Move all the numbers once clockwise around the 5, and then put the even numbers in the opposite corners. Reflect and rotate the square to get all possibilities. The total of all 9 numbers is 45. Divide this by 3 rows makes 15. The possible ways to create 15 are 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, and 4+5+6. All other combinations don't add up to 15. The central square must fit in 4 different combinations. This requires the number in the center to be 5. This occupies the combinations 1+5+9, 2+5+8, 3+5+7, and 4+5+6, leaving the combinations 1+6+8, 2+4+9, 2+6+7, and 3+4+8. All of the corner squares must fit two of these combinations, which makes 2, 4, 6, and 8 the corners. You can fill the square based on this fact, and you will reveal there is only one real answer for a 3x3 magic square, with variations based on rotation or reflection.
there is an easy way I learned years ago. and i actually applied it on 10*10 square, the trick is you have to start on cell which is the last cell on right of the 2nd row from bottom. and after that the next no to put in will be 1 down and 1 right 1 2 3 A B C you are on the cell B3 in this cell you will start and write 1 in it. now you will go 1 down and one left so one down will be C3, then 1 right will be C1 (as there is no cell, you have to go to the left most cell.) now in C1 write 2, and follow this rule, when you come to a cell which is already occupied then you must select the cell which is on the top of it. Assume you come to cell B2 which is already occupied then you select the cell B1. Follow this pattern and you can solve it easily. Once you understand the formula you can use this to solve any magic square
Another interesting note is that all the pairs you can make out of 7, 8, and 9 sum to 15 or more with only two numbers. This means that 7, 8, and 9 must never be in the same row, column, or diagonal as each other. This leaves only 3 possibilities for where 7 8 and 9 can go. Also, 6 and 9 cannot be in the same row, column, or diagonal because they also add up to 15 with just two numbers. This set of restrictions was the first observation I made when trying to solve the question on my own. Unfortunately it's not enough information to solve the whole puzzle and it basically left me in a position of trying to do sudoku to figure out where to place the other numbers by process of elimination. The missing link for me was counting the number of times each other digit shows up in the possible calculations.
When I was about 13, I saw this puzzle without explanation. I took me a whole day to figure it out. Now I know the logic and must teach it to my grandkids.
Great video! I figured out a quicker way: if we want to fill the number 9 in, we have a total of 6 left for the other 2 squares that are in row/column with 9(bcuz 9+6 =15, its the sum)since there are only 2 ways to add up to 6(1&5, 2&4), we can garantee that 9 has to be at where the T shape intersects. Then according to the middle number is 5, we can fill out the other number in row with 9 - which is 1. Then we take the other pair of numbers that add up to 6 - whcih is 2&4, we can fill out the column that has 9 in there. Then we just subtract the numbers we fill out and can easily fill out the dquare.without finding all 8 variations that gives 15. Sorry for my English.
Hi @mindyourdecision I got a question for you. You have a 3x3 grid, your goal is to add up to 15 and the center square is 3. You can only use the numbers one time. I've tried it myself and I don't think it's possible. Do you can solve it? Anybody can answer it.
I studied this magic square. I'm wondering if the problem has originated from "Knapsack problem"? In resource allocation where the decision makers have to choose from a set of non-divisible projects or tasks under a fixed budget or time constraint, respectively, and they want to deliver a max value.
I watched a similar digital root video earlier, so I decided to follow that a bit. All the odd #'s have 2 of each, The even #'s have 3 of each, and the (5) has 4. So there's 4 odd #'s - 4 even #'s and the 1 (5) So I took 234,441 / 9 = 26,049 reversed it and did the same 441,234 / 9 = 49,026 Now I took 49,026 - 26,049 = 22,977 the digital root of this is 7+2 , 7+2, 9 = 999
This is actually a very interesting approach to this kind of puzzle. You are more or less tackling some of the fundamentals of group theory in without explicitly saying so.
This is the best explanation of this magic square that I have been able to find on RUclips. Thank you for taking time out of your day to provide free knowledge to the public! Good work.
