instead of x1. Then all grids are shaded the same.

[@zacharyfeldman1776]

Rather than looking at squares as being possibly correct choices, use the shaded positions that you know must be correct to rule out candidates. A,B,D, and E all have four common positions (a2,c1,c4,d2) so C must definitely be incorrect. But since C can only have exactly one incorrectly shaded tile, the correctly shaded fifth tile must either be a3 or b4 (comparing it to the remaining grids). Suppose b4 is correct: no other grid however has b4 shaded so a3 MUST be the correctly shaded 5th tile. The only grid satisfying these conditions (a2,a3,c1,c4,d2) is E.

[@grimswell]

I basically did the overlap method. I drew my own grid and wrote the grid letter for all shaded squares. Any squares with two or more letters got filled in leaving four squares with a single letter; ABC & D...meaning E had all of its squares overlapped by other grids so therefor for was the "correct" grid.

[@Gruuvin1]

I disagree with this solution being the only correct solution. There is another way to look at this puzzle. If you bisect each grid with a diagonal line from top right to bottom left, then you will notice a symmetry that only exists with B, and all others violate symmetry by exactly one shaded square. I saw this solution in about 20 seconds and checked to make sure it does indeed satisfy the given directions and question.

[@miguelcardenas7758]

Yes! I noticed this in the thumbnail, but I was disappointed the correct answer wasn't B. Nice to see I wasn't the only one who saw this tho.

[@Gruuvin1]

@@miguelcardenas7758 awesome! I'm glad you saw it too!

[@goddosyourself7970]

I saw it too but figured that was probably not the answer

[@Gruuvin1]

Also, I can now imagine that those who would disagree with this alternate solution would argue the semantics of what does "correct" mean. I would suggest that 1. there is no objective meaning of the word 'correct', with respect to this puzzle's question, 2. since "correct" is open for interpretation, then symmetry is 'correct' for those OCD of us who appreciate the beauty of symmetry (non-symmetry is incorrect), and 3. the proposed solution's grounds for the definition of 'correct' are very weak (actually nonexistent); you could as well have used the word 'bananas' instead of the word 'correct' and it would have made no less sense.

[@vermillionreaper]

you are wrong, if it wasn't bc the problem says that the rest of the grids have exactly 1 square wrong, then in this case C would have two squares that are wrongly positioned, which still excludes B into being one possible answer

[@crackerjaq9974]

This reminds me of one "odd one out" puzzle that I thought was really clever. I dont remember the exact perameters, but basically, every item on the list was different from one another and there was no real "one difference" to point out. The answer, as it turns out, was the one that was MOST similar to the rest! Because it "fit in" when the others didnt, that meant that the odd one out was the one that was, ironically, the most fitting item in the list.

[@MrMayhem010]

I got it through trial and error. I assumed that one was correct and compared it to the other ones and eventually I got the letter E as the correct one.

[@bdot02]

Same but with the addition that if I couldn't get from one to another then both of them were out.

[@MrMayhem010]

@@bdot02 I didn't mention it‚ but I also did the same thing you did.

[@denvercheddie]

I am pretty sure there are more approaches you could take. For example if you look at the diagonal (SW to NE), you will see that only B is symmetrical about this diagonal. All of the others would have been symmetrical except for one colored square. So why can't B be the answer based on this approach?

[@AidanHomewood]

I used a graph of five nodes and 10 weighted edges, where the weights corresponded to "differing squares". I then just found the node that had all four of its edges being weight 1.

[@TaladrisKpop]

Similar idea. The number of differing squares is a distance between grids (the distance induced by the 1-norm precisely).

