Create n x 2n "mirror" matrix of this type:
For n = 2
m = [ 1 2 2 1
1 2 2 1 ]
For n = 3
m = [ 1 2 3 3 2 1
1 2 3 3 2 1
1 2 3 3 2 1 ]
This problem was fun. I was able to find a better way to solve beat my own solution. Not the best way though. But lot better.
Find the sum of all the numbers of the input vector
Getting the row and column location from a matrix
Sum the Digits of a Number
Returning a "greater than" vector
Length of the hypotenuse
Sum of series IV
Sum of series VII
Sum of series II
Sum of series III
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