Great but how do you use the same method for 4x4 matrix? Row, column and diagonal add up to 34. I tried using the same method and had a problem. According to my list, there are 7 possible additions involving the number 1 but that can’t be right because the most the number could be used is 3 times. I got 1+16+15+2, 1+15+14+4, 1+14+13+6, 1+13+12+8, 1+12+11+10, 1+10+9+14, 1+9+8+16
Here's how I did: 1+2+3+4+5+6+7+8+9 = 45, so the sum of the elements in each triplet has to be 45/3 = 15. If the lowest number in a triplet is 1, then the second lowest number has to be at least 5, giving us the highest number as 9. If we then increase the middle number by one, giving us 6, we have to decrease the highest number by one (in order to maintain the sum of 15), giving us 8. Repeating this process, the middle number becomes equal to the highest number, which is of course not an option, so all-in-all there are two possible triplets containing the number 1. In the same manner we can show that there are only eight unique triplets: (1, 5, 9), (1, 6, 8), (2, 4, 9), (2, 5, 8), (2, 6, 7), (3, 4, 8), (3, 5, 7) and (4, 5, 6) Seeing as there are 3 rows, 3 columns and 2 diagonals (3+3+2 = 8), all eight triplets must be used in order to complete the magic square. Let's start with the middle square. This square is connected to the middle row, the middle column and the two diagonals (2+2+1 = 4), so any number that goes here has to be present in at least four of the eight triplets, and the only number that applies to is 5. So the middle square can only contain the number 5. Now, the number 1 can only be put in the upper-left square or the upper-middle square (all other possibilities are either rotations or mirrors of these two scenarios). If we choose to put it in the corner square, this gives us that the lower-right square has to contain the number 9, and the upper row has to be either (1, 6, 8) or (1, 8, 6): 1 6 8 1 8 6 5 5 9 9 Both scenarios make the sum of the numbers in the rightmost column to exceed 15, which means that the number 1 can only go into the upper-middle square. This gives us that the lower-middle square has to contain the number 9, and the upper row has to be either (6, 1, 8) or (8, 1, 6), but these two scenarios are mirror images of each other, which leaves us with but one possibility: 6 1 8 5 9 This gives us that the diagonals have to be (6, 5, 4) and (8, 5, 2): 6 1 8 5 2 9 4 Then, the leftmost column has to be (6, 7, 2) and the rightmost column has to be (8, 3, 4), thus completing our magic square: 6 1 8 7 5 3 2 9 4 This can be rotated up to three times. It can also be mirrored and then that mirror image can be rotated up to three times. This gives us a total of 4×2 = 8 possible solutions: 6 1 8 2 7 6 4 9 2 8 3 4 7 5 3 9 5 1 3 5 7 1 5 9 2 9 4 4 3 8 8 1 6 6 7 2 8 1 6 4 3 8 2 9 4 6 7 2 3 5 7 9 5 1 7 5 3 1 5 9 4 9 2 2 7 6 6 1 8 8 3 4
There is another way to find the center no. Assuming the sum of each row/col/diagonal to be S(where S=15) then 4S= 45(sum of 1-9)+ 3*centre (as shown in video we can pass 4 lines from center and counting the sum of numbers traced by lines equals sum of all numbers plus 3*centre no) now on evaluating the above eqn you can find the centre no.
I just found this and realized a math competition I entered in had this same problem but used 5 in the middle like always and x above that and x+1 in the bottom right corner and asked for the sum of all possible x values. I just skipped that one :p
Here's a purely mathematical way to show the middle square has to be 5: Going from left to right in each case, (a) let the numbers on the top row be a, b, and c; (b) let the numbers on the middle row be d, e, and f; (c) let the numbers on the bottom row be g, h, and i. Knowing that every row, column, and diagonal must total 15, we have: a + e + i = 15 d + e + f = 15 g + e + c = 15 Adding those three equations together and rearranging the terms a bit gives us: a + d + g + c + f + i + 3e = 45 But we also know a + d + g = 15, as it is the left column. Similarly, c + f + i = 15 because it is the right column. So the above equation simplifies to: 15 + 15 + 3e = 45 3e = 15 e = 5 ...and we're done!
By the way, once you have the 5 in the middle, the rest is simple. There are only two ways to use the 1 to total 15, either 1 + 5 + 9 or 1 + 6 + 8, which means the 1 can't go in the corner. So, without losing generality due to rotations and reflections, put the 1 in the middle square on the top row, the 8 and 6 on either side of the 1, and the rest of the magic square fills in automatically.
If someone knows the Agrippa or Durer methods to solve 3 x 3 square can make even 7 x 7, 11 x 11, 21 x 21 and so on. A new square odd number method 2018 is introduced by MH.
