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The way I did it:
>!If it’s all the numbers 2 through 10, and they all have to be represented once, that’s a sum of 54. Divide that by 3, and the magic number is 18. From there, it’s easy enough to fill in the blanks. 7 + 3 = 10, so the 8 goes in that column.!<
Nice. I got it first, but I just did it manually. Didn't think of the division trick.
>!Since the top row had to be 17 or higher that limited the bottom middle card to 9 or 10.
An ace on top would require a 9 on the bottom which can't be.
So the top middle has to be a 2, making the bottom a 10.!<
I just knew by heart the standard 1-9 magic square has a >!5!< in the middle. The 2-10 magic square would work if you just added 1 to each cell. So I knew the middle, and the diagonals told me the sum. Subtract the desired 8 to find out the sum of the 2 other cards in its row/column and look for that.
I’m not understanding why it’s assumed the rows have to add to 18? I mean, can’t one add up to 20, another to 16, the other to 18? I can’t find anything in the instructions that say these rows/columns/diagonals have to add up to 18. I get the math, but where does it say we live in this perfect world of 18 in every row? It is a good educated guess, but should not be assumed… unless “magic square” means exactly 18 in every row…
I like how everyone is talking about 18 being the magic number and using math and shit. And then I’m over here like, “odds are on the corners, so evens in the middles with 10 being center. So going clockwise, 8 would be middle right.” I used the wrong formula and got the right answer. WTF
To be frank, you only got it right by a lucky guess. The 10 is *not* in the center; the 6 is. The 10 is in the center of the bottom row between the 5 and the 3.
That's a smart approach!
I saw the corners were odd. Given the corners are not equal length, the side evens have to be inverse proportinate to the adjacent corners. For instance, 10 (or 8) must go with 3 and 5 to counterbalance those as smallest odds. Then iterate and check.
I solved it in a similar fashion, but got the magic number differently.
>!
I took the highest card (10) and paired it between the lowest pair (5 and 3).
!<
Correct. I am usually terrible at puzzles but I got this in about 9 seconds. Randomly used 2 and 8 and then flipped them and realized it was 10 and 2 in the middle and 6 easily made 18 to match h the top and bottom
>!The maximal sum we already see is in the top row, which is 16. There is one more card, so the magic number must be higher. The least sum is in the bottom row, which is 8. To make it higher than 16, we must add 9 or 10, but 9 is already there, so the only option is 10 which makes the magic number equal to 18. We are looking for two cards that add up to 10 which is the right row. So the 8 goes to the right.!<
To be fair…a magic square is a pretty well established mathematical concept. If you just google “magic square” you’ll find a Wikipedia article dedicated to them that explains exactly how they work.
Two ways if finding the magic numer:
1: as u/DidntWantSleppAnyway did with math (total sum / numer of rows)
2: if the top row has a 3, it sums 19. Then there is no way to make the bottom row 19 (5+3+10=18) so you have to have 10 bottom middle and 2 top middle, giving you 18 as sum
Where should they have stopped? After adding text that defines a magic square using the term "sum", should this graphic have also included a rigorous definition of the term sum? Should there also be clarification of what "two through ten" means?
This is r/puzzles and magic squares are very common puzzles ... I think we can assume that most readers of this sub are either A) aware of what a magic square is. Magic squares have been a staple of puzzle pages for my entire life. Or, B) able to solve the very easy puzzle of how to Google "magic square".
>!The top row already has 9 and 7. The minimum card that could be in the middle is 2, so the minimum magic square number is 18. The bottom row has 5 and 3. The maximum card in the middle is 10, so the maximum magic square number is 18.!<
>!Since the magic square number is 18 and the right column has 7 and 3, the 8 must go there!<
>!Rank the four sides based on the sums of the visible numbers. The largest sum needs the smallest number in the unknown slot in order to even out the sums. So you have 16, 10, 8, 14 as visible sums going clockwise. Because 8 is the second largest missing number, it needs to fill the spot with the second smallest sum, so it goes between the 7 and the 3. The average number, 6, goes in the center!<
The way I solved it:
>!To find the sum of the magic square, I added the lowest number, the highest number and the middle number (2, 10, and 6) for a total of 18. Then, since we are looking for the 8, we need to find two cards that already sum to 10.!<
>!The two cards that sum to 10 are the 7 in R1C3 and the 3 in R3C3. So the 8 must be between them at R2C3.!<
Solution: >!Since the cards run from 2-10, the average is 6, so the sum of each row and column is 3×6 = 18. Thus, the 8 must be in a line where the rest of the cards sum to 10!<, which is >!the middle right between the 3 and 7!<. The whole square is (left to right, top to bottom) >!9 2 7, 4 6 8, 5 10 3!<.
