You can also try plugging in an odd integer and an even integer to see what happens.

Start by plugging in an EVEN integer. The easiest even value to plug in is x = 0
We get...
A: 0² - (0) - 1 = -1 = ODD (ELIMINATE)
B: 0² - 4(0) + 6 = 6 = EVEN (KEEP)
C: 0² - 5(0) + 5 = 5 = ODD (ELIMINATE)
D: 0² + 3(0) + 8 = 8 = EVEN (KEEP)
E: 0² + 2(0) + 10 = 10 = EVEN (KEEP)

We're down to B, D and E
Now plug in an ODD integer. The easiest odd value to plug in is x = 1
B: 1² - 4(1) + 6 = 3 = ODD (ELIMINATE)
D: 1² + 3(1) + 8 = -1 = EVEN (KEEP)
E: 1² + 2(1) + 10 = 13 = ODD (ELIMINATE)



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and

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Yes, picking the number is the best way to go.

Using O*E = E, E*E = E, O*O = O, E+O = O, O+O = E, If you assume x to be odd, you can straight away eliminate all answer choices except D.

How do you approach answer choice A?

X^2 - X - 1

If X is even:
E * E - O * E - E
= E


If X is odd:
O * O - O * O - O
= O

Am I wrong?

X^2 - X - 1

If X is even:

E * E - E - O = O (Since 1 is odd)

If X is odd:

O * O - O - O = O

We first must recognize that when squaring an odd integer, the result will still be odd and when squaring an even integer, the result will still be even. That fact will help when analyzing the answer choices.

It will also be helpful to use our addition, subtraction, and multiplication rules, for odd and even numbers, when analyzing the answer choices.

Addition Rules

E + E = Even
O + O = Even
E + O = Odd

*the same rules apply for subtraction

Multiplication Rules

E x E = Even
E x O = Even
O x O = Odd

Now let's analyze each answer choice for when x is odd and when x is even.

A) x^2-x-1

When x is even: E - E - 1 = odd

Since we have found an odd result, answer A is not correct.

B) x^2 - 4x + 6

When x is even: E - E + 6 = even

When x is odd: O - E + 6 = odd

Since we have found an odd result, answer B is not correct.

C) x^2 - 5x + 5

When x is even: E - E + 5 = odd

Since we have found an odd result, answer C is not correct.

D) x^2 + 3x + 8

When x is even: E - E + 8 = even

When x is odd: O + O + 8 = even

Since we only get an even result, answer D is correct.

E) x^2 + 2x + 10

When x is even: E + E + 10 = even

When x is odd: O + E + 10 = odd

Since we have found an odd result, answer E is not correct.

Answer: D
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If x is an integer, which of the following must be an even integer?

A) \(x^2\) — x — 1

B) \(x^2\) — 4x + 6

C) \(x^2\) — 5x + 5

D) \(x^2\) + 3x + 8

E) \(x^2\) + 2x + 10
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Explanation

Again the best strategy is plugging values. Take an odd integer say 1. For simplicity and put the values


A) \(x^2\) — x — 1 = 1 - 1 -1 = -1

B) \(x^2\) — 4x + 6 = 1 -4 + 6 = 3

C) \(x^2\) — 5x + 5 = 1 -5 +5 = 1

D) \(x^2\) + 3x + 8 = 1 + 3 + 8 = 12

E) \(x^2\) + 2x + 10 = Don't need to evaluate this. You already have your answer.


Hence D is correct.


Note: During plugging values if you have found an answer the do not continue plugging in other options. This is very important time saving tactic.
Also keep in mind when a question has very generic statements like all integers satisfy "Expression..." It is highlly likely that plugging values is the best strategy.
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X can be any integer (i.e. either odd or even) so even though your answer is correct the procedure is incomplete. You would have to check:

1) if D) fits for X as an even integer and than you are finished.

OR

2) if E) fits for x = 1, which it does not [so D) is the only possible option].

You are correct I should have plugged an even and an odd integer!
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The question isn't mine but thanks!

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

A. \(x^2−x−1\)

B. \(x^2−4x+6\)

C. \(x^2−5x+5\)

D. \(x^2+3x+8\)

E. \(x^2+2x+10\)
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fastest way for me was to plug in odd to those X and only one will have even..which is D
you can just do it in ur head w/o writing down bunch of E,O stuff which i don't even know what that is.

x is an integer-
why did we not consider 0 ?

Because this question ask you which of the following is an EVEN integer.

This imply that you should use the strategy of EVEN/ODD numbers, integer.

This imply, turns out, that the number must be different from zero even though is not specified.

If you use zero you will end up with two answers which are even but the question is a must be true NOT a could be true or even.

So, zero is not possible. The question would not have sense, whatsoever.

I hope now is clear.

Norice how this is a key aspect to infer these things. They are number properties.

Best Regards


See here more on properties of the numbers https://greprepclub.com/forum/gre-quant ... tml#p51913
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Hey, thanks.
Yes it is clear now

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