Thank you so much! Home schooling my child and this video has been a godsend! Extremely helpful and so very easy to understand. Your other book resources look awesome too! So glad I "stumbled" across your channel whilst performing a mathematical search
I love your videos, I got the last few, this one I got, but I loved the way that you solved it, and I had no idea how you could prove there were no others. LOVE YOUR VIDEOS
My grandfather done this back in the mid 1980’s. Then kept going. The last puzzle that was never finished was a either 21x21 or 20x20. But he went threw 3x3 all the way up over a span of 5 years. Daily puzzle for him to keep his 80 year old brain going. All were writing down. Not sure I can find them any more but. You solution answer of 1 can’t be in a corner is wrong. If you rotate each number 1 clock rotation left or right all the numbers will still add to 15. As long as 5 stays on the middle. Just letting you know. I was thinking of picking up where gramps stopped and start at 21x21 and work my way up from there. I could use a puzzle in the corner like a chess set to start and let it wait when I’m busy. I was just looking up how he got the 15 all ways and the 25 all ways for the 4x4 puzzle if I recall correctly. Nice job.
That is not true. Odd numbers cannot be in a corner. If 1 was in a corner, then there have to be 3 different sums, one including a 5. So, there is 1+5+9, 1+8+6... the third one would have to be 1+7+7 which obviously is not possible.
@@thomaslohrschall9410 I relooked and I don't have a clue what I was thinking.. It looked like it could.. Sorry.. Your right.. I goofed up.. Thanks .. :P...
now to tell you a fact about magic squares i discovered many years ago and hoped to weite a paper about,but never got it together. in any magic square of consecutive numbers then, if you subtract the lowest value, so that the numbers run from 0 to 8, now convert the numbers to base 3 in two digit form ie 00,01,02,10,11,12,20,21,22 you will see a pattern. in the firstdigit and a different pattern in the second digit.this fact applies to every magic square no matter its size. these patterns can be used in the 4 x4 and higher squares to generate differnet types of magic square( apart from the rotation and reflection. for uniqueness put 00 in the lowest possible cell number then put 01 in the lowest possible cell number. there are several interesting transformations.t one is to swap the digit positions ie 21 is transformed to 12 another is to swap cell number and cell value. anther thing you cna do with the patterns is to multiply them by coprime number eg 4 and 5. this will give a magic square of non consecutive numbers.
Just like there are 8 ways for a 3x3 matrix from the number 1 to 9 then If there is 1,2,3,4 till 16 in the 4x4 matrix then what are the number of ways (by forming different matrix) of achieving the total of 34 into rows, columns, and diagonals.
this is cool i love squares, i make all kinds of magic squares i can use any numbers to make a 3x3 or 4x4 5x5 etc i love these i also make magic circles and stars its the best thing for a technical mind.
That's easy there's 2 diagonal squares joined together and there apart of a ball This is how they can use the pelican with nas nose with the dino hiding behind it to go to the iftari people to use the same ball as a fidget club spinner with an dark nose ready to fire down for an iftari meal 😂
the easy step is put in middle 5..... 2 step: put odd numbers up,down,right,and left,,(1,9,3,7) 3 step: there are four corners,, put any one even numbers at one corner and try add those two numbers to get third one... than its sloved...
A blackjack magic square is a 3x3 array of cards where for each row/column/diagonal the magic sum is exactly 21 as in blackjack. The proof that an ace can never be 1 in any sum is very straightforward, and the proof that the center card has to be a seven immediately follows.
What I'm getting is, 5 must go in the center. 1-8-6 must go along some side (top side, without loss of generality), in some order. 1 must go in the center of said side. Without loss of generality, place 6 on the left side of 1 (TL corner) and 8 on the right side of 1 (TR corner). Of course 9 goes opposite 1; 2 goes opposite 8; yada, yada, yada, it solves itself from that point. So this is the only solution, along with its reflections and rotations. You can get 4 rotations. Then flip it, and you get 4 rotations there too. 8 solutions, but only 1 if you discount rotations and reflections.
The Square of the Sun (6 x 6) is also the Square of the Beast. That is the total sum of all the Numbers ='s '666' (Each row, column and/or diagonal ='s 111) ! For who is like unto the Beast....for 'Ye Shalt Know Them By Their Tales or is it Tails' !
