Question 265387: Will n! +2, n! +3,..., n! +n for n>=2 always be a sequence of n-1 composite numbers? Why?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Take note that 2 is a factor of n! if n>=2. Why? Let's say that n=5, then n!=5!=5*4*3*2*1.
For any other value of n>=2, 2 will always be a factor of n!
Likewise, 3 is a factor of n! if n>=3. Notice that if n=2, then you only have one term. Similarly, 4 is a factor of n!+4 for n>=4 (if n<4 then you have at most two terms). This idea generalizes to the fact that n is a factor of n!. This is trivial since n!=n(n-1)...(2)(1) for any value of n.
Because of the facts stated above, we can factor out the GCFs to get the new sequence:
2((n-1)!+1), 3((n-1)!+1), 4((n-1)!+1), ... n((n-1)!+1),
Since each term in the sequence is a product of two factors (other than 1 or itself), this means that each term is a composite number for any value of n.
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