bigp0ppa1046
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If n≥ 2, show that all numbers
n!+2, n!+3, n!+4, .... n!+(n-1), n!+n
are composite numbers.[/code][/tex]
n!+2, n!+3, n!+4, .... n!+(n-1), n!+n
are composite numbers.[/code][/tex]
Chapter: Counting Primes
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Here's sort of an informal idea...
Since n>=2, we know that 2 is a factor of n!, so n!+2 also has two as a factor (just factor out a two). Two is even better though since it takes car5e of all numbers of the form n!+even#.
Similarly, if n >= k we know k is a factor of n! thus n!+k also has a factor of k.
Since n>=2, we know that 2 is a factor of n!, so n!+2 also has two as a factor (just factor out a two). Two is even better though since it takes car5e of all numbers of the form n!+even#.
Similarly, if n >= k we know k is a factor of n! thus n!+k also has a factor of k.
bigp0ppa1046
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I see what your saying...so for example if n=4
we have n!+2 n!+3 and n!=4, but the numbers 2,3, and 4 are already included in n! so if we add them to n! then they are still a different factor than 1 and n itself, making it composite. do i kinda have it?
we have n!+2 n!+3 and n!=4, but the numbers 2,3, and 4 are already included in n! so if we add them to n! then they are still a different factor than 1 and n itself, making it composite. do i kinda have it?
Sorta... but try something more rigorous. Try to prove the following statement: Let n be a natural number >= 2. Then all numbers of the form n! + k with 1<k<=n are composite numbers.
Since n>k, we know n! looks something like this:
n! = 1*2*3*...*(k-1)*k*(k+1)*...(n-1)(n).
So n! + k = (1*2*3*...*(k-1)k(k+1)*...(n-1)(n)) + k = k(1*2*...(k-1)(k+1)*...*(n-1)(n) + 1). I.e. k is a factor of both n! and k.
Further, k cannot be 1, nor can it be n! so any factor k you can take out makes that number composite.
Since n>k, we know n! looks something like this:
n! = 1*2*3*...*(k-1)*k*(k+1)*...(n-1)(n).
So n! + k = (1*2*3*...*(k-1)k(k+1)*...(n-1)(n)) + k = k(1*2*...(k-1)(k+1)*...*(n-1)(n) + 1). I.e. k is a factor of both n! and k.
Further, k cannot be 1, nor can it be n! so any factor k you can take out makes that number composite.
Hello, bigp0ppa1046!
The numbers are of the form: \(\displaystyle N\:=\:n!\,+\,k\), where \(\displaystyle k\:=\;2,\,3,\,4,\,\cdots\,n\)
Since \(\displaystyle 2\,\leq\,k\,\leq\,n,\;k\) is one of the factors of \(\displaystyle n!\)
Then: \(\displaystyle \:N\;=\;\left(1\cdot2\cdot3\cdot4\cdots k\cdots n\right)\:+\:k\)
. . which can be factored: \(\displaystyle \:N\;=\;k\left(1\cdot2\cdot3\cdot4\cdots n\:+\:1\right)\)
Therefore, \(\displaystyle N\) is a composite number.
The numbers are of the form: \(\displaystyle N\:=\:n!\,+\,k\), where \(\displaystyle k\:=\;2,\,3,\,4,\,\cdots\,n\)
Since \(\displaystyle 2\,\leq\,k\,\leq\,n,\;k\) is one of the factors of \(\displaystyle n!\)
Then: \(\displaystyle \:N\;=\;\left(1\cdot2\cdot3\cdot4\cdots k\cdots n\right)\:+\:k\)
. . which can be factored: \(\displaystyle \:N\;=\;k\left(1\cdot2\cdot3\cdot4\cdots n\:+\:1\right)\)
Therefore, \(\displaystyle N\) is a composite number.