I have been watching your videos for half a decade and I still didn't catch up what do you say your name is on beginning of videos. Something like Presh Toloker??
I'm pretty sure that It's Presh Talwalkar, an Indian surname from the Midwestern region of India where the "kar" suffix occurs often in surnames such as Sachin Tendulkar
I did not use that step 2 at all. Here is how I did it: Step 1: Prove that each sum is 15 Step 2: Diagonals + middle row + middle column is 4*15, which is 60. Subtract 45 (the whole square) from 60 to get three times of the middle square. Deduce from that the center is 5. Step 3. Prove the number 1 cannot be at the corner. If 1 is at the corner, the opposite corner must be 9. Suppose a remaining corner is odd, since odd+odd+even is not 15, the numbers between that corner and both 1 and 9 are odd. There are only five odd numbers in all, so this is impossible. Suppose a remaining corner is even, all the remaining squares can be shown even, which is also impossible. Therefore, 1 cannot be at the corner. Step 4: If 1 is in the top spot, 9 must be at the bottom. 1 plus two numbers yield 15 means that the sum of the two numbers next to 1 is 14. Since 9 is used up already, only 6+8 works. Similarly for 9, the other two numbers at the bottom are 2 and 4. Since 5 is in the middle, 2 must oppose 8, and 4 must oppose 6. This provides two different arrangements as 2 can be on the left or right. The number 3 and 7 must be filled in according to the position of 2, so they do not provide further arrangements. Step 5: The same thing can be done with 1 in the other three side positions. Thus, the total number of ways to construct a 3X3 magic square is 2*4, which is 8.
How about seeing the square as a ball. Start with 1 top middle and then go 1 right and 1 up place 2, right and up place 3, right and up now is occupied by 1 then go 1 down and so on. Works die every odd squared puzzel.
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
Move all the numbers once clockwise around the 5, and then put the even numbers in the opposite corners. Reflect and rotate the square to get all possibilities. The total of all 9 numbers is 45. Divide this by 3 rows makes 15. The possible ways to create 15 are 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, and 4+5+6. All other combinations don't add up to 15. The central square must fit in 4 different combinations. This requires the number in the center to be 5. This occupies the combinations 1+5+9, 2+5+8, 3+5+7, and 4+5+6, leaving the combinations 1+6+8, 2+4+9, 2+6+7, and 3+4+8. All of the corner squares must fit two of these combinations, which makes 2, 4, 6, and 8 the corners. You can fill the square based on this fact, and you will reveal there is only one real answer for a 3x3 magic square, with variations based on rotation or reflection.
Nicely explained 😊
Take a prize 🥈
there is an easy way I learned years ago. and i actually applied it on 10*10 square, the trick is you have to start on cell which is the last cell on right of the 2nd row from bottom. and after that the next no to put in will be 1 down and 1 right
1 2 3
A
B
C
you are on the cell B3 in this cell you will start and write 1 in it. now you will go 1 down and one left so one down will be C3, then 1 right will be C1 (as there is no cell, you have to go to the left most cell.) now in C1 write 2, and follow this rule, when you come to a cell which is already occupied then you must select the cell which is on the top of it. Assume you come to cell B2 which is already occupied then you select the cell B1.
Follow this pattern and you can solve it easily.
Once you understand the formula you can use this to solve any magic square
Hey can you pls provide yhe proof why diagonals sum is 15?
Pls can anyone prove why diagonals add up to 15?
Hoe sbout using this in chess rook moves
Another interesting note is that all the pairs you can make out of 7, 8, and 9 sum to 15 or more with only two numbers. This means that 7, 8, and 9 must never be in the same row, column, or diagonal as each other. This leaves only 3 possibilities for where 7 8 and 9 can go. Also, 6 and 9 cannot be in the same row, column, or diagonal because they also add up to 15 with just two numbers. This set of restrictions was the first observation I made when trying to solve the question on my own. Unfortunately it's not enough information to solve the whole puzzle and it basically left me in a position of trying to do sudoku to figure out where to place the other numbers by process of elimination. The missing link for me was counting the number of times each other digit shows up in the possible calculations.
When I was about 13, I saw this puzzle without explanation. I took me a whole day to figure it out. Now I know the logic and must teach it to my grandkids.
Great video!
