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If $x = (a-1)(b-2)(c-3)(d-4)(e-5)$, where $a,b,c,d,e \in \mathbb{N}$ are distinct natural numbers less than 6. If x is a non zero integer, then the how to count the no of sets of possible values of $(a,b,c,d,e)$?

which is the fastest (pencil-paper) solution of this problem?

$ \mathbb{N} = 1,2,3,\cdots $

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    A thing you can use to start is factore x and put it in the form $x=p_1^{\alpha_1} p_2^{\alpha_2} p_3^{\alpha_3} \cdots$. Then, necessarily, your factors $(a-1), (b-2), \cdots$ must be a combination of this factors of x ($p_1^{\alpha_1}, \cdots$). Remember the negatives products too.2011-11-02

5 Answers 5