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If M is a finite set of negative integers, is the total numb
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18 Aug 2012, 09:35
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If M is a finite set of negative integers, is the total number of integers in M an odd number? (1) The product of all integers in M is odd (2) The product of all integers in M is negative My answer was A which is wrong but I don't understand why? though I tried to solve the problem again ...
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Re: If M is a finite set of negative integers, is the total numb
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18 Aug 2012, 10:27
Sunnysmiley wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative
My answer was A which is wrong but I don't understand why? though I tried to solve the problem again ... (1) We can have an even number of negative odd numbers. Their product is odd. We can also have an odd number of negative odd numbers, their product is still odd. The product is odd if all the numbers are odd, regardless how many numbers are there. Nothing is stated about the sign of the product. Not sufficient. (2) The product being negative, we must necessarily have an odd number of negative integers, as all the numbers are negative. Now we had information about the sign of the product Sufficient. Answer B.
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Re: If M is a finite set of negative integers, is the total numb
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18 Aug 2012, 13:46
Sunnysmiley wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative
My answer was A which is wrong but I don't understand why? though I tried to solve the problem again ... i> take any 3 negative numbers, say 3, 9,  13 Product of any pair is odd and product of all the three is also odd. Can you conclusively say that an odd product necessarily means an odd number of factors? Since you get the same result (i.e. odd product albeit different signs), the info is INSUFFICIENT ii>do the same test, this time check for signs. 3*9, 9*13 etc will have +ve signs, but 3*9*13 will have a ve sign, you can use this to differentiate odd vs even number of factor. SUFFICIENT
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Re: If M is a finite set of negative integers, is the total numb
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18 Aug 2012, 15:12
(1) Do we know definitively that the total number of integers in M is an odd number?
Well let's try some examples {1, 3} The product of all integers in M is odd... TRUE. Total # of integers is EVEN. {1, 3, 5} The product of all integers in M is odd... TRUE. Total # of integers is ODD.
THus, we know we cannot DEFINITIVELy say that the total # of integers in the set is ODD. Why? Because we had one example where the total # of integers was EVEN while satisfying the conditions for (1).
(2) The product of all integers in M is negative. Well, if all the integers in M are negative and the product of all the integers is negative, then negative * negative = positive. Times another negative number is going to be negative.
Thus, if there are even # of integers  then the product of all integers in the set will always be positive. If there an odd # of integers  then the product of all integers in the set will always be odd.
Key here is ALWAYS odd when (2) is satisfied. That's why (B) alone is sufficient to answer the question  we can definitively say that the answer to the question " is the total number of integers in M an odd number?" is going to be YES when (2) is satisfied.



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Re: If M is a finite set of negative integers, is the total numb
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19 Aug 2012, 04:17
Legendaddy wrote: Sunnysmiley wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative
My answer was A which is wrong but I don't understand why? though I tried to solve the problem again ... i> take any 3 negative numbers, say 3, 9,  13 Product of any pair is odd and product of all the three is also odd. Can you conclusively say that an odd product necessarily means an odd number of factors? Since you get the same result (i.e. odd product albeit different signs), the info is INSUFFICIENT ii>do the same test, this time check for signs. 3*9, 9*13 etc will have +ve signs, but 3*9*13 will have a ve sign, you can use this to differentiate odd vs even number of factor. SUFFICIENT Thank you everyone, This is how I considered to infer the answer A: the product of 1 and 2 is an odd the product of 1, 2, and 3 is an even but looking at your answers I figure out that I might not work on the solution inclusively, thus I have an inadequate answer. I should be more careful next time.