I figured out a quicker way: if we want to fill the number 9 in, we have a total of 6 left for the other 2 squares that are in row/column with 9(bcuz 9+6 =15, its the sum)since there are only 2 ways to add up to 6(1&5, 2&4), we can garantee that 9 has to be at where the T shape intersects. Then according to the middle number is 5, we can fill out the other number in row with 9 - which is 1. Then we take the other pair of numbers that add up to 6 - whcih is 2&4, we can fill out the column that has 9 in there. Then we just subtract the numbers we fill out and can easily fill out the dquare.without finding all 8 variations that gives 15.
Sorry for my English.
Pls can anyone prove why diagonals add up to 15?
Hi @mindyourdecision I got a question for you. You have a 3x3 grid, your goal is to add up to 15 and the center square is 3. You can only use the numbers one time. I've tried it myself and I don't think it's possible. Do you can solve it? Anybody can answer it.
I studied this magic square. I'm wondering if the problem has originated from "Knapsack problem"? In resource allocation where the decision makers have to choose from a set of non-divisible projects or tasks under a fixed budget or time constraint, respectively, and they want to deliver a max value.
Very nice explanation.
I watched a similar digital root video earlier, so I decided to follow that a bit.
All the odd #'s have 2 of each, The even #'s have 3 of each, and the (5) has 4.
So there's 4 odd #'s - 4 even #'s and the 1 (5)
So I took 234,441 / 9 = 26,049
reversed it and did the same
441,234 / 9 = 49,026
Now I took 49,026 - 26,049 = 22,977 the digital root of this is 7+2 , 7+2, 9 = 999
This is actually a very interesting approach to this kind of puzzle. You are more or less tackling some of the fundamentals of group theory in without explicitly saying so.
This is the best explanation of this magic square that I have been able to find on RUclips. Thank you for taking time out of your day to provide free knowledge to the public! Good work.
I agree. Thanks for taking the time out. I guess it's the love for maths.
Yes
Pls can anyone prove why diagonals add up to 15?
Great but how do you use the same method for 4x4 matrix? Row, column and diagonal add up to 34. I tried using the same method and had a problem. According to my list, there are 7 possible additions involving the number 1 but that can’t be right because the most the number could be used is 3 times. I got 1+16+15+2, 1+15+14+4, 1+14+13+6, 1+13+12+8, 1+12+11+10, 1+10+9+14, 1+9+8+16
Pls can anyone prove why diagonals add up to 15?
Here's how I did:
1+2+3+4+5+6+7+8+9 = 45, so the sum of the elements in each triplet has to be 45/3 = 15.
If the lowest number in a triplet is 1, then the second lowest number has to be at least 5, giving us the highest number as 9.
If we then increase the middle number by one, giving us 6, we have to decrease the highest number by one (in order to maintain the sum of 15), giving us 8.
Repeating this process, the middle number becomes equal to the highest number, which is of course not an option, so all-in-all there are two possible triplets containing the number 1.
In the same manner we can show that there are only eight unique triplets:
(1, 5, 9), (1, 6, 8), (2, 4, 9), (2, 5, 8), (2, 6, 7), (3, 4, 8), (3, 5, 7) and (4, 5, 6)
Seeing as there are 3 rows, 3 columns and 2 diagonals (3+3+2 = 8), all eight triplets must be used in order to complete the magic square.
Let's start with the middle square. This square is connected to the middle row, the middle column and the two diagonals (2+2+1 = 4), so any number that goes here has to be present in at least four of the eight triplets, and the only number that applies to is 5.
So the middle square can only contain the number 5.
Now, the number 1 can only be put in the upper-left square or the upper-middle square (all other possibilities are either rotations or mirrors of these two scenarios).
If we choose to put it in the corner square, this gives us that the lower-right square has to contain the number 9, and the upper row has to be either (1, 6, 8) or (1, 8, 6):
1 6 8 1 8 6
5 5
9 9
Both scenarios make the sum of the numbers in the rightmost column to exceed 15, which means that the number 1 can only go into the upper-middle square.