[@regiondeltas]

I enjoyed this one - I did it similar to your example, and worked grid square by grid square. I.e., C is the only one without a colour in A2, so we can eliminate C. D is the only one with something in B1, so we can eliminate that etc

[@archivewarrior8535]

See I originally thought “shaded differently” meant a tile was incorrectly shaded or not shaded, not that a shaded tile was in a different spot. I quickly discovered no answer is correct with that logic, and eventually found the answer

[@somewhatblankpaper1423]

By the way, it is not that no answer is correct. The question itself allows too many combinations of 4 incorrect and 1 correct. Since this is a logical question, if the answer is unique, then it has to be derivable from the premises. the question should state that the correct grid only differs from others by one. As to whether it "makes sense" or not, I think that's another question since one can justify other possible correct combinations of 4 incorrect and 1 correct from other perspectives with other extra assumptions added to the question.

[@bartolhrg7609]

B is correct (because of symmetry over diagonal). Each of the other grids has exactly one shaded square that doesn't have pair. I said this sentence in the exact form that the question is asked to prove that this is valid answer.

[@seanfraser3125]

My guess: E differs from each of the other grids by exactly one square. The others all differ by 2 compared to at least one other. Since the four incorrect grids have only one incorrect shade, E is the correct grid.

[@jaymercer4692]

I did a Cambridge maths entrance exam a few months ago in November and the test was actually pretty easy. It's the step exams that happen in the summer that are quite difficult. I have to sit mine in less than a month.

[@hypedmaniac8444]

Might’ve lucked out when I got this right - here’s the reasoning: Immediately, when I read the question, I looked at D. I imagined giving it another row downwards, shifting the blue square in (3,4) to (3,5) and putting a blue square in (3,3) to express my confusion at what the hell this question was asking me to do. Then I actually got started. My gut instinct was to try ‘matching’ individual blue squares on each grid to the same squares on another. If no other grid had a blue square in that spot, then I would rule that one out. I didn’t try overlaying them or assuming any of them were correct. In A’s case, the (3,3) square eliminated it. In B’s case, the (1,4) square eliminated it. In C’s case, the (2,4) square eliminated it. In D’s case, the (2,1) square eliminated it. So by now, I’m about ready to perform activities that the KZbin comment filter would probably put my comment up for review for. I’d gone 4/5, and each had been wrong. Oh boy. But when it came time to look at E, each of its squares matched at least one other grid. In my head, that meant that each other grid must differ from each of the others by 1 blue square and therefore were incorrect. But since each of E’s blue squares could be paired up to at least one other grid, and therefore it had no blue squares exclusive to it, it had to be correct. What do you think? Did I get lucky or was this just an unrefined way of thinking about the problem?

[@newBreedOfInsect]

This is the first time I got a question from your videos right. But I had real fun solving it. Thanks for the videos!

[@JhettJones]

It's faster to not check every possibility - any grid that differs from another grid by 2 squares or more can't be correct because then the others wouldn't be "only 1 square away." A&C are obviously two squares off of each other just by checking the left side. Same is true for B&C, and D&C. Therefore E must be correct without even getting to it in the comparisons.

[@Grassmpl]

The overlap is nice. We can program it. Just use integer matrices instead of color shades.

[@yuvrajsaxena9601]

My approach: I wrote all the shaded block's Rows vs Column number that is the matrix element (Aij / ith row and jth column ), then started to strike off all the similar sets like (1,3) , (2,4) and so on for the grid E all the elements were cut off, that meant E was correct and all the other just had exactly one element left.

[@ocfrey9706]

I used a strategy that you did not go over, instead of ruling out grids i first found the 4 squares that were shared enough to determine they were correct, then i looked at C since that had 2 unproven squares and looked at other grids to see if any shared any of the 2 c had. The only one that shared one was square e, meaning that that had to be the correct one

[@francomiranda706]

Got this one in about a minute. Easiest way to solve it is to look at C. It's the only one that doesn't have that (0,1) square filled, but it also differs from the set of the squares in in the (1,3) square. This means that the misplaced (0,1) square must be its defect, and the (0,2) square below it must be correct. The only square that has both (0,1) and (0,2) filled is E, so E must be the correct square.

[@basharathhussainmohammed5585]

Its like E is the average of all other grids. Finally, one question that i got right from ur channel.

