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Re: Is the product of two integers A and B odd?
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21 Aug 2015, 04:27
PerptualKnite56 wrote: I have one doubt! Is 1 not considered a perfect square in GMAT? 1^2 = 1 If 1 is considered as a perfect square then Option A also does not yield single solution.
If N is 1 then A =1 so B = 1^3 1 =0 A.B = 1.0 = 0 not odd. For all other perfect squares A.B is definitely odd. So option A is also insufficient. Hence answer should be E. Can you please clarify?
Thanks, Pk56 You are partially correct there. 1 is indeed a perfect square in the realm of maths not just GMAT. For any perfect square N, N = \(k^{2p}\), where k,p \(\in\)integer, The number of factors of such a perfect square = 2p+1. As 2p is always even for all 'p', 2p+1 will always be odd. This is true for ALL perfect squares = 25 = \(5^2\) > Number of factors of 25 = 2+1 = 3 (odd number). Thus, A = odd > B = odd 1 = even for all A. Thus, out of A and B , B will always be even and hence AB = even for all A,B You get a "sufficient" yes for statement 1.



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Re: Is the product of two integers A and B odd?
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14 Nov 2015, 10:04
Part 2 of this question says " A is a product of two consecutive prime numbers...."
the only two consecutive prime numbers are 2 and 3, so why even consider the possibility of 3*5 or 5*7??
Please tell if you agree.



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Re: Is the product of two integers A and B odd?
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14 Nov 2015, 10:43
kamni30 wrote: Part 2 of this question says " A is a product of two consecutive prime numbers...."
the only two consecutive prime numbers are 2 and 3, so why even consider the possibility of 3*5 or 5*7??
Please tell if you agree. Consecutive primes are primes between which there are no other primes. 2,3 3,5 5,7 7,11 etc are consecutive prime pairs. 2,3 is a particular case of "consecutive prime numbers" that are also 'consecutive' on the number line.



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Re: Is the product of two integers A and B odd?
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23 Nov 2015, 10:38
Statement two clearly states that A = product of two consecutive prime numbers....and according to your prime 1 concept file, the only consecutive primes mean 2 and 3. thus A = 6. so statement 2 is sufficient to answer the questions .... please help me ....



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Re: Is the product of two integers A and B odd?
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28 Mar 2016, 09:07
Tmoni26 wrote: Damm, it really is a knockout question
No of factors in a perfect square is odd, therefore B would be even Test it Say 36....no of factors is A = 9, then B = 9^3  = even integer Therefore odd x even must be even, so the answer is sufficiently no .........Sufficient
Stmt 2. A could be 2*3 = even or, 3*5 = odd If A is even, then B must be even for (B+ 3^11) to be odd, and for the total sum of A+ (B+ 3^11) to be odd, therefore A*B is even
If A is odd, then B must be odd for (B + 3 ^11) to be even, and for the total sum of (B+ 3^11), to be odd, therefore A*B is odd Stmt 2 is insufficient
Gosh, I need to get neater in my paperwork You actually dont need to do any calculations St:1 Lets assume A can be odd or odd . we know that B=A^3 + 1 (Power does not affect even odd) so If A is odd B is even and vice versa!!!! WE got the answer without thinking what would be the value of A



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Re: Is the product of two integers A and B odd?
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22 Apr 2016, 10:07
Hey the question actually confused me. The second statement could have been more clear. What does two consecutive prime numbers mean? Two prime numbers that are consecutive or Two prime numbers taken consecutively??? This has hampered my approach. My Approach: Statement one : N is perfect square. Then the number of factors of N are always odd. Hence A is ODD. A^3 is odd...So (A^3)  1 is even. So B is even and hence value can be determined.
Statement two: (No clarity at all or my understanding was poor): only prime numbers that are consecutive are 2, 3 So A is even. B + 3^11(Odd) + A(Even) gives Odd .. So B must be Even and product can be determined.
I may be wrong but my approach was the same.



