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If $V$ is a vector space over the field $\mathbb{Z}/5\mathbb{Z}$ and $\dim_{Z/5\mathbb{Z}}(V)=3$ then $V$ has.

  1. 125 elements
  2. 15 elements
  3. 243 elements
  4. None of the above
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Field $\mathbb{Z}/5\mathbb{Z}$ means set of integers modulo 5 i.e. this field is $\{0,1,2,3,4\}$.

So vector over this field will contain elements from set $\{0,1,2,3,4\}$ only.

Dimension of each vector is 3 i.e. a vector is a 3-tuple where each element is from $\{0,1,2,3,4\}$.

So for each element in 3-tuple, we have 5 choices, hence total number of vectors is $5*5*5=125$.
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modulo will be 0,1,2,3,4 how 5 will be remainder.
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Oh Sorry. Corrected now. Thanks :)
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