If A and B are positive integers, is the product AB even?
(1) The sum A + B is odd.
(2) A is even.
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- MartyMurray
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What a cool little question.
Statement 1 tells us that the sum of the two integers is odd.
If we add two odd integers, we get an even integer. 3 + 5 = 8
If we add two even integers, we get an even integer. 2 + 166 = 168
The only way to get an odd integer by adding two integers is by having one of them be odd and the other even. 5 + 8 = 13
So either A or B has to be even and so the product AB must be even, and Statement 1 is sufficient.
Statement 2 tells us that A is even, and so the product AB must be even, and Statement 2 is sufficient.
Choose D.
Statement 1 tells us that the sum of the two integers is odd.
If we add two odd integers, we get an even integer. 3 + 5 = 8
If we add two even integers, we get an even integer. 2 + 166 = 168
The only way to get an odd integer by adding two integers is by having one of them be odd and the other even. 5 + 8 = 13
So either A or B has to be even and so the product AB must be even, and Statement 1 is sufficient.
Statement 2 tells us that A is even, and so the product AB must be even, and Statement 2 is sufficient.
Choose D.
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For this question, you need to know the following:
1. ODD +/- ODD = EVEN
2. ODD +/- EVEN = ODD
3. EVEN +/- EVEN = EVEN
4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN
For more on ODD and EVEN integers, watch our free video - https://www.gmatprepnow.com/module/gmat- ... /video/837
Cheers,
Brent
1. ODD +/- ODD = EVEN
2. ODD +/- EVEN = ODD
3. EVEN +/- EVEN = EVEN
4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN
For more on ODD and EVEN integers, watch our free video - https://www.gmatprepnow.com/module/gmat- ... /video/837
Cheers,
Brent
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
If A and B are positive integers, is the product AB even?
(1) The sum A + B is odd.
(2) A is even.
There are 2 variables (A,B), but only 2 equations are given by the conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, A=even, B=odd, so it is a 'yes' and the condition is sufficient, making the answer seem like (C), but this is a commonly made mistake;
if we look at them separately,
For condition 1, A+B=odd --> (A,B)=(odd,even),(even,odd). The answer always becomes 'yes' and the condition is sufficient.
For condition 2, in A=even --> AB=even*B=even, the answer becomes 'yes' and the condition is sufficient as well. The answer is (D).
For cases where we need 2 more equations, such as original conditions with "2 variables", or "3 variables and 1 equation", or "4 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
If A and B are positive integers, is the product AB even?
(1) The sum A + B is odd.
(2) A is even.
There are 2 variables (A,B), but only 2 equations are given by the conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, A=even, B=odd, so it is a 'yes' and the condition is sufficient, making the answer seem like (C), but this is a commonly made mistake;
if we look at them separately,
For condition 1, A+B=odd --> (A,B)=(odd,even),(even,odd). The answer always becomes 'yes' and the condition is sufficient.
For condition 2, in A=even --> AB=even*B=even, the answer becomes 'yes' and the condition is sufficient as well. The answer is (D).
For cases where we need 2 more equations, such as original conditions with "2 variables", or "3 variables and 1 equation", or "4 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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