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Re: If M is a finite set of negative integers, is the total numb
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19 Aug 2012, 07:11
Sunnysmiley wrote: Legendaddy wrote: Sunnysmiley wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative
My answer was A which is wrong but I don't understand why? though I tried to solve the problem again ... i> take any 3 negative numbers, say 3, 9,  13 Product of any pair is odd and product of all the three is also odd. Can you conclusively say that an odd product necessarily means an odd number of factors? Since you get the same result (i.e. odd product albeit different signs), the info is INSUFFICIENT ii>do the same test, this time check for signs. 3*9, 9*13 etc will have +ve signs, but 3*9*13 will have a ve sign, you can use this to differentiate odd vs even number of factor. SUFFICIENT Thank you everyone, This is how I considered to infer the answer A: the product of 1 and 2 is an odd the product of 1, 2, and 3 is an even but looking at your answers I figure out that I might not work on the solution inclusively, thus I have an inadequate answer. I should be more careful next time. the product of 1 and 2 is an odd NO! It is even, because (1)*(2) = 2 is even. A product of integers is even if and only if at least one of the factors is even. A product of integers is odd if and only if all the integers are odd. The product of an even number of negative numbers is positive. The product of an odd number of negative numbers is negative.
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Re: If M is a finite set of negative integers, is the total numb
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26 Feb 2013, 06:23
since we are given finite set of negative integers so it can be 3 9 5 =  135 I can't find examples to prove it is not true can someone help out with statement A..
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Re: If M is a finite set of negative integers, is the total numb
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26 Feb 2013, 06:52
fozzzy wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative
since we are given finite set of negative integers so it can be
3 9 5 =  135
I can't find examples to prove it is not true can someone help out with statement A.. The stem says that M consists of negative integers and the first statement says that the product of all integers in M is odd. The question asks whether the total number of integers in M is odd. If M={1, 3, 5} > M consists of negative integers and the product of all integers in M is odd (1*(3)*(5)=15=odd) > the total number of integers in M is odd. If M={1, 3} > M consists of negative integers and the product of all integers in M is odd (1*(3)=3=odd) > the total number of integers in M is even. Thus we can have an YES as well as a NO answer, which means that the first statement is not sufficient. Hope it's clear. P.S. You could just exclude one of the elements from your list ({3, 5, 9}) to get a NO answer.
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Re: If M is a finite set of negative integers, is the total numb
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27 Nov 2013, 20:26
For statement 2, why can't I have a M set (1, 2, 1, 1)? There's no rule that says you cannot times it by 1 repeatedly...



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Re: If M is a finite set of negative integers, is the total numb
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28 Nov 2013, 05:50



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Re: If M is a finite set of negative integers, is the total numb
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29 Dec 2015, 22:14
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 M is a finite set of negative integers, is the total number of integers in M an odd number? (1) The product of all the integers in M is odd (2) The product of all the integers in M is negative There are numerous variables (because we have to know the total number of integers in M) in the original condition. In order to match the number of variables and the number of equations, we need numerous equations as well. Since the condition 1) and 2) each has 1 equation, there is high chance that E is going to be the answer. Using both the condition 1) and 2), we can see that the total number of integers in M has to be an odd number in order for the product of all the integers in M to be odd and negative. Therefore, the answer is ‘yes’ and the correct answer is C. This is an integer question, which is one of the key questions. This means we have to apply the Common Mistake Type 4(A). In case of the condition 1), it states that the product of all the integers in M is odd and there is no way we can know if the total number of integers in M is an odd number. In other words if M={1, 3}, then the answer is ‘no’. If M={1, 3, 5}, then the answer is ‘yes’. Therefore, the condition is not sufficient. In case of the condition 2), in order for the product of all the integer in M to be negative, the total number of integers in M has to an odd number. Therefore, the answer is ‘yes’ and the condition is sufficient. The answer is, then, B. 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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Re: If M is a finite set of negative integers, is the total numb
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26 Jan 2017, 21:37
Statement 1 deals with the issue of even/odd  which has set rules for addition, multiplication and subtraction. if the product of all the integers is odd that means that there are no even numbers in the set  but doesn't give us any info as to how many numbers there are. Statement 2 deals with the rules of neg/pos  if you have a set of negative numbers whose product is negative then there has to be an odd number as an even number of negatives multiplied together will always give you a positive number. So B.
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Re: If M is a finite set of negative integers, is the total numb
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11 Dec 2017, 06:52
ivyph wrote: If M is a finite set of negative integers, is the total number of integers in M an odd number?
(1) The product of all integers in M is odd (2) The product of all integers in M is negative We are given that M is a finite set of negative integers and we need to determine whether the total number of integers in M is an odd number. Statement One Alone: The product of all the integers in M is odd. The information in statement one is not sufficient to answer the question. For instance, there could be 2 numbers in M, 1 and 3, and their product would be odd, or there could be 3 numbers in M, 1, 3, and 5, and their product would also be odd. Statement one alone is not enough information to answer the question. Statement Two Alone: The product of all the integers in M is negative. To analyze the information provided in statement two, we can use our multiplication rules for negative numbers. We know that when an even number of negative numbers are multiplied together, the product is positive, and when an odd number of negative numbers are multiplied together, the product is negative. Since the product of all the integers in M is negative, we know that the number of integers in M must be odd. Answer: B
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