# How to calculate product 1*2*3*...7 in the ring Z8 with two defined operations?

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### 1 Answer

You should remember that the abstract algebraic structure called ring is a set of integres, hence, the ring `Z_8` is the following set such that:

`Z_8 {hat 0 , hat 1 , hat 2 , hat 3 , hat 4 , hat 5 , hat 6 , hat 7}`

Since you need to evaluate the given product, you should know that if `x in Z_8` and `y in ` `Z_8` , hence `x*y in ` `Z_8` (x*y mod 8 represents the reminder when x*y is divided by 8) such that:

`hat 1*hat 2*hat 3*...*hat 7 = (hat 1*hat 2)*hat 3*...*hat 7`

`hat 1*hat 2*hat 3*...*hat 7 = (hat2*hat 3)*...*hat 7`

`hat 1*hat 2*hat 3*...*hat 7 = (hat 6*hat 4)*hat 5*hat 6*hat 7`

`hat 1*hat 2*hat 3*...*hat 7 = (hat 0*hat 5)*hat 6*hat 7`

`hat 1*hat 2*hat 3*...*hat 7 = (hat 0*hat 6)*hat 7`

`hat 1*hat 2*hat 3*...*hat 7 = (hat 0*hat 7)`

`hat 1*hat 2*hat 3*...*hat 7 = hat 0`

**Hence, evaluating the given productin the ring `Z_8` yields `hat 1*hat 2*hat 3*...*hat 7 = hat 0` .**

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