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If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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If positive integer \(N = A – B\), where \(A\) is a positive integer and \(B\) is a prime number, is \(N\) odd? (1) \(B\) and \(X\) are the only prime factors of \(A\), and \(B –X = 4\) (2) \(A\) is divisible by 9 numbers in total, one of which is \(B^2\) This is Ques 3 of The EGMAT Number Properties Knockout Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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EgmatQuantExpert wrote: If positive integer N = A – B, where A is a positive integer and B is a prime number, is N odd? (1) B and X are the only prime factors of A, and B –X = 4 (2) A is divisible by 9 numbers in total, one of which is\(B^2\) This is Ques 3 of The EGMAT Number Properties Knockout Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts! 1) from this statement we know that B and X are odd numbers because we don't have any prime  2 = 4. As A = B*X than A is odd too. So odd  odd = even. Sufficient. 2) if A divisible by 9 numbers than it's have 2 prime factors in second power and one of which is b^2. so B can be equal 2 and in this case we will have odd  even = odd or B can be equal to any other prime and in this case we will have odd  odd = even. Insufficient And answer is A
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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08 Apr 2015, 23:03
Dear Harley Thank you for posting your analysis. Let us now use your analysis to clarify a few conceptual nuances 1. In your analysis of Statement 1, you wrote: Quote: "from this statement we know that B and X are odd numbers because we don't have any prime  2 = 4. As A = B*X than A is odd too. So odd  odd = even. Sufficient." My question: Is the expression A = B*X correct? If no, then what should be the correct expression?2. In your analysis of Statement 2, you wrote: Quote: "if A divisible by 9 numbers than it's have 2 prime factors in second power and one of which is b^2. so B can be equal 2 and in this case we will have odd  even = odd or B can be equal to any other prime and in this case we will have odd  odd = even. Insufficient"
My question: If A is divisible by 9 numbers, can you infer that it has 2 prime factors in second power? If yes, how? If no, then why?It is important to understand these nuances because it is such nuances that the GMAT question architects use to set up traps for students. Look forward to hearing from you!  Japinder
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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08 Apr 2015, 23:45
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EgmatQuantExpert wrote: Dear Harley Thank you for posting your analysis. Let us now use your analysis to clarify a few conceptual nuances 1. In your analysis of Statement 1, you wrote: Quote: "from this statement we know that B and X are odd numbers because we don't have any prime  2 = 4. As A = B*X than A is odd too. So odd  odd = even. Sufficient." My question: Is the expression A = B*X correct? If no, then what should be the correct expression?2. In your analysis of Statement 2, you wrote: Quote: "if A divisible by 9 numbers than it's have 2 prime factors in second power and one of which is b^2. so B can be equal 2 and in this case we will have odd  even = odd or B can be equal to any other prime and in this case we will have odd  odd = even. Insufficient"
My question: If A is divisible by 9 numbers, can you infer that it has 2 prime factors in second power? If yes, how? If no, then why?It is important to understand these nuances because it is such nuances that the GMAT question architects use to set up traps for students. Look forward to hearing from you!  