I would like to share with you some simplifications about even and odd number, that you can use in solving GMAT problems.

1. Don't care about positive integral powers.Obviously, if you take an even number in any power, that is positive integer, you will get an even number, and the same for odd numbers. So, you can just cross out the powers.

For example, if you have \(a^n\)-even and \(n\) is a positive integers, it means just that \(a\) is even.

2. When you add numbers, don't care about even numbers.Even numbers don't change the pairness when add:

even+

even=even

odd+

even=odd

Again, you can cross them out.

3. When you multiply numbers, don't care about odd numbers. The same reason as for the previous:

even\(\times\)

odd=even

odd\(\times\)

odd=odd

Again, you can cross them out.

Let's take a look for one GMAT problem

if-a-and-b-are-both-positive-integers-is-b-a-1-ba-b-118598.htmlIf \(a\) and \(b\) are both positive integers, is \(b^{(a+1)}- ba^b\) odd?

(1) \(a + (a + 4) + (a - 8) + (a + 6) + (a - 10)\) is odd

(2) \(b^3 + 3b^2 + 5b + 7\) is oddSolution:First, according to the first rule, simplify the question. Since \(a\) and \(b\) are both positive integers, we can cross out the powers and the simplified question will be:

Is \(b- ba\) odd?Secondly, according to the second rule, simplify first statement. We can cross out all even numbers (and \(a+a\) too)

(1) a + (

~~a~~+

~~4~~) + (

~~a~~-

~~8~~) + (

~~a~~+

~~6~~) + (

~~a~~ -

~~10~~) is odd

And at the end we have just the following statement:

(1) \(a\) is oddCome back to the question. \(b- b\times odd=b-b=0\) is even (we omit odd using 3rd rule). (1) is sufficient.

Thirdly, according to the first and second rule, simplify second statement:

\(b^3 + 3b^2 + 5b + 7\) is the same as \(b+3b+5b+7\), and as \(b+7\), which is odd

And at the end we have just the following statement:

(2) \(b\) is evenCome back to the question. \(even- even\times a\) is even. (2) is sufficient.

The correct answer is D.

One more

is-the-positive-integer-x-odd-1-x-y2-4y-6-where-y-32691.html:

Is the positive integer \(x\) odd?

(1)\(x=y^2 +4y+6\), where \(y\) is a positive integer.

(2) \(x = 9z^2 + 7z - 10\), where z is a positive integer.Immediately after simplification:

Is the positive integer \(x\) odd?

(1)\(x=y\), where \(y\) is a positive integer.

(2) \(x =16z\), where z is a positive integer.The correct answer is B.

Hope it helps for somebody)))