[@aspiringscientist1053]

It wasn't that hard, I based my answer on the most common coloured squares, like you did in the first part of the video. Now my dreams can come true so I can get into Cambridge (too bad I don't have money and I'm not in the UK). 😔

[@Thriller_Author]

As someone who did graduate from Cambridge University, the entrance exam questions were much, much harder than this

[@two697]

@@Thriller_Author yeah. I feel like presh always takes the easiest questions from the admissions test and makes it seem like every question will be this easy

[@aspiringscientist1053]

@@Thriller_Author I know, I was just joking. It made me feel a bit more optimistic (especially since I had a maths test today), so yeah...

[@zoraackerman1684]

Yeah, same.

[@mrosskne]

@@two697 no of the"admissions test" problems are actually from admissions tests

[@Buttered.Coffee]

The way I solved it was by noticing that 3 squares were consistent between all 5 grids, and ignoring those squares. I then saw that 4 of the grids shared another square, so the one without that square could not be the answer. Finally, I compared the other 4 grids to the one without the common square. Only one grid had one discrepancy from that grid, which revealed the correct answer. I later noticed after solving that the correct grid used all the 5 most common squares between the grids!

[@paulcooper8818]

I did it by counting how many times a square was shaded, same as the second solution.

[@fguo79]

My answer is that choice B is the correct grid. Note only grid B is symmetrical along the 45 degree line. The other 4 grids are only one shaded square away from reaching the same symmetry. I believe this explanation is simpler and more straightforward than the solution given in this video.

[@ricardokosovic6824]

I got a very much easy way, just notice that the 3 right side squares of every grid are in the same place, all of them. Now, notice that the squares that are different on every grid are always close to the middle left section of every grid (that's what every grid has in common), and the only grid that has that entire middle section shaded is e. Now you can simply start assuming that's the correct answer and compare to the others, and notice once again that every grid has only 1 shaded square misplaced compared to grid e :)

[@ryxsenl3018]

Same solution, felt natural.

[@multicrogamer]

I used colors and paint. A,B,C,D,E have 3 red squares. ABDE Have left green square. C and E have Yellow square. ABCD have 4 blues which don't match with anything. Those blues are things that are individual to each square. So ABD have red green blue colors, C has red yellow blue color and E has red yellow colors. The fact that green is same in every color expect C has no green but a yellow and blue but E has also yellow but no blue means that answer must be E. If you move every blue square to get a missing yellow piece in A,B,D and move blue in C to get green missing piece then every grid is exactly 1 piece away from E. Really messy stuff but it works.

[@markgearing]

Without watching: B is correct. It is already diagonally symmetrical bottom left to top right. All the others can be made diagonally symmetrical bottom left to top right by moving one block to that centreline.

[@waldtrautwald8499]

We want to know the positions of 5 blue squares. The bottom right blue square is the same for all 5 grids so there's one. By focusing only on the top 2 rows of each grid, 3 more blue squares can easily be deduced visually as they are the same for 3 out of 5 grid. This leaves just one more blue square. When we compare grid C with what we know so far, we can deduce that one the of the 2 blue squares of the bottom left quadrant of that grid must be the remaining blue square. This excludes A, B and D, so E must be the correct solution.

[@alexgibb1084]

I just looked at any specific shaded square that was present in multiple options. If it was in all five, i didn't worry about it. If it was in 4, the remaining option can be eliminated. Also, if I noticed a shaded square that only appeared in one option, i could just eliminate that option right away, probably would have been quicker just to use that approach from the beginning. I quickly eliminated down to the correct answer

[@elobez]

Bearing the 4 Colour Map Theorem in mind, B is the only 'map' where there are no bordering countries shaded blue. All the others have one pair of countries with a border.

Classical example of thinking outside the box

[@billyd78]

I had picked b

[@mrosskne]

what in the problem instructed you to imagine arbitrary rules?