Japinder That's interesting questions, thank you Is the expression A = B*X correct? If no, then what should be the correct expression?It's not exactly correct because B and X can be in any power. But we can omit this information because powers don't have any influence on parity of numbers. That's why I write this equation in such manner. If A is divisible by 9 numbers, can you infer that it has 2 prime factors in second power? If yes, how? If no, then why?I made such infer by using the rule about calculating factors of number: We should find all primes of number. take theirs powers plus 1 and calculate a product all this numbers. For example number 24 has prime factors \(2^3\) and \(3^1\) so number of factors equal (3+1) * (1+1) = 8 And by applying this rule in reversal mode we sometimes can find number of primes from information about how many factors number has: When we have information that number has 9 factors, we can make infer that product of some integers should give us 9 in answer So it can be product only of two odd integers (because of rule we can't have 1 in this product). So we don't know meaning of primes but we know that this number has exactly two primes in second power: (2 + 1) * (2 + 1) = 9 Hope I've answered on your questions correctly )
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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09 Apr 2015, 00:02
#1 Since B and X are the only prime factors of A, we can conclude that \(A = B^n*X^k\) B  prime. X is prime factor so X > 1, moreover, B and X ain't same coz of the "X = B 4" equation and also X != 2 coz in this case B = 6  not prime, so X is odd prime and so is B coz if B = 2, X turns negative, thus A = odd*odd coz \(odd^n = odd\) regardless of n Now we got our question: N = A (odd)  B(odd), odd  odd = even, so the answer is NO #1  sufficient
#2 Lets say B is different from all numbers I've written down there, lets give 2 examples 1) \(A = B^2*3*5*7*11*13*17*19*2\)  even, so N = even  odd = odd 2) \(A = B^2*3*5*7*11*13*17*19*23\)  odd, so N = odd  odd = even insufficient
A is the answer



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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09 Apr 2015, 02:36
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Dear Harley Thank you for your response. 1. I'm glad that you do understand conceptually that the expression A = B*X is Quote: . . . not exactly correct because B and X can be in any power. However, I would advise you against the following part: Quote: But we can omit this information because powers don't have any influence on parity of numbers. That's why I write this equation in such manner. Reason: In this particular question, the powers didn't matter because this is an evenodd question. But in a primes question, it may become critical to write the expression for A as \(B^{n}*X^{k}\) (Good job, Zhenek! ) So, as a best practice, I would suggest that you always write the expression in full. As they say, it's easier to follow a rule 100% of the times than 99% of the times 2. Your logic is absolutely correct. However, if the total number of factors of A is 9, does it necessarily mean that A is of the form \(B^{2}*X^{2}\)? Is this the only form for which the total number of factors comes out to be 9? Think about it: 3*3 = 9 [This is the reverse logic you used] But 9*1 is also 9 That is (8+1)(0+1) = 9 So, A can also be an expression of the form \(B^{8}\) In this particular question, you could get the answer correct even without considering this alternate expression for A. However, what if the question had been: Is positive integer A divisible by \(B^3\), where B is a prime number? (1) Some Statement (2) A is divisible by 9 numbers in total, one of which is \(B^2\)In such a question, if you were not clear about this nuance, you would have concluded that since A is a number of the form \(B^{2}*X^{2}\), the answer to the posed question is a definite NO and so, St. 2 alone is sufficient. However, in the light of our discussion now, you can see that the correct conclusion to draw from St. 2 is that A may or may not be divisible by \(B^3\). Therefore, St. 2 is NOT SUFFICIENT.This is why understanding this nuance is very important. Hope our discussion helped you clarify it! I enjoyed our discussion  Japinder
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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10 Apr 2015, 06:11