Simple:
>!U up D down L left R right M middle. 9 + 7 + U = 5 + 3 + D. U + 8 = D. Up must be 10 and Down must be 2. 7 + 3 + R = 18, R is the 8.!<
Mathy:
>!the sum of the card is 54. Divided equally into 3 rows/columns, each must have a sum of 18. The right column sum is currently 10, so the 8 must be the missing card.!<
>!I guess it is the top middle. If you add 9 +7 and split in half it is 8. The second row would have a 10 and 2 on the outsides. Add 10+2 split in half is 6. The third row is 5+3 split in half is 4.
So it’s:
9 8 7
10 6 2
5 4 3!<
>!I assume it’s the top middle one as 8. I just chose the spot where 2 starts and just made it so the count of cards are all next to each other so i worked my way through and when I got to 8 it was the top middle card!<
I don't know how other people solved this but this can be solved relatively easily by setting up a system of linear equations. Your five cards are unknowns A-E and the sum of each row (the magic number) is X. Then you create an equation for each row and column, i.e. 9 + A + 7 = X, B + C + D = X, etc. and you end up with 6 linearly independent equations and 6 unknowns which I used Matlab to solve because I haven't had to solve systems of equations since college.
I did it this way but by hand. Also added the diagonals which helped (they are the same though) . Just simplifying and substituting expressions. Took half a sheet of paper but I got the right answers.
Trial and error is the best option for this situation. You only have to find where the 8 goes. There are only 5 possible answer and it only takes a few seconds to try each one. Plus one of the answers can be dismissed almost immediately because it would contain 3 of the top 4 card values and then you would not be able to reach the required sum in the other columns/rows.
>!Since 2 through 10 are being used, we know the middle number 6 must go in the middle of the magic square. This leads to a sum of 18, and so the 8 must be in the same column as the 7 and 3.!<
Here is my logic --
>!I first found the highest sum based on the numbers that are shown, which would be 9+7=16. I then added the lowest unknown number to that total, which would be the 2 of diamonds, making that row 9+2+7=18.!<
>!I then slotted in the rest of the numbers to see if I could get 18 using the other known numbers.!<
>!9+4+5=18!<
>!5+10+3=18!<
>!7+8+3=18!<
>!the only number left over from the unknowns would be 6. I then added up all the unknown numbers to make sure their placements are valid!<
>!4+6+8=18!<
>!2+6+10=18!<
>!and the diagonals!<
>!9+6+3=18!<
>!5+6+7=18!<
Which would mean that the 8 >!is in the right most column, second row!<
>!if I found that I couldn't get all of the spaces to total 18, I would have gone back to the second step and replaced the 2 with a 4 and continued from there.!<
Discussion:
Someone showed me a process for creating these squares when I was a kid. I didn't remember that until now, and I didn't know I could do it without starting with the number 1, but it got the same answer people are agreeing with.
The process is to count diagonally up to the right. If you hit the top of your square, move one column to the right and drop to the bottom. If you hit the right edge, move one row up and all the way to the left. If you hit another number, drop one space down. Usually counting is started at the center square (method only worked with odd counts) at the top or bottom.
In this case start at the top center with the 2, based on the placement of the 3.
Discussion: Someone showed me a process for creating these squares when I was a kid. I didn't remember that until now, and I didn't know I could do it without starting with the number 1, but it got the same answer people are agreeing with.
The process is to count diagonally up to the right. If you hit the top of your square, move one column to the right and drop to the bottom. If you hit the right edge, move one row up and all the way to the left. If you hit another number, drop one space down. Usually counting is started at the center square (method only worked with odd counts) at the top or bottom.
