3x3 magic square 25 total
Yes, it is not that difficult.
use the inverse square method, it works the fastest
If you are familiar with Pythagoras's theorem you will know that it is 5x5 + 5x5 => 25 + 25=50 find the square root of fifty and you will have your answer
all verticals, horazontals and diagonals must add up to one common number
3x3 magic square 25 total
You don't
Simply,square roots are solved like this; example: 5 2 = 5x5 square roots are simply multiplying the number itself.
Yes, it is not that difficult.
A normal 3x3 magic square has a sum of 15. So you subtract 3 from each number in the square.
use the inverse square method, it works the fastest
No, not really - at least not for me (I'm 52yo) I think that the differences between the 4x4 and 5x5 are slight. It's one of those things where "if you can solve a 3x3 you can solve a 4x4 and if you can solve a 3x3 and a 4x4 you can solve a 5x5. The hardest part comes with the edges - there's some weird manuevers you have to do to get the edges right and oddly enough the 5x5 is easier on the edges than the 4x4 because the 5x5 has a center edge for each color and that makes it easier to get matches. It did for me anyway. But once I learned the 4x4, the 5x5 wasn't too hard to pick up.
The fastest 5x5 solve in official WCA competitions is 44.74 seconds, set by Max Park from the USA in 2021.
This skills that someone has to solve a Rubik's cube can be used on any size Rubik's cube, whether it be a 2x2 or a 5x5.
25 it is the square of 5 (5x5=25)
If you are familiar with Pythagoras's theorem you will know that it is 5x5 + 5x5 => 25 + 25=50 find the square root of fifty and you will have your answer
all verticals, horazontals and diagonals must add up to one common number