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Detailed SolutionStepI: Given InfoThe statement tells us about a positive integer \(N = A – B\) and we are told that \(A\) is a positive integer and \(B\) is a prime number. We are asked to find if \(N\) is odd. StepII: Interpreting the Question StatementThe expression \(N\) is expressed as a difference of two numbers, for \(N\) to be odd the nature of these two numbers would have to be opposite i.e. one has to be odd and other has to be even. We are given that \(B\) is prime number, we know that except for 2, all the prime numbers are odd. So, if we can establish if \(B\) is greater than 2, we would be able to say with certainty that \(B\) is odd. StepIII: StatementIStatementI tells us that \(B\) and \(X\) are prime numbers such that \(BX = 4\), the difference of two prime numbers is even, which would mean that either both are odd or both are even. Since there is only even prime number possible (i.e. 2), we can say with certainty that both \(B\) and \(X\) are odd. Since \(A\) has only \(B\) and \(X\) as its prime factors, this would imply that \(X\) is a product of two odd numbers. Thus A would also be odd. We now know the even/odd nature of both \(A\) and \(B\), thus we can determine with certainty the even/odd nature of \(N\). Hence, StatementI is sufficient to answer the question. StepIV: StatementIIStatementII tells us that \(A\) has 9 factors. Since 9 can factorized only as (9*1) or (3*3) this would imply that \(A\) can be written as: \(A = P_{1}^8 or P_{1}^2*P_{2}^2\), where \(P_{1}*P_{2}\) are the prime factors of \(A\). Since we are told that \(A\) is divisible by \(B^2\), this would mean that \(P_{1} = B\). Please note here that \(B^2\) is a divisor of \(A\) in both the cases where \(A = P_{1}^8 or P_{1}^2 * P_{2}^2\). So, we can’t say with certainty that \(A\) can be expressed in one of the ways. Let’s evaluate both the cases now: • CaseI: \(A= B^8\) In this case \(A\) will have the same even/odd nature as that of \(B\). We know that the difference of two even numbers or two odd numbers would always be even. Hence, we can say that \(N\) will be even. • CaseII: \(A= B^2*P_{2}^2\)In this case if \(B\) is an even prime number (i.e. 2) then \(A\) would also be even and subsequently \(N\) would also be even. However if \(B\) is odd, the nature of \(A\) would depend on the nature of \(P_{2}\). If \(P_{2}\) is even, then \(A\) would be even and \(N\) would be odd ( since \(B\) is odd) If \(P_{2}\) is odd, then \(A\) would be odd and \(N\) would be even ( since \(B\) is odd) So, we see here that we can’t predict with certainty the exact even/odd nature of \(N\). Hence, StatementII is insufficient to answer the question. StepV: Combining Statements I & IISince, we have received our unique answer from StatementI, we don’t need to combine the inferences from StatementI & II. Hence, the correct Answer is Option A.Key Takeaways1. In evenodd questions, simplify complex expressions into simpler expressions using the properties of evenodd combinations. 2. Know the properties of EvenOdd combinations to save the time spent deriving them in the test 3. Remember that there is only 1 even prime number i.e. 2. 4. If 2 is a prime factor of the number, the number would be even else it would be odd. 5. If the total number of factors of a number is given to be T, evaluate the possible ways in which T can be expressed as a product of 2 or more numbers.Regards Harsh
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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30 Jul 2015, 13:01
EgmatQuantExpert wrote: If positive integer \(N = A – B\), where \(A\) is a positive integer and \(B\) is a prime number, is \(N\) odd?
(1) \(B\) and \(X\) are the only prime factors of \(A\), and \(B –X = 4\)
(2) \(A\) is divisible by 9 numbers in total, one of which is \(B^2\)
I arrived at the answer by following a process similar to the one explained by Harsh. Statement 1: \(B\) and \(X\) being prime, and \(B  X = 4\) implies that \(B\) and \(X\) are both odd. Therefore, \(A\) is odd as it only has two prime factors \(B\) and \(X\). Statement 2: \(A\) has a total of 9 factors including \(1\), \(B\), \(B^2\) and \(A\) itself. I have no information that tells me whether \(2\) is a factor of \(A\) or not. Therefore, I cannot conclude whether \(A\) is even or odd. Therefore, the answer is A, statement 1 alone is sufficient.



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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09 Nov 2015, 02:08
EgmatQuantExpert wrote: If positive integer \(N = A – B\), where \(A\) is a positive integer and \(B\) is a prime number, is \(N\) odd?
