If n is an even integer, which of the following must also be an even

are there any way I needn't try one by one

Yes.

N being even means

N=2K where K is any integer.

So plugging in to each option:

I. (4K^2-4)/4 = K^2-1. If K is an even number so is K^2 making the expression odd, so eliminate I.


II. (N^2+2N)/4 = (4K^2+4K)/4

= K^2 +K. If K is odd, so is K^2 and adding two odd numbers is even. If K is even, so is K^2, so the expression is always even.


III. (N^2+4N)/4 =

(4K^2+8K)/4 = K^2 + 2K =

K(K+2). If K is odd, so is K+2 and two odd numbers multiplied is always odd, so eliminate III.

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I. (n^2 - 4)/4 , let n = 4

So, (4^2 - 4)/4 = 3

II. (n^2 + 2n)/4 , let n = 4

So, (4^2 + 2*4)/4 = 6

III. (n^2 + 4n)/4 , let n = 2

So, (2^2 + 4*2)/4 = 3

Thus, Answer must be (A) II only

­If n is an even integer, which of the following must also be an even integer?

I. \(\frac{(n^2 - 4)}{4}\)

The square of any n that is a multiple of 4 is 4 × an even number.

So, if n is a multiple of 4, \((n^2 - 4)\) will produce a number that's 4 × an odd number.

Thus, \(\frac{(n^2 - 4)}{4}\) can be an odd number.

Eliminate.

II. \(\frac{(n^2 + 2n)}{4}\)

The square of any even n that is a multiple of 4 is 4 × an even number.

Also, 2 × any n that is a multiple of 4 is 4 × an even number.

So, \((n^2 + 2n)\) is always 4 × an even number +  4 × an even number when n is a multiple of 4.

So, since even + even = even, if n is a multiple of 4, then \((n^2 + 2n)\) is always 4 × even.

Thus, \(\frac{(n^2 + 2n)}{4}\) is always an even number when n is a multiple of 4.

The square of any even n that is not a multple of 4 is 4 × an odd number. For example \(6^2 = 36 = 4 × 9\).

Also, 2 × any even n that is not a multiple of 4 is 4 × an odd number. For example \(2 × 6 = 12 = 4 × 3\).

So, since odd + odd = even, if n is not a multiple of 4, then \((n^2 + 2n)\) is always 4 × odd + 4 × odd = 4 × even.

Thus, \(\frac{(n^2 + 2n)}{4}\) is always even when n is not a multiple of 4.

So, for any even n, \(\frac{(n^2 + 2n)}{4}\) is even.

Keep.

III.\(\frac{(n^2 + 4n)}{4}\)

The square of an even number n that is not a multiple of 4 is 4 × an odd number.

4n is always 4 × an even number.

So, since odd + even = odd, \((n^2 + 4n)\) can be 4 × an odd number.

Thus, \(\frac{(n^2 + 4n)}{4}\) can be odd.

Eliminate.

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III­

Correct answer: A­