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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## a, b, and c are positive integers. Is a+b+c an odd number? tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 1 member reply ### GMAT/MBA Expert ## a, b, and c are positive integers. Is a+b+c an odd number? ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [GMAT math practice question] a, b, and c are positive integers. Is a + b + c an odd number? 1) ab is an odd number 2) c is an odd number _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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Max@Math Revolution wrote:
[GMAT math practice question]

a, b, and c are positive integers. Is a + b + c an odd number?

1) ab is an odd number
2) c is an odd number
Given: a, b, and c are positive integers.

Target question: Is a + b + c an odd number?

Statement 1: ab is an odd number
NOT SUFFICIENT

Statement 2: c is an odd number
No information about a or b.
NOT SUFFICIENT

Statements 1 and 2 combined
If ab is ODD (statement 1), then we know that a and b are both odd
Statement 2 tells us that c is odd
So, a + b + c = odd + odd + odd = ODD
So, the answer to the target question is YES, a+b+c IS an odd number
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent

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Odd + Odd = even
(2n+1) + (2n+1) = 4n+2
If n = 1 ; 4n+2 = 6
If n =2 ; 4n+2 = 10
Hence,
Odd + Odd = even number
Odd + even = Odd number
even + even = even number
If a + b + c = (Odd + Odd) + Odd = even + Odd = Odd number
If a + b + c = (Odd + Odd) + even = even + even = even number
If a + b + c = (Odd + even) + Odd = Odd + Odd = even number

In other words;
Odd + Odd + Odd = Odd number
Odd + Odd + even = even number
Odd + even + even = Odd number
even + even + even = even number
even + Odd + Odd = even number
even + even + Odd = Odd number

Statement 1
ab is an odd number; bearing in a method that
Odd * Odd = odd number
even * Odd = even number
even *even = even number
This means that a and b are odd numbers if c = odd
Odd + Odd + Odd = Odd number but, if c = even
Odd + Odd + even = even number ;
Information given is not enough, statement 1 is INSUFFICIENT.

Statement 2
c is an odd number ; if a = odd and b = odd
Odd + Odd + Odd = Odd number but if either of a or b = even number, then
even + Odd + Odd = even number, if both a and b = even number, then
even + even + Odd = Odd number
Statement 2 is INSUFFICIENT.

Both statements together
ab = Odd , hence a = odd and b = odd , from statement 2 ; c = odd , hence
Odd + Odd + Odd = Odd number
Both statements together are SUFFICIENT.

$$answer\ is\ Option\ C$$

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (a, b, x and y) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables with the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Condition 1), ab is an odd integer, is equivalent to the statement that both a and b are odd numbers.
As c is also an odd number, a + b + c is an odd number since it is the sum of three odd numbers.

Both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If a = 1, b = 1, and c = 1, then a + b + c = 3 is an odd number, and the answer is ‘yes’.
If a = 1, b = 1, and c = 2, then a + b + c = 4 is not an odd number, and the answer is ‘no’.
Thus, condition 1) is not sufficient.

Condition 2)
If a = 1, b = 1, and c = 1, then a + b + c = 3 is an odd number, and the answer is ‘yes’.
If a = 2, b = 1, and c = 1, then we have a + b + c = 4 is not an odd number, and the answer is ‘no’.
Thus, condition 2) is not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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