- #1
smiddy
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Homework Statement
Prove that for every k >= 2 there exists a number with precisely k divisors.
I know the solution, but don't fully understand it, here it is;
Consider any prime p. Let n = p^(k-1). An integer divides n if and only if it has the form p^i where 0<= i <= (k-1). There are k choices for i, therefore n has exactly k divisors.
Could someone fully explain the thought process involved in finding the solution, I understand p^i etc, just don't know where p^(k-1) comes from.