If we use 9 distinct numbers from 1 to 27 to make a 3 by 3 magic square, what is the most number of primes that we can use?
well, there are 10 primes, but I have no idea whether you can make a magic square with just primes.
My basic 3 by 3 magic square looks like this using the numbers form 1 to 9:
8 1 6
3 5 7
4 9 2 for a sum of 15 for all its rows, columns, and diagonals
Without changing the sum, I can use symmetry to have different versions of that
e.g
2 7 6
9 5 1
4 3 8
There is also a magic square where the numbers form a fixed product of the numbers in the rows, columns, and diagonals
eg.
http://mathworld.wolfram.com/MultiplicationMagicSquare.html
but I don't think you were thinking of that.
For a "normal" magic square, the numbers form an arithmetic sequence.
so with a common difference of 1 we could use
1-9 , number of primes = 4 ,
2-10 , 4 primes
3-11 , 4primes
4-12 , 3 primes
....
19-27, 2 primes
There will be 19 of these magic squares, with the maximum number of primes in any one of them being 4
As soon as I make the common difference in the arithmetic sequence other than 1, e.g. 3
the numbers in the square would be multiples of that difference, hence no primes!
So primes can exist only if we have consecutive numbers.
I would say the max number of primes in any 3by3 magic square is 4
Steve mused if there was a magic square using only primes,
my guess is no, since the numbers must be in arithmetic sequence.
it is also wrong. plss check it again
Can not a magic square be just a collection of random numbers that work magically? I was not aware that they had to be in arithmetic sequence.
If we require an arithmetic sequence, then just lay a sliding scale over the numbers from 1-27 and see which string of 9 contains the most primes.
Steve, back in the 80's when I was teaching a computer science course using GW-Basic on a "Trash -80" , I recall an assignment where using any number from 1 to 50, they were to find a magic square using a random number generator of size 5by5.
Of course the processor was so slow that we actually had to run the machines all night and wait to next day to get any results. We observed that for all results, the numbers were always consecutive numbers.
This does not prove it, but indicates strong indication it might be true.
You also stated that I was wrong in my reply, I can't find my error.
Note that the assertion of error, came from "steve" not from "Steve"
Identity Theft!!!!
Back in the 60's when I was in HS, the CS class had to punch up their FORTRAN programs on cards, send them to Olympia for processing, and wait several days for the results. Now THAT's slow turnaround!
Luckily, I was a math geek back then, and had no real interest in computers. How times change...
so atlast what is the answer of my question? ???????
To find the most number of primes that can be used to create a 3 by 3 magic square using 9 distinct numbers from 1 to 27, we need to consider a few things.
First, let's understand what a magic square is. A magic square is an arrangement of numbers in a square grid, where the sum of each row, column, and diagonal is the same.
Since we need to use 9 distinct numbers, we can rule out any duplicates. Now, let's consider the range of numbers from 1 to 27. To maximize the number of primes, we should try to use as many prime numbers as possible.
Let's list the prime numbers within the given range:
2, 3, 5, 7, 11, 13, 17, 19, 23
Now, let's analyze the possible combinations to find the most number of primes. We can start by selecting one prime number and then filling the other cells in the magic square with non-prime numbers.
If we choose 2 as our prime number, we are left with 8 remaining numbers to choose from (3, 5, 7, 11, 13, 17, 19, 23). However, if we choose 2 as our prime number, we can't use any of the odd numbers (3, 5, 7, 11, 13, 17, 19, 23) in the magic square.
Similarly, if we choose 3 as our prime number, we can't use any other odd prime numbers. So, this approach won't help us achieve the maximum number of primes.
Therefore, we need to consider a different approach. Let's start by selecting the largest prime number, 23, for one of the corners of the magic square. Now, there are 8 remaining numbers to choose from.
To maximize the number of primes, we'll continue to select the largest prime numbers for the other two corners of the magic square. This will leave us with 5 remaining numbers to fill the remaining 6 cells.
To maximize the number of primes, we'll select the remaining candidates from the list of prime numbers (2, 3, 5, 7, 11, 13, 17, 19). After careful consideration, we can choose the following arrangement:
23 | x | 2
--------------
x | 11 | x
--------------
x | x | 7
In this arrangement, we have used 4 prime numbers (23, 2, 11, 7) to fill the magic square. Therefore, the most number of primes we can use is 4.
Note: While this solution provides the highest number of primes, it does not guarantee that there isn't a solution with more primes. Finding the absolute maximum requires an exhaustive search of all possible combinations.