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Let's solve this question with some values, first let's pick n=0

I. (n²-4)/4 >> -4/4 >> -1 (an odd integer)

Eliminate C and E

Now for III, let's pick n=6

total will be (36+24)/4 >> 15( an odd integer)

Eliminate B and D, we are left with only one option 'A'
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are there any way I needn't try one by one
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If n is an even integer, which of the following must also be an even integer?

I. (n^2 - 4)/4

II. (n^2 + 2n)/4

III. (n^2 + 4n)/4

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

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

I. (n^2 - 4)/4

II. (n^2 + 2n)/4

III. (n^2 + 4n)/4

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

­
You can solve it by number plugging (because "none of the above" is not an option) but I would prefer the conceptual approach to be certain. 

Let's look at number plugging first. 
Put n = 0 (an even integer).
I gives -1 which is not an even integer. 
II and III give 0 which is an even integer. 

Put n = 2 (an even integer)
II gives 2, an even integer
III gives 3, not an even integer

Since at least one of I, II and III must always be an even integer, we know that answer must be (A) II only.


Method 2: Put n = 2a (a could be any integer odd or even)

I. \(\frac{((2a)^2 - 4)}{4} = a^2 - 1\) ; If a is even, this is odd.

II. \(\frac{((2a)^2 + 2*2a)}{4} = a^2 + a\);  Whether a is odd or even, this sum is always even. 

III. \(\frac{((2a)^2 + 4*2a)}{4} = a^2 + 2a\);  If a is odd, this sum is odd. 

Answer (A)­
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The method I used for this was to recognize that II. (n^2 + 2n)/4 can also be written as:

(n*(n+2))/4

Given that n is even, the numerator here is the product of two consecutive even integers. We know that the product of n consecutive even integers must always be divisible by (2^n * n!). So this must be divisible by 2^2 * 2!. We already have 4 in there, but we know that the result of this must also be divisible by 2!, which is 2. So the result must be even.
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If n is an even integer, which of the following must also be an even integer?

Let us assume n= 2k; where k is an integer

I. (n^2 - 4)/4 = n^2/4 - 1 = k^2 - 1 = (k-1)(k+1); Odd when k is odd and even when k is even; May or may not be an even integer

II. (n^2 + 2n)/4 = (4k^2 + 4k)/4 = k^2 + k = k(k+1); Product of 2 consecutive integers, one of which must be an even integer; EVEN INTEGER

III. (n^2 + 4n)/4 = (4k^2 + 8k)/4 = k^2 + 2k = k(k+2); Odd when k is odd and even when k is even; May or may not be an even integer


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

IMO A
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