Like for 75 we have (5,5,3), (25,3),(15,5),(75,1). Similarly how to do this for any given number if that number is given in its prime-factorised form?
Finding the set of all possible divisor products to form a number
Like for 75 we have (5,5,3), (25,3),(15,5),(75,1). Similarly how to do this for any given number if that number is given in its prime-factorised form?
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