fozzzy wrote:
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. \(\frac{n}{2}\)
B. \(2n\)
C. \(n + 2\)
D. \(n^2\)
E. none of the above
What are examples of square-free numbers?
kanusha wrote:
sir, pls explain this problem
Dear
Kanusha,
I'm happy to respond to your p.m.
First of all, let's think about this --- the perfect squares greater than 1 are
4, 9, 16, 25, 36, 49, etc. Any numbers divisible by these would not be square-free, and any number that doesn't not happen to be a multiple of these would be square-free. Notice, in particular, all prime numbers are automatically square-free. The first few square-free numbers would be
1, 2, 3, 5, 6, 7, 10, 11, 13, 14 ,15, 17, 19, 21, 22, 23, 26, 29, 30, ....
The problem tells us that n is an even square-free numbers, so candidate could include
2, 6, 10, 14, 22, 26, 30 --- all of these are even numbers that are NOT divisible by 4, so they are the odd multiples of 2, and not even all of those, because some (such as 18) are divisible by other squares (18 is divisible by 9).
Let's think about these .....
(A) n/2 --- using our example even square-free numbers above, this leads to 1, 3, 5, 7, 11, 13, 15 --- all square free --- hmmmm, this is promising
(B) Any even number time 2 is divisible by 4 --- this always leads to a number divisible by a square, i.e. NOT square-free. This is
incorrect.
(C) Any odd multiple of 2, plus two, will equal an even multiple of two --- i.e. a multiple of 4, a number that is NOT square-free. This is
incorrect.
(D) n^2 would be even times even, which is always divisible by four, i.e. NOT square-free. This is
incorrect.
(E) hmmm.
It's very easy to eliminate
(B) &
(C) &
(D). In order to choose
(A) over
(E), we have to be sure that
(A) works, not just for some examples that we pick, but for all possible even square-free numbers. Here's an argument, involving factors.
Suppose n is an even square-free number. As we have seen, it must be an odd multiple of two. This means n = 2*q, where q is some odd number. Because n has no factors that are squares, this necessarily means that q has no factors that are squares. (If P = R*Q, then any factor of Q or R must be a factor of P.) Now, n/2 = q, and q has no square factors, so if n is square-free, then n/2 = q must be square-free. That's why
(A) must be the correct answer.
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)