document.write( "Question 265387: Will n! +2, n! +3,..., n! +n for n>=2 always be a sequence of n-1 composite numbers? Why? \n" ); document.write( "

Algebra.Com's Answer #195142 by jim_thompson5910(35256) $\"\"$ $\"About$

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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.\r
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\n" ); document.write( "\n" ); document.write( "For any other value of n>=2, 2 will always be a factor of n!\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "Because of the facts stated above, we can factor out the GCFs to get the new sequence:\r
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\n" ); document.write( "\n" ); document.write( "2((n-1)!+1), 3((n-1)!+1), 4((n-1)!+1), ... n((n-1)!+1),\r
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\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "

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