[@reformed_attempt_1]

sorry you can't go to Harvard, you need to know English

[@CosmicNewbie]

The correct answer is B. All other grids have exactly ONE square wrong sharing a side with another. Grid B is the only one where no two squares share sides.

[@MEothain]

This is the first time I dont understand the question at all.I am curious how the question is formulated at Cambridge. Seeing how you start the solution made me understand the question then its relatively simple to understand that the question is simple fiind the same paterns...

[@homebooster]

All other grids except C has exect one shaded square in 1st column 2nd row. Assuming grid is correct if cell in 1st column 2nd row is not shaded.

[@Joshuaskehan-mk8cj]

I took a different approach, I coloured all the squares each grid had in common one colour, then each square with 4 in common, 3 in common and so on. I then drew a blank square and filled in all of the ones in common. I then I looked at which squares had 4 in common and put that into the new square and checked if that worked, and worked my way down and was able to eliminate the squares that didn't suit until I got E

[@und3rgroundman865]

There are multiple answers to this. For example, grid A is correct and all the others should have their one out of place square added to the third row to form a one step progression from left to right, except for grid C where the one wrong square - 2nd from left on the bottom - should be added to the left hand side of of row 2. In this way we have 5 identical grids with one square moving on a regular pattern along row 3.

[@bloxio3001]

I got it by making a weighted graph where edges are the differences between the 2 vertices so that way C has 2 differences from everyone except from E, making E the correct answer as anything else would end up with C having 2 wrong squares.

[@ziad4744]

I solved this in a different way. B is the correct answer simply because if you connect the blue squares you will get a symmetrical pentagon. while for the other shapes you have to move one blue square to get a symmetrical pentagon.

[@spronkthesprinter273]

Easiest way to do it is looking at what squares are common among all of them. Every one had 4 tiles except c which only had 3. So I looked for which other square had 1 of the two other shaded tiles that was the same as c (excluding the 3 tiles that all squares had), and then u know it has to be E

[@YindigoYangMonk]

I got it. I approached it by looking at the position of the shaded squares. I looked for the shaded squares that didn't repeat in any of the others. The one that all the shaded squares inside were repeated at least once in at least one of the other grids is the correct one. That is E.

[@curtisclement5467]

The top right 4 square are always the same so number the others 1-12 then that makes each comparison easier. One of the numbers appears 4 times except for C and so E is the only one that differs from C by exactly 1 square

[@9adam4]

"One grid is correct. Each of the other grids has four shaded squares in their correct locations and one shaded square moved to an incorrect location. Which grid is correct?"

[@VincentPeer]

Nailed this one, but got it a slightly different way. I started with 2 different (overly complex) approaches before it dawned on me to find a grid that didn’t contain any uniquely lit squares. E was the only one that qualified.

[@bix64000]

what if the definition of "the grid is correct" was "it has 5 squares shaded, and only one on each sides (or on the perimeter)". Then grid A is correct and the four others are not with only one shaded square wrong. So ...?

[@ulrap1202]

This is one of those where you just look at each one as if it were the answer and then see if it satisfies the requirements. I actually started with E for whatever reason. After comparing the others with E I thought it was the answer. But I felt I still had to do the same with each of the others to make sure. It's a lot of work. But really that's all this one is. Work, focus, discipline. It's like proofing.

[@williamniver6063]

Not really hard or time-consuming at all. See my answer above.

[@nb8205]

I did it differently. Perhaps wrongly, but still got E. I got to E by going line by line. When I had ruled out first D and then C based on lines 1 and 2, I just compared the bottom lines and saw that since A, D and E all had the same bottom rows I could then rule out B, leaving me with only A and E as options. I then saw that only C and E had the 3rd line in common, and since C had been ruled out already that left me with E as the only option.

