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27 Jul 2014, 05:36
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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.html
If \(a\) 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 odd
Solution:
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 odd
Come 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 even
Come 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)))
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.html
If \(a\) 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 odd
Solution:
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 + (
And at the end we have just the following statement:
(1) \(a\) is odd
Come 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 even
Come 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)))
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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18 Aug 2014, 07:29
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This is awesome. Thank you very much. I'd really appreciate if someone could help me find similar posts that deal with simplifications of other quant topics .
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