(1) \(B\) and \(X\) are the only prime factors of \(A\), and \(B –X = 4\)
(2) \(A\) is divisible by 9 numbers in total, one of which is \(B^2\)
My take is Option A. Took 4 mins. to solve this question Explanation: Stmt I: B, X are the only prime factors, and BX = 4. BX = 4 implies that both of them are odd, since prime number "2" can't give the difference of 4 (Odd  Odd = Even concept) Now, A must be Odd, since "2" can't be its prime factor. Hence N = A  B = Odd  Odd = Even. Sufficient Stmt II: A is divisible by 9 numbers in total, one of which is B^2. Hmm...we can assume that A = B^2 * K, where K = 1, 2, 3, 4.... (Factor concept) Now substitute this value of A in original equation: N = B^2 * K  B = B(BK1)..Now let's focus on the values in the bracket i.e. (BK1) = BK  Odd Case 1: K is odd => N = Odd  Odd = Even Case 2: K is even => N = Even  Odd = Odd Not Sufficient. P.S. you may account for values of B, but it will give you only 2 cases i.e. Odd, and Even Hit kudos if you like the explanation. Thanks, Chanakya



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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12 Nov 2015, 23:13
Hi, If positive integer N=A–B, where A is a positive integer and B is a prime number, is N odd? Let's analyze the question stem first; for N to be Odd, we have two ways Odd = Odd  Even (B can be 2) or Odd = Even  Odd (prime numbers apart from 2) (1) B and X are the only prime factors of A, and B–X=4 Looking at this statement: For B  X to be 4; we are definite that X can't be 2. So that leaves us with X to be Odd. So B has to be Odd. We now arrive at A; as B and X are the only prime factors of A, we can deduce that A will always be an Odd number. As Odd ^ any number * Odd ^ any number = Odd So we have the nature of A and B. A is odd and B is Odd. Thus we arrive at the solution. Statement 1 is sufficient. (2) A is divisible by 9 numbers in total, one of which is B^2 Here, we can arrive at a situation where A has 9 factors out of B ^ 2 is one. So we know that 9 = 3 * X => X = 3. Thus the other number is also a square. But statement 2 does not give us a lead to identify the nature of A and B. The squares can be even or odd. So the number can be even or odd. So there's an ambiguity. We can not arrive to a definite conclusion with Statement 2. Hence Statement 2 is insufficient.. Thus with Statement 1 we are able to identify whether N is odd. Thus Option A. Any comments EgmatQuantExpertRegards



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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28 Mar 2016, 08:33
Copy to  Bunuel hey Team @ egmat and EgmatQuantExpert i have an important question to ask here Now i have solved this question twice and i got it correct both the times. But the question's difficultly level is bothering me.. the first time it took me 4 minutes to solve this question and the next time too about 4 minutes . The Question i want to ask here is that is it really a replica of what GMAT may ask us? Don't get me wrong this is an excellent question for someone who wants to understand the concept but is it really a question that GMAT will Put us through? I seriously Doubt that. This Question isn't really difficult But is Really Time Consuming ... Maybe chetan2u can Shed some light here Thanks and regards S.C.S.A
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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28 Mar 2016, 09:11
Chiragjordan wrote: Copy to  Bunuel hey Team @ egmat and EgmatQuantExpert i have an important question to ask here Now i have solved this question twice and i got it correct both the times. But the question's difficultly level is bothering me.. the first time it took me 4 minutes to solve this question and the next time too about 4 minutes . The Question i want to ask here is that is it really a replica of what GMAT may ask us? Don't get me wrong this is an excellent question for someone who wants to understand the concept but is it really a question that GMAT will Put us through? I seriously Doubt that. This Question isn't really difficult But is Really Time Consuming ... Maybe chetan2u can Shed some light here Thanks and regards S.C.S.A Hi, It may not be the replica of ACTUAL gmat but does test on CONCEPTS that are a part of GMAT .. The concepts are clearly NUMBER PROPERTIES.. It may have taken anything from 1 min to 45 minutes See where you went wrong in TIME.. If positive integer N=A–B, where A is a positive integer and B is a prime number, is N odd? (1) B and X are the only prime factors of A, and B–X=4The moment you see this, BX=4 should immediately tell us that B and X are ODD prime..since A consists of ONLY B and X, A is ODD B is already seen to be ODD so N= ODDODD= EVEN.. Suff .. NOT >4050secs (2) A is divisible by 9 numbers in total, one of which is \(B^2\)Here an ODD factors should strike this means A is PERFECT SQUARE, But it should have B^2 as Factor and B is PRIME so A should be = B^2*(some other prime number)^2 so N= B^2*P^2B = B(B*P^21)We do not know about P and B If B is even, N is even If B is ODD, N is EVEN if P is ODD and N is ODD if P is EVEN Insuff.. may be 11:20 in Normal circumstances.. The timing part is only to say that it is possible in normal circumstances to do it in within 2 minutes.. But if any one of us gets struck in any of the following concepts, It can be very longish method 1) If difference of TWO prime number is EVEN, both have to be ODD and If difference is ODD, the smaller has to be 2.. 2) ODD number of factors means a PERFECT SQUARE..