[@williamniver6063]

I haven't read all 600+ comments. Maybe someone else thinks as I do. What I DO know is that I did this in my head without paper, calculations, or overlays in less than 60 seconds. All grids share three positions: the upper and lower squares in the third column and the second square from the top in the fourth column. The remaining two squares are the ones that are positioned differently. Only one pair shares a side with the other, contiguous square: E. Grids A-D display these paired squares as separated by other spaces or only touching diagonally at a vertex. Therefore, E is the outlier and must be the solution. When the goal is to come up with the correct answer, the quickest, often "outside the box" method is superior to sophisticated mathematical proofs. Scantron doesn't give a damn about showing your work. Neither should you.

[@yavuzyilmaz28]

I found the correct answer by only checking the choices B and C. When you look at the choices B and C, you will see that 3 cells are common and the other two are different. The correct answer should include the common cells and one different cell from each choice. The only choice suits this condition is E.

[@priyanshujain5597]

I solved it by the 2nd method! It's rare for me to solve the problems you give us

[@kmp8563]

Each grid has a unique shaded spot. Except E. It shares every spot with at least 1 other grid. Because each has at most 1 wrong spot, they can't have more than 1 unique spot and they can't share a "wrong" spot with anyone else without being completely identical. I'm going with E

[@QwDragon]

Why is it 1 square when 2 cells differ?

[@nephalm5357]

2 cells dont differ from E and C, only 1...

[@ChristophWerner1975LX]

@@nephalm5357 In C the first cell of the second row and the second cell of the last row are differently colored as in E

[@nephalm5357]

@@ChristophWerner1975LX no, look again

[@RedaKadem]

Only square C has two shades different from the other squares so if you can make it like any of the other squares by moving only one shade you will have the answer

[@humanl4903]

I didn’t get the correct answer because I didn’t know what the problem was saying

[@jgoog7442]

If you look at the diagonals, answer B has a symmetry 0,1,1,1,1,1,0 shaded box in each diagonal (low to high, left to right). Each other answer can have a similar symmetry by changing one box.

[@nicolasgaio]

I found the same

[@arunkennedy9267]

0:23, E because if you start observing the difference between E and other squares, swapping just one shaded square is enough to interchange the grid.

[@puliverius]

Simple rule. The correct one can be transformed into the incorrect one by changing exactly one shaded square to white and exactly one white square to shaded. This must be true for every such transformation. This way if you consider one by one each of these possibilities as a correct one, you can eliminate those, that must shade two or more white squares in order to achieve the transformation. That way the E is the correct one.

[@aidancheng7927]

Got the correct answer E :) - first I established the squares that were commonly shaded on all the grids. Then, I assumed a singular grid was correct, seeing if the other grids had a singular error given that assumption (repeated this with each grid). Got E by elimination. Seems like I did this the same way as another person in the comments

[@scottlawson1464]

Your solution assumes that each 'incorrect' grid is a copy of the 'correct' grid, with one square shaded wrong. I don't think this (that the grids are damaged _copies_) can be inferred by the wording of the problem. Grid B is diagonally symmetrical, with the symmetry line from bottom left to top right. All the other grids have a broken symmetry, where moving a single square to repairs the diagonal symmetry. Thus B is correct.

[@justsomebody6037]

Is it really necessary to stack them on to one another? I didn't necessarily find it difficult to find that E would be correct

[@two697]

Yh the first method he used Is probably the easiest

[@shashwatgupta4005]

hey ... I've been following the channel for a long while now. thanks for the quizzes it always helps me gain some confidence with some practice. also i would appreciate it if you mention the average difficulty level of the problem as you did in some of your videos. LOVE FROM INDIA ❤️

[@kenhaley4]

Problem with the question statment: When you say that a grid has exactly one square shaded wrong, that means it should have one more or one less shaded square than the correct grid. Since all 5 grids have the same number of shaded squares, there can be no answer to the question as stated. Instead the question should say something like, "each of the other grids has only one shaded square IN A DIFFERENT PLACE than the correct gird." Since there's no solution to the problem as stated, I assumed the second meaning, and arrived at the given answer...E. I know....picky, picky, picky. But I think math puzzles should be precisely and clearly stated to avoid confusion or disagreement about the solution.