This gives us that the lower-middle square has to contain the number 9, and the upper row has to be either (6, 1, 8) or (8, 1, 6), but these two scenarios are mirror images of each other, which leaves us with but one possibility:
6 1 8
5
9
This gives us that the diagonals have to be (6, 5, 4) and (8, 5, 2):
6 1 8
5
2 9 4
Then, the leftmost column has to be (6, 7, 2) and the rightmost column has to be (8, 3, 4), thus completing our magic square:
6 1 8
7 5 3
2 9 4
This can be rotated up to three times. It can also be mirrored and then that mirror image can be rotated up to three times. This gives us a total of 4×2 = 8 possible solutions:
6 1 8 2 7 6 4 9 2 8 3 4
7 5 3 9 5 1 3 5 7 1 5 9
2 9 4 4 3 8 8 1 6 6 7 2
8 1 6 4 3 8 2 9 4 6 7 2
3 5 7 9 5 1 7 5 3 1 5 9
4 9 2 2 7 6 6 1 8 8 3 4
good job, nerd
Jerry Nilsson solve the magic puzzle sum 12
Salman Mehmood don't say nerd
Jerry Nilsson ...
Good ...!
This is given by my teacher as a holiday assignment ,
Excellent Communication!
You can also form 3 * 3 squares with trios of constant ratio numbers
Example: 2 3 4
6 7 8
10 11 12
11 2 8
4 7 10
6 12 3
Very nicely explained. Was able to understand how to place the numbers in the matrix very easily. Good job.
Pls can anyone prove why diagonals add up to 15?
There is another way to find the center no.
Assuming the sum of each row/col/diagonal to be S(where S=15)
then
4S= 45(sum of 1-9)+ 3*centre
(as shown in video we can pass 4 lines from center and counting the sum of numbers traced by lines equals sum of all numbers plus 3*centre no)
now on evaluating the above eqn you can find the centre no.
This is really expected from you only Sir.
Wow, Brilliant explanation thank you so much
please continue making this kind of Math puzzles or do a course of Discrete Math
Hi you are a good teacher hope you have a blessed day thank you dear
Soooo, what about a 4*4 Magic square? Or a 5*5? Could you do a video about these ones?
i can do 3×3 5×5 7×7 9×9 and so on..
3*3,4*4,5*5 is easy
6*6~ hard
I am korea people
@@SP-ti8kz all are easy
@@ravigupta5899 lolol
No
I just found this and realized a math competition I entered in had this same problem but used 5 in the middle like always and x above that and x+1 in the bottom right corner and asked for the sum of all possible x values. I just skipped that one :p
Really Nice.Everywhere I got trick and shortcuts.But here I got mathematical approach.Wonderful Sir.Very nicely explained.
Very nice explaining
Thank you! This was really helpful for my assignment.
Me too
Cheaters
@@Violet_Evergarden Its not cheating its just like a formula.
Me toooo
Mee to
I think this can help me to do homework
Here's a purely mathematical way to show the middle square has to be 5:
Going from left to right in each case, (a) let the numbers on the top row be a, b, and c; (b) let the numbers on the middle row be d, e, and f; (c) let the numbers on the bottom row be g, h, and i. Knowing that every row, column, and diagonal must total 15, we have:
a + e + i = 15
d + e + f = 15
g + e + c = 15
Adding those three equations together and rearranging the terms a bit gives us:
a + d + g + c + f + i + 3e = 45
But we also know a + d + g = 15, as it is the left column. Similarly, c + f + i = 15 because it is the right column. So the above equation simplifies to:
15 + 15 + 3e = 45
3e = 15
e = 5
...and we're done!
By the way, once you have the 5 in the middle, the rest is simple. There are only two ways to use the 1 to total 15, either 1 + 5 + 9 or 1 + 6 + 8, which means the 1 can't go in the corner. So, without losing generality due to rotations and reflections, put the 1 in the middle square on the top row, the 8 and 6 on either side of the 1, and the rest of the magic square fills in automatically.
If someone knows the Agrippa or Durer methods to solve 3 x 3 square can make even 7 x 7, 11 x 11, 21 x 21 and so on. A new square odd number method 2018 is introduced by MH.
If you look at higher odd ones, like 5X5 or 7X7 or greater, you can immediately see the genertal solution.
This is very interesting
Thank you so much! Home schooling my child and this video has been a godsend! Extremely helpful and so very easy to understand. Your other book resources look awesome too! So glad I "stumbled" across your channel whilst performing a mathematical search
very easy explanation
very helpful video
thanks,🙂 it is very needed for my coding problem in 2022
I love your videos, I got the last few, this one I got, but I loved the way that you solved it, and I had no idea how you could prove there were no others. LOVE YOUR VIDEOS
THERE ARE EIGHT MORE WAYS TO DO IT THE EIGHT HE MIST ARE THE MOST IMPORTANT . 16 WAYS
Pls can anyone prove why diagonals add up to 15?