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Re: Is the product of two integers A and B odd?
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22 Aug 2016, 06:49
Here we need to get if AB is even or oddStatement 1 => A = number of factors of any perfect square But the number of factors of any perfect square is odd
Hence A=odd and B=A^31 => B=OO=even => B is even Hence AB= Even . The answer to the question is No AB is not odd Suff Statment 2 => Here A = product of two consecutive primes. NoTe => The product of two consecutive prime numbers ≠ The product two of consecutive numbers that are prime
Hence A can be even if one of those primes is 2 else it would be odd E.G => A=2*3=6 and A=3*5=15 So A can e even or Odd Hence we Don't know the character of A But we need to check B too as if B is even then we wont need A Now B+3^11+A=odd any odd^n =odd ; so 3^11 = odd
hence A+B=oddodd=even So B can be even or odd depending on A and we dont have A Hence A,B can be both even or both odd If both even the answer would be NO if both odd the answer would be YES Hence we dont have a unique answer. Insuff SMASH that A
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Re: Is the product of two integers A and B odd?
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06 Dec 2016, 09:35
EgmatQuantExpert wrote: Detailed Solution
StepI: Given Info
The question tells us about two integers \(A\) & \(B\) and asks us if their product is odd.
StepII: Interpreting the Question Statement
To find if the product of two numbers is odd/even, we need to establish if either of the number is even or not. In case either of the number is even, the product would be even else the product would be odd.
StepIII: StatementI
Statement I tells us that \(A\) is the number of factors of a perfect square, since the no. of factors of a perfect square is odd, we can deduce that \(A\) is odd. Since \(A\) is not even, to find the nature of product of \(A\) & \(B\), we need to find if \(B\) is odd/even.
It’s given that \(B= A^3 1\), since we have established that \(A\) is odd, \(A^ 3\) will also be odd. Subtracting 1 from an odd number will give us an even number, hence we can deduce that \(B\) is even. Since, we know now that \(B\) is even it is sufficient for us to deduce that the product of \(A\) & \(B\) would be even.
Thus StatementI is sufficient to get the answer.
StepIV: StatementII
Statement II tells us that \(A\) is a product of two consecutive prime numbers, so \(A\) may be even if one of the prime number is 2 and may be odd if none of the prime number is 2. So, we can’t establish with certainty the even/odd nature of \(A\). The statement also tells us that \(A+ B + 3^{11} =\) odd, since 3 is an odd number, \(3^ {11}\) would also be odd and subtracting an odd from an odd number would give us an even number. So, we can rewrite \(A+ B =\) even which would imply that either both \(A\), \(B\) are even or both are odd. In both the cases the nature of product of \(A\) & \(B\) can’t be established with certainty, it will be even if both \(A\) & \(B\) are even and will be odd if both \(A\) & \(B\) are odd.
So, Statement II is not sufficient to answer the question.
StepV: Combining Statements I & II
Since, we have a unique answer from Statement I we don’t need to be combine Statements I & 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.The number of factors of a perfect square would always be odd. 3.Know the properties of EvenOdd combinations to save the time spent deriving them in the test
Regards Harsh Hi Harsh, According to Statement 1: (1) "A is the number of factors of N, where N is a perfect square" In contrast to the answer you have provided, I have interpreted this to be the absolute number of factors of N (not the unique number) . Hence, if interpreted in this way, A can be Even. For example if N = 4^2 = 16 (a perfect square) the factors would be 1,16,2,8,4,4. Hence A would be = 6, Even. On the other hand, the number of unique factors would be 5, and in this case A would Odd. Please correct me if this interpretation is wrong. The same answer, statement 1 is sufficient, is achieved. Many thanks Ronak



