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A positive integer is called "square-free" if it has no
factor that is the square of an integer greater than 1.
If n is an even square-free integer, which of the following
must also be square-free?
A)n/2
B)2n
C)n+2
D)n^2
E)None of the above
Archived Topic
Hi there,
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First let us see what a "square free"(SF) integer means. By definition it means that the SF ineteger doesn't has a factor which is sqaure of any integer greater than 1. What that means is that when SF integer written in terms of prime factors than none of the prime factors have power greater than 1.
Example of SF: 2*3*5=30, 5*7=35, 3*7*2=42
Example of Not SF: 2^2*3=12, 2*3^3*5=90
* Now its evident that if any integer n is even SF (must contain 2 as one of its prime factors), n/2 will also be.
some other stuff to ponder over: * If an interger is even SF, "2n" will NOT be SF. (2^2 factor will come)
* If an integer >1 is odd SF integer, n/2 will not be an Integer.
- Vicks
if N is a even square-free integer then N/2 has to be odd other wise
it would have a factor which is 4 or square of 2
So we can represent N as 2*m * ( 2n+ 1 )
m is odd and n is either odd or even
so N/2 = m * (2n+1 )
m != 2n+1 because N would then have a square of some odd number
So N/2 is also a square-free integer
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.