My grandfather done this back in the mid 1980’s. Then kept going. The last puzzle that was never finished was a either 21x21 or 20x20. But he went threw 3x3 all the way up over a span of 5 years. Daily puzzle for him to keep his 80 year old brain going. All were writing down. Not sure I can find them any more but. You solution answer of 1 can’t be in a corner is wrong. If you rotate each number 1 clock rotation left or right all the numbers will still add to 15. As long as 5 stays on the middle. Just letting you know. I was thinking of picking up where gramps stopped and start at 21x21 and work my way up from there. I could use a puzzle in the corner like a chess set to start and let it wait when I’m busy. I was just looking up how he got the 15 all ways and the 25 all ways for the 4x4 puzzle if I recall correctly. Nice job.
That is not true. Odd numbers cannot be in a corner. If 1 was in a corner, then there have to be 3 different sums, one including a 5. So, there is 1+5+9, 1+8+6... the third one would have to be 1+7+7 which obviously is not possible.
@@thomaslohrschall9410 I relooked and I don't have a clue what I was thinking.. It looked like it could.. Sorry.. Your right.. I goofed up.. Thanks .. :P...
Thanks for this magic game.......
my teacher gave us this
Me too
Same
@@mr.orange2404 did you figure out how to solve it
Mee tooooo
This teacher is so asome ✌
any good discrete math course or book do you recommend for beginners
Good job my bro, yo my ji!😀
Need bad
now to tell you a fact about magic squares i discovered many years ago and hoped to weite a paper about,but never got it together. in any magic square of consecutive numbers then, if you subtract the lowest value, so that the numbers run from 0 to 8, now convert the numbers to base 3 in two digit form ie 00,01,02,10,11,12,20,21,22 you will see a pattern. in the firstdigit and a different pattern in the second digit.this fact applies to every magic square no matter its size. these patterns can be used in the 4 x4 and higher squares to generate differnet types of magic square( apart from the rotation and reflection. for uniqueness put 00 in the lowest possible cell number then put 01 in the lowest possible cell number. there are several interesting transformations.t
one is to swap the digit positions ie 21 is transformed to 12 another is to swap cell number and cell value.
anther thing you cna do with the patterns is to multiply them by coprime number eg 4 and 5. this will give a magic square of non consecutive numbers.
Just like there are 8 ways for a 3x3 matrix from the number 1 to 9 then If there is 1,2,3,4 till 16 in the 4x4 matrix then what are the number of ways (by forming different matrix) of achieving the total of 34 into rows, columns, and diagonals.
this is cool i love squares, i make all kinds of magic squares i can use any numbers to make a 3x3 or 4x4 5x5 etc i love these i also make magic circles and stars its the best thing for a technical mind.
My tuition teacher is gave it to me . Thank you
U made it so easy to understand
Genius 🤔👌👌very much interesting
How do you find the sum of 4 corners and determine the key number
Hr chez bht achy sy explain krny Wali sb comments ma apeel kry alishey ko youtube ki tarf sy prize gift Kary
That's easy there's 2 diagonal squares joined together and there apart of a ball
This is how they can use the pelican with nas nose with the dino hiding behind it to go to the iftari people to use the same ball as a fidget club spinner with an dark nose ready to fire down for an iftari meal 😂
Thank u, I got what I came for.
Helpful video
Nice explanation
Good job bro
why does the 5 always ends in the center from it all??? Is a half, 5 is representing a half?
Is it the same method for a 4x4 or 5x5 cube?
the easy step is put in middle 5.....
2 step: put odd numbers up,down,right,and left,,(1,9,3,7)
3 step: there are four corners,, put any one even numbers at one corner and try add those two numbers to get third one... than its sloved...
can you slove sum of puzzle 12 plz
this guy is simply awesome
A blackjack magic square is a 3x3 array of cards where for each row/column/diagonal the magic sum is exactly 21 as in blackjack. The proof that an ace can never be 1 in any sum is very straightforward, and the proof that the center card has to be a seven immediately follows.
The number in a row or colloum or diaginal equals to the same number
Mcdonald French fries amazing, how did you come up with this?
Thank you so much
Did anyone else note the Konami code in 0:40 when Presh shows the 8 possible arrangements?