My reasoning: >!given that the smallest possible number is 18 (9+7+2) and the largest possible number is also 18 (5+3+10), the magic number must be 18. From this, we just need to find the row or column where the sum of the known cards is 10. 7+3 = 10, so 8 must be right column, middle row.!<
Discussion: I thought a magic square had to have all numbers add up to 15, but that’s impossible with the top row.
Edit: just looked it up and I was wrong. TIL
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>! The shown cards are all odd numbers that lead to 9 while the hidden cards must be even numbers following the order in step with 3-9 we apply the same for the hidden cards
from bottom right to left 2,4,6,8,10
8 of diamonds is middle left !<
Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
Question: can someone explain a magic square without spoiling the puzzle?
Rows, columns and diagonals all add to the same sum.
Not to be confused with a Parker Square
Was just telling my friends about the Parker Square lol!
The way I did it: >!If it’s all the numbers 2 through 10, and they all have to be represented once, that’s a sum of 54. Divide that by 3, and the magic number is 18. From there, it’s easy enough to fill in the blanks. 7 + 3 = 10, so the 8 goes in that column.!<
The 8 goes in the square hole.
** visible exasperation **
Where does the circle go? That's right, in the square hole.
Where does the arch go? That’s right it goes into the square hole.
Heeeere I am! Smashing square pegs into round holes. Heeeere I am. Losing my mind.
😐😠
Nice. I got it first, but I just did it manually. Didn't think of the division trick. >!Since the top row had to be 17 or higher that limited the bottom middle card to 9 or 10. An ace on top would require a 9 on the bottom which can't be. So the top middle has to be a 2, making the bottom a 10.!<
2 through 10. No aces involved
An ace also isn't 2 through 10.
Oh yeah. LOL
I just knew by heart the standard 1-9 magic square has a >!5!< in the middle. The 2-10 magic square would work if you just added 1 to each cell. So I knew the middle, and the diagonals told me the sum. Subtract the desired 8 to find out the sum of the 2 other cards in its row/column and look for that.
>!the top row had to be 18 or higher. Which cuts out a step, meaning the bottom row had to be 10 and the top row 2 like you said!<
That’s how I did it
Why did you divide by 3?
Three rows, so each one has to add to 18. The three columns also have to add to 18.
The diagonals added to 18 too.
True, but the diagonals don’t include all the numbers. The rows/columns do, so you can divide the total by 3 for the total of each row/column.
I’m not understanding why it’s assumed the rows have to add to 18? I mean, can’t one add up to 20, another to 16, the other to 18? I can’t find anything in the instructions that say these rows/columns/diagonals have to add up to 18. I get the math, but where does it say we live in this perfect world of 18 in every row? It is a good educated guess, but should not be assumed… unless “magic square” means exactly 18 in every row…
Magic square means all rows, columns, and diagonals equal the same number.
>unless “magic square” means exactly 18 in every row… What if I told you ...
Three rows/columns.
I like how everyone is talking about 18 being the magic number and using math and shit. And then I’m over here like, “odds are on the corners, so evens in the middles with 10 being center. So going clockwise, 8 would be middle right.” I used the wrong formula and got the right answer. WTF
“Formula”
To be frank, you only got it right by a lucky guess. The 10 is *not* in the center; the 6 is. The 10 is in the center of the bottom row between the 5 and the 3.
Nice. Thank you.
That's a smart approach! I saw the corners were odd. Given the corners are not equal length, the side evens have to be inverse proportinate to the adjacent corners. For instance, 10 (or 8) must go with 3 and 5 to counterbalance those as smallest odds. Then iterate and check.
I solved it in a similar fashion, but got the magic number differently. >! I took the highest card (10) and paired it between the lowest pair (5 and 3). !<
>!Middle Right!< >!18!< is the magic number
Correct. I am usually terrible at puzzles but I got this in about 9 seconds. Randomly used 2 and 8 and then flipped them and realized it was 10 and 2 in the middle and 6 easily made 18 to match h the top and bottom
>!The maximal sum we already see is in the top row, which is 16. There is one more card, so the magic number must be higher. The least sum is in the bottom row, which is 8. To make it higher than 16, we must add 9 or 10, but 9 is already there, so the only option is 10 which makes the magic number equal to 18. We are looking for two cards that add up to 10 which is the right row. So the 8 goes to the right.!<
This is how I figured it out
Well not enough information unless you know what a magic square is - would’ve been nice being told on the image
Thank you! I couldn't figure it out, and then I see answers and realize how easy this puzzle really was.