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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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09 Jan 2017, 03:16
EgmatQuantExpert wrote: If positive integer \(N = A – B\), where \(A\) is a positive integer and \(B\) is a prime number, is \(N\) odd? (1) \(B\) and \(X\) are the only prime factors of \(A\), and \(B –X = 4\) (2) \(A\) is divisible by 9 numbers in total, one of which is \(B^2\) This is Ques 3 of The EGMAT Number Properties Knockout Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Statement 1: B and X are the only prime factors of A, and B–X=4 If prime was not mentioned here then A would be a prime number. Is it right?



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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18 Jan 2018, 23:20
Shiv2016 wrote: EgmatQuantExpert wrote: If positive integer \(N = A – B\), where \(A\) is a positive integer and \(B\) is a prime number, is \(N\) odd? (1) \(B\) and \(X\) are the only prime factors of \(A\), and \(B –X = 4\) (2) \(A\) is divisible by 9 numbers in total, one of which is \(B^2\) This is Ques 3 of The EGMAT Number Properties Knockout Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Statement 1: B and X are the only prime factors of A, and B–X=4 If prime was not mentioned here then A would be a prime number. Is it right? Hi Yes, if the statement read 'B and X are the only factors of A, and B  X = 4' then that would have meant the following: A has only two factors, and thus A is prime The two factors have a difference of 4, and since one of the factors has to be 1, then the other factor must be 5 So 1 and 5 are the only factors of A, thus A must be 5. So, in that case, we could conclude the value of A also



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Re: If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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07 Apr 2018, 05:00
Hi EgmatQuantExpert thanks for the detailed explanation, I understood the reasoning behind the first statement but I couldn't follow the reasoning behind statement 2. This seems to be a new concept to me and I am not able to understand why are we splitting numbers of factors of A which is 9 into factors. And why are we writing 9 as a permutation form? Can you please help me with the concept of statement 2 which follows below. EgmatQuantExpert wrote: StepIV: StatementII
StatementII tells us that \(A\) has 9 factors. Since 9 can factorized only as (9*1) or (3*3) this would imply that \(A\) can be written as:
\(A = P_{1}^8 or P_{1}^2*P_{2}^2\), where \(P_{1}*P_{2}\) are the prime factors of \(A\).
Since we are told that \(A\) is divisible by \(B^2\), this would mean that \(P_{1} = B\). Please note here that \(B^2\) is a divisor of \(A\) in both the cases where \(A = P_{1}^8 or P_{1}^2 * P_{2}^2\). So, we can’t say with certainty that \(A\) can be expressed in one of the ways. Let’s evaluate both the cases now:
• CaseI: \(A= B^8\) In this case \(A\) will have the same even/odd nature as that of \(B\). We know that the difference of two even numbers or two odd numbers would always be even. Hence, we can say that \(N\) will be even.
• CaseII: \(A= B^2*P_{2}^2\) In this case if \(B\) is an even prime number (i.e. 2) then \(A\) would also be even and subsequently \(N\) would also be even. However if \(B\) is odd, the nature of \(A\) would depend on the nature of \(P_{2}\).
If \(P_{2}\) is even, then \(A\) would be even and \(N\) would be odd ( since \(B\) is odd) If \(P_{2}\) is odd, then \(A\) would be odd and \(N\) would be even ( since \(B\) is odd)
So, we see here that we can’t predict with certainty the exact even/odd nature of \(N\).
Hence, StatementII is insufficient to answer the question.
StepV: Combining Statements I & II
Since, we have received our unique answer from StatementI, we don’t need to combine the inferences from StatementI & II. Hence, the correct Answer is Option A.
Key Takeaways
1. In evenodd questions, simplify complex expressions into simpler expressions using the properties of evenodd combinations. 2. Know the properties of EvenOdd combinations to save the time spent deriving them in the test 3. Remember that there is only 1 even prime number i.e. 2. 4. If 2 is a prime factor of the number, the number would be even else it would be odd. 5. If the total number of factors of a number is given to be T, evaluate the possible ways in which T can be expressed as a product of 2 or more numbers.