[@fyshman152]

I took a more logic-based approach. I could see that grids B and C were different by two immediately, so I ruled both of them out. I then looked at the square on the left edge of C. I noted that it was the only of 2 grids who has a square in the third row, and all the others had a square in the second row. I assumed that the square in the third row was incorrect. For that to be true, there would need to be a second grid that has two squares at the bottom of the middle columns, of which there isn’t one. Therefore, in grid C, the incorrect square has to be one of the squares at the bottom of the two middle columns. If that is the case, the square in the third row has to be correct, and so the correct grid needs a square in both the second and third rows of the first column, therefore E.

[@qwang3118]

B & C have 3 different squares in the left column, so B & C are wrong. Both A & D have 4 blank squares at the lower left corner. But C has 2 shaded squares at the lower left corner. So A & D are wrong. Then E is correct (one can compare with A, B, C, D).

[@RioM8z]

Square B has all colored squares symmetrically distributed along the diagonal. All other squares have exactly one incorrectly colored box. B is correct.

[@mtthsh5301]

stopped the vid at 0:17 I looked at it and found that only E can be correct. But I know this channel, there will me surely more to this problem.

[@miltoninsipid8108]

As there is in no further information in the question as to what constitutes "correct", it could also mean "colored grids do not share a border". Then the unique answer is "B". The question is not worded specifically enough.

[@sujanaryal833]

When doing it on a paper I think comparing would be faster. You don't have to compare all coloured square. There are multiple coloured square fixed on same spots so just comparing the moving parts will take less time then visualizing in the head which is done by comparing anyways. While doing it on a computer software yeah this way is faster.

[@Akronox]

Problem should start by one grid is "exactly" correct, I thought (hoping I'm not one of the few ones) that the grid has to follow a set of rules and all the other where one off of doing that.

[@ed1726]

These are so satisfying when you get them on your own :)

[@yehiaelyamani6943]

I got a different answer: I started looking for grids short of 1 square to complete similarity and these were A,B,D & E. By elimination the result would be C. Is this line of reasoning acceptable?

[@SamLyndonShow]

I thought it was B, because each grid would be symmetrical along the diagonal (top right to bottom left) if not for one square. Only B is correctly symmetrical along that diagonal

[@davspa6]

E&C are different by TWO squares. E differs by two squares with all of them.. For instance on E,C, the last two columns are the same but E has two and three of column 1 but C has only 2. C has 1 of column two but E has none on column 2..

[@NightTiger26]

I got it with different logic looking at right 3 squares on c they are identical on all. I then noticed left most square on a is shared by b d and e but not c. C shared a square with e that none of the others have. So logically e is the answer

[@davidblack9071]

Paused at 00:18. Guess (E) as it 3 dots are universal so can be ignored the two middle rows in the left most column are occupied by at least one of the remaining 2 blocks in all cases then that leaves E being the only one with both those occupied and everything else off by one.

[@FoxMcCloudV2]

The clue to solve this is grid C. This grid has two differences with each of the other grids except grid E, with which it has only one. Therefore, C cannot be the correct grid and it must be E.

[@oneoranota]

Thanks god for grid C !

[@Meitner-109]

I Got it correct through 1st method! but 2nd is amazing! overpowered!

[@electricninja7493]

I got this one pretty easily, so maybe I stand a chance at applying to Cambridge next year

[@abhay7760]

Solved it!

[@txmanx3304]

The deterministic approach is to fill out the matrix (ABCDE x ABCDE). In each cell, count the number of squares that are different. Here, it is 0 (A on A), 1 cell different, or 2 cells different. The matrix is symmetric about the 0 diagonal. The answer is the row with all 1's (except for the diagonal that is zero).

[@ArielLorusso]

You cold also put the numbers of same state squares then sum up every square for an option and choose the one with Biggest number. It only works because they guaranteed a correct opinion exist.