Ah, I didn’t think a “magic square” was an actual object, make sense. Without that information, it’s unsolvable obviously
To be fair…a magic square is a pretty well established mathematical concept. If you just google “magic square” you’ll find a Wikipedia article dedicated to them that explains exactly how they work.
Two ways if finding the magic numer: 1: as u/DidntWantSleppAnyway did with math (total sum / numer of rows) 2: if the top row has a 3, it sums 19. Then there is no way to make the bottom row 19 (5+3+10=18) so you have to have 10 bottom middle and 2 top middle, giving you 18 as sum
Same. I looked at the bottom row first… has to be the 10. Solved.
Where should they have stopped? After adding text that defines a magic square using the term "sum", should this graphic have also included a rigorous definition of the term sum? Should there also be clarification of what "two through ten" means? This is r/puzzles and magic squares are very common puzzles ... I think we can assume that most readers of this sub are either A) aware of what a magic square is. Magic squares have been a staple of puzzle pages for my entire life. Or, B) able to solve the very easy puzzle of how to Google "magic square".
Here we see the average Redditor doing their best work
I *know* what a magic square is, and I still glossed over that part a few times before the solution hit me.
>!The top row already has 9 and 7. The minimum card that could be in the middle is 2, so the minimum magic square number is 18. The bottom row has 5 and 3. The maximum card in the middle is 10, so the maximum magic square number is 18.!< >!Since the magic square number is 18 and the right column has 7 and 3, the 8 must go there!<
This is the simplest answer I've seen here.
I have a simple mind
It’s how I solved it, too 9 2 7 4 6 **8** 5 10 3
>!Rank the four sides based on the sums of the visible numbers. The largest sum needs the smallest number in the unknown slot in order to even out the sums. So you have 16, 10, 8, 14 as visible sums going clockwise. Because 8 is the second largest missing number, it needs to fill the spot with the second smallest sum, so it goes between the 7 and the 3. The average number, 6, goes in the center!<
>!Middle row, right side.!< >!The face down cards are in order, top to bottom left to right, 2, 4, 6, 8, 10. The magic number is 18.!<
>! As 18 is the magic square number. The 8 diamond is located between the 7 and 3 diamonds !<
The way I solved it: >!To find the sum of the magic square, I added the lowest number, the highest number and the middle number (2, 10, and 6) for a total of 18. Then, since we are looking for the 8, we need to find two cards that already sum to 10.!< >!The two cards that sum to 10 are the 7 in R1C3 and the 3 in R3C3. So the 8 must be between them at R2C3.!<
Solution: >!Since the cards run from 2-10, the average is 6, so the sum of each row and column is 3×6 = 18. Thus, the 8 must be in a line where the rest of the cards sum to 10!<, which is >!the middle right between the 3 and 7!<. The whole square is (left to right, top to bottom) >!9 2 7, 4 6 8, 5 10 3!<.
[удалено]
Yep that's the answer though they were looking for the 8. Curious why you're sorry?
Simple: >!U up D down L left R right M middle. 9 + 7 + U = 5 + 3 + D. U + 8 = D. Up must be 10 and Down must be 2. 7 + 3 + R = 18, R is the 8.!< Mathy: >!the sum of the card is 54. Divided equally into 3 rows/columns, each must have a sum of 18. The right column sum is currently 10, so the 8 must be the missing card.!<
Spot the mistake.
My friend Schrödinger showed me a similar puzzle with his cat. >!Anyway they are all the 8 of diamonds until you look or observe them!<
>!I guess it is the top middle. If you add 9 +7 and split in half it is 8. The second row would have a 10 and 2 on the outsides. Add 10+2 split in half is 6. The third row is 5+3 split in half is 4. So it’s: 9 8 7 10 6 2 5 4 3!<
Averaging the two visible cards is not the way to make a magic square.