What I'm getting is, 5 must go in the center. 1-8-6 must go along some side (top side, without loss of generality), in some order. 1 must go in the center of said side. Without loss of generality, place 6 on the left side of 1 (TL corner) and 8 on the right side of 1 (TR corner). Of course 9 goes opposite 1; 2 goes opposite 8; yada, yada, yada, it solves itself from that point. So this is the only solution, along with its reflections and rotations. You can get 4 rotations. Then flip it, and you get 4 rotations there too. 8 solutions, but only 1 if you discount rotations and reflections.
thanks for the magic square
The Square of the Sun (6 x 6) is also the Square of the Beast. That is the total sum of all the Numbers ='s '666' (Each row, column and/or diagonal ='s 111) ! For who is like unto the Beast....for 'Ye Shalt Know Them By Their Tales or is it Tails' !
Thanks it really helped me
outstanding 👍💝👏👏
I have been watching your videos for half a decade and I still didn't catch up what do you say your name is on beginning of videos. Something like Presh Toloker??
I'm pretty sure that It's Presh Talwalkar, an Indian surname from the Midwestern region of India where the "kar" suffix occurs often in surnames such as Sachin Tendulkar
This is really helpful and interesting thank you 😊
I did not use that step 2 at all. Here is how I did it:
Step 1: Prove that each sum is 15
Step 2: Diagonals + middle row + middle column is 4*15, which is 60. Subtract 45 (the whole square) from 60 to get three times of the middle square. Deduce from that the center is 5.
Step 3. Prove the number 1 cannot be at the corner. If 1 is at the corner, the opposite corner must be 9. Suppose a remaining corner is odd, since odd+odd+even is not 15, the numbers between that corner and both 1 and 9 are odd. There are only five odd numbers in all, so this is impossible. Suppose a remaining corner is even, all the remaining squares can be shown even, which is also impossible. Therefore, 1 cannot be at the corner.
Step 4: If 1 is in the top spot, 9 must be at the bottom. 1 plus two numbers yield 15 means that the sum of the two numbers next to 1 is 14. Since 9 is used up already, only 6+8 works. Similarly for 9, the other two numbers at the bottom are 2 and 4. Since 5 is in the middle, 2 must oppose 8, and 4 must oppose 6. This provides two different arrangements as 2 can be on the left or right. The number 3 and 7 must be filled in according to the position of 2, so they do not provide further arrangements.
Step 5: The same thing can be done with 1 in the other three side positions. Thus, the total number of ways to construct a 3X3 magic square is 2*4, which is 8.
Thanks this helped alot
why so many dislikes? i thought that explanation was amazing
doing this for fun :3
Brillant idea thank you sir
Nice sum
洛書
戴九履一，左三右七
二四為肩，六八為足
五在中央。
四奇為陽居正
四偶為陰在隅
四正逆行與四隅相減
則數皆返於中五
由後天逆行回返先天之象
How to do for sum =18
??
3 8 7
10 6 2
5 4 9
How about seeing the square as a ball. Start with 1 top middle and then go 1 right and 1 up place 2, right and up place 3, right and up now is occupied by 1 then go 1 down and so on. Works die every odd squared puzzel.
8 3 4
1 5 9
6 7 2
Sum of all row , all diagonal , all columns is 15
Thank you This was really helpful for my assignment
Nice video
As a 5th grader we did this in school and it was a lot easier there
Very nice
Well Done :)
The centre square has to be 5, the 1 can not be in a corner, the solution follows.
could this problem be solved: APRIL + FOOLS = PRANK where letters are one-one mapped to digits {0, 1, 2, ..., 8, 9}
can you elaborate ?
3:50 look at the numbers on the side and you would figure out that they would add to 15 in that way to
A true magick square is embedded with commands that spirits must follow.
Nice method
making 15 is easy peasy place them in number order
8:02 I managed to figure out what you explain here, but I wish you wouldn't have shot past it the way you did.
Vulcan Viper g
thanks
Patrick reid
I swear, your voice gets higher and higher every video...
b/c get excited when problems have been clearly solved.
You spoiled the solution in the thumbnail
Awesome sir...thanks
Please do the magic square of 3×3 by keeping the centre as 7
You are joking right? It might be with different set of number, not number from 1 to 9.
very very nice
Thanks for this bro...This is awesome
Amazing...solution