Regards Harsh
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If positive integer N = A – B, where A is a positive integer and B is [#permalink]
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07 Apr 2018, 20:39
ujjwal80 wrote: Hi EgmatQuantExpert thanks for the detailed explanation, I understood the reasoning behind the first statement but I couldn't follow the reasoning behind statement 2. This seems to be a new concept to me and I am not able to understand why are we splitting numbers of factors of A which is 9 into factors. And why are we writing 9 as a permutation form? Can you please help me with the concept of statement 2 which follows below. EgmatQuantExpert wrote: StepIV: StatementII
StatementII tells us that \(A\) has 9 factors. Since 9 can factorized only as (9*1) or (3*3) this would imply that \(A\) can be written as:
\(A = P_{1}^8 or P_{1}^2*P_{2}^2\), where \(P_{1}*P_{2}\) are the prime factors of \(A\).
Since we are told that \(A\) is divisible by \(B^2\), this would mean that \(P_{1} = B\). Please note here that \(B^2\) is a divisor of \(A\) in both the cases where \(A = P_{1}^8 or P_{1}^2 * P_{2}^2\). So, we can’t say with certainty that \(A\) can be expressed in one of the ways. Let’s evaluate both the cases now:
• CaseI: \(A= B^8\) In this case \(A\) will have the same even/odd nature as that of \(B\). We know that the difference of two even numbers or two odd numbers would always be even. Hence, we can say that \(N\) will be even.
• CaseII: \(A= B^2*P_{2}^2\) In this case if \(B\) is an even prime number (i.e. 2) then \(A\) would also be even and subsequently \(N\) would also be even. However if \(B\) is odd, the nature of \(A\) would depend on the nature of \(P_{2}\).
If \(P_{2}\) is even, then \(A\) would be even and \(N\) would be odd ( since \(B\) is odd) If \(P_{2}\) is odd, then \(A\) would be odd and \(N\) would be even ( since \(B\) is odd)
So, we see here that we can’t predict with certainty the exact even/odd nature of \(N\).
Hence, StatementII is insufficient to answer the question.
StepV: Combining Statements I & II
Since, we have received our unique answer from StatementI, we don’t need to combine the inferences from StatementI & II. Hence, the correct Answer is Option A.
Key Takeaways
1. In evenodd questions, simplify complex expressions into simpler expressions using the properties of evenodd combinations. 2. Know the properties of EvenOdd combinations to save the time spent deriving them in the test 3. Remember that there is only 1 even prime number i.e. 2. 4. If 2 is a prime factor of the number, the number would be even else it would be odd. 5. If the total number of factors of a number is given to be T, evaluate the possible ways in which T can be expressed as a product of 2 or more numbers.
Regards Harsh Hi Ujjwal, Let us understand statement 2 first, then we will come to the analysis part. Statement 2: A is divisible by 9 numbers in total, one of which is \(B^2\) If A is divisible by 9 number then we can conclude A has 9 factors in total. Thus, Total factors of A=9 Let us assume A= \(B^a*P2^b*P3^c\)....... where P2,P3... = other prime numbers and a,b,c... = Nonnegative integer. Then, total factors of A= (a+1)(b+1)....... Now, can you think in how ways (a+1)(b+1).......or multiplication of few numbers can be equal to 9?? It can happen in 2 ways only, (9*1) and (3*3). Case1) When a=8 and b, c...=0(a+1)(b+1)(c+1)......= (8+1)(0+1)(0+1)...........=9 Hence, N= \(B^8\) Case2) When a=2 and b=2 and c....=0(a+1)(b+1)(c+1)......= (2+1)(2+1)(0+1).. .= 9 Hence, \(N= B^2*P2^2\)Now, you can proceed to solve the question. Hope this helps you.
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If positive integer N = A – B, where A is a positive integer and B is
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07 Apr 2018, 20:39