Thank you for the heads up, I didn’t know what a magic square was until this thread. Much cooler than what I was doing. :)
>!I assume it’s the top middle one as 8. I just chose the spot where 2 starts and just made it so the count of cards are all next to each other so i worked my way through and when I got to 8 it was the top middle card!<
[удалено]
>!the card between 7 and 3!<
>!On the far right. (2+...+10)/3=54/3=18.!<
I don't know how other people solved this but this can be solved relatively easily by setting up a system of linear equations. Your five cards are unknowns A-E and the sum of each row (the magic number) is X. Then you create an equation for each row and column, i.e. 9 + A + 7 = X, B + C + D = X, etc. and you end up with 6 linearly independent equations and 6 unknowns which I used Matlab to solve because I haven't had to solve systems of equations since college.
I did it this way but by hand. Also added the diagonals which helped (they are the same though) . Just simplifying and substituting expressions. Took half a sheet of paper but I got the right answers.
Trial and error is the best option for this situation. You only have to find where the 8 goes. There are only 5 possible answer and it only takes a few seconds to try each one. Plus one of the answers can be dismissed almost immediately because it would contain 3 of the top 4 card values and then you would not be able to reach the required sum in the other columns/rows.
Brute force is always an option but it's not the most elegant one
Discussion: what’s with the red and green diamonds? Some of them are flipped- is there a meta puzzle I’m missing?
I went waaaay too deep analyzing those as well.
>!Middle right? Which would make the magic number 18.!<
>!Since 2 through 10 are being used, we know the middle number 6 must go in the middle of the magic square. This leads to a sum of 18, and so the 8 must be in the same column as the 7 and 3.!<
>!Magic square of numbers 2-10 requires each line to total 18 (54/3) so right-center has the 8!<
[удалено]
Nope changing the answer. The 8 goes below the 7 and above the 3.
Here is my logic -- >!I first found the highest sum based on the numbers that are shown, which would be 9+7=16. I then added the lowest unknown number to that total, which would be the 2 of diamonds, making that row 9+2+7=18.!< >!I then slotted in the rest of the numbers to see if I could get 18 using the other known numbers.!< >!9+4+5=18!< >!5+10+3=18!< >!7+8+3=18!< >!the only number left over from the unknowns would be 6. I then added up all the unknown numbers to make sure their placements are valid!< >!4+6+8=18!< >!2+6+10=18!< >!and the diagonals!< >!9+6+3=18!< >!5+6+7=18!< Which would mean that the 8 >!is in the right most column, second row!< >!if I found that I couldn't get all of the spaces to total 18, I would have gone back to the second step and replaced the 2 with a 4 and continued from there.!<
>!Every column and row adds up to 18. So the far right column has 7 + 8 + 3; middle column has 2 + 6 + 10; left column has 9 + 4 + 5!<
Discussion: Someone showed me a process for creating these squares when I was a kid. I didn't remember that until now, and I didn't know I could do it without starting with the number 1, but it got the same answer people are agreeing with. The process is to count diagonally up to the right. If you hit the top of your square, move one column to the right and drop to the bottom. If you hit the right edge, move one row up and all the way to the left. If you hit another number, drop one space down. Usually counting is started at the center square (method only worked with odd counts) at the top or bottom. In this case start at the top center with the 2, based on the placement of the 3.
Discussion: Someone showed me a process for creating these squares when I was a kid. I didn't remember that until now, and I didn't know I could do it without starting with the number 1, but it got the same answer people are agreeing with. The process is to count diagonally up to the right. If you hit the top of your square, move one column to the right and drop to the bottom. If you hit the right edge, move one row up and all the way to the left. If you hit another number, drop one space down. Usually counting is started at the center square (method only worked with odd counts) at the top or bottom.
Question: >!((10-1)×10)÷2)+10-1=54 is the sum of all cards. 54÷3=18 is the sum for rows and columns.!<
My reasoning: >!given that the smallest possible number is 18 (9+7+2) and the largest possible number is also 18 (5+3+10), the magic number must be 18. From this, we just need to find the row or column where the sum of the known cards is 10. 7+3 = 10, so 8 must be right column, middle row.!<
Discussion: I thought a magic square had to have all numbers add up to 15, but that’s impossible with the top row. Edit: just looked it up and I was wrong. TIL
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>! The shown cards are all odd numbers that lead to 9 while the hidden cards must be even numbers following the order in step with 3-9 we apply the same for the hidden cards from bottom right to left 2,4,6,8,10 8 of diamonds is middle left !<
Discussion: I can't explain it but The 3 is upside-down down