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Re: Is the product of two integers A and B odd?
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Updated on: 07 Aug 2018, 00:19
ronakk29 wrote:
Hi Harsh,
According to Statement 1: (1) "A is the number of factors of N, where N is a perfect square"
In contrast to the answer you have provided, I have interpreted this to be the absolute number of factors of N (not the unique number) . Hence, if interpreted in this way, A can be Even.
For example if N = 4^2 = 16 (a perfect square) the factors would be 1,16,2,8,4,4. Hence A would be = 6, Even. On the other hand, the number of unique factors would be 5, and in this case A would Odd.
Please correct me if this interpretation is wrong. The same answer, statement 1 is sufficient, is achieved.
Many thanks Ronak Hey Ronak, Thanks for posting your query. When we are finding the number of factors of any number, we consider only the unique factors and there is no need to double count any numbers . The example that you have considered, i.e. N= \(4^2\), in this case, we will write N as \(2^4\) and say that the factors of N are \(2^0\),\(2^1\),\(2^2\),\(2^3\),\(2^4\). We wont be writing \(2^2\) twice! You can understand it through this logic also...if the reasoning that you have suggested would have been valid that while writing N= \(4^2\), we should have counted 2 four times..since N=\(2^4\) and 2 is occurring 4 times.. I hope this clears your doubt. Therefore, please remember, when we are counting the number of factors of any number, we only consider each factor once. Thanks, Saquib Quant Expert, eGMAT
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Re: Is the product of two integers A and B odd?
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08 Jan 2017, 21:57
EgmatQuantExpert wrote: riyazgilani wrote: EGmat number properties concept file mentions that " 2 & 3 are the only consecutive prime numbers" so when the que says consecutive prime nos, what are we supposed to assume..? whether we are talikng about 2 & 3 here or anything else like say 5 & 7 or 11 & 13..? Dear riyazgilaniPlease note that there is a difference between ' consecutive numbers that are prime' and 'consecutive prime numbers.'Our concept file mentions that 2 and 3 are the only 'consecutive numbers that are prime'On the other hand, examples of 'consecutive prime numbers' are many: 2, 3, 5, 7, 11, 13, 17, 19 . . . all these prime numbers are consecutive prime numbers, because they come one after the other in the list of prime numbers. So, when this question says that two numbers are consecutive prime numbers, they can be (2,3) or (5, 7) or (7,11) etc. I hope this clarification helped. Best Regards Japinder Hello egmat team This surely is a knockout question. I marked D very confidently assuming that I read somewhere in egmat file that only 2 and 3 are consecutive prime numbers. Thanks for explaining the importance of every word and construction of the question. Great post.



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Re: Is the product of two integers A and B odd?
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02 Apr 2017, 08:03
(1) Perfect squares have odd number of factors, thus A is ODD. B=A^31=A1=EVEN Basing on aforementioned, product of A & B is EVEN Suff. (2) A can be EVEN or ODD: 2*3=EVEN 3*5=ODD B+3^11+A=ODD 3^11 is ODD (as power of 11 ends with 7 which is ODD) B+ODD+A=ODD B+A=ODDODD B+A=EVEN, thus A and B can be even or odd Suff.
Answer A.



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Re: Is the product of two integers A and B odd?
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12 May 2017, 06:24
EgmatQuantExpert wrote: riyazgilani wrote: EGmat number properties concept file mentions that " 2 & 3 are the only consecutive prime numbers" so when the que says consecutive prime nos, what are we supposed to assume..? whether we are talikng about 2 & 3 here or anything else like say 5 & 7 or 11 & 13..? Dear riyazgilaniPlease note that there is a difference between ' consecutive numbers that are prime' and 'consecutive prime numbers.'Our concept file mentions that 2 and 3 are the only 'consecutive numbers that are prime'On the other hand, examples of 'consecutive prime numbers' are many: 2, 3, 5, 7, 11, 13, 17, 19 . . . all these prime numbers are consecutive prime numbers, because they come one after the other in the list of prime numbers. So, when this question says that two numbers are consecutive prime numbers, they can be (2,3) or (5, 7) or (7,11) etc. I hope this clarification helped. Best Regards Japinder This is actually not true! I made a mistake in this question only because i followed your first concept on PN to the word. And there is a mistake in it, you can check it out by yourself, here is the proof. The audio is also wrong.
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PN error.png [ 210.52 KiB  Viewed 692 times ]



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Re: Is the product of two integers A and B odd?
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08 Jun 2018, 22:11
EgmatQuantExpert wrote: Is the product of two integers \(A\) and \(B\) odd? (1) \(A\) is the number of factors of \(N\), where \(N\) is a perfect square and \(B = A^3  1\) (2) \(A\) is a product of two consecutive prime numbers and when \(B + 3^{11}\) is added to \(A\), the sum is an odd number. Source: eGMATIs \(A*B\)  Odd?, which means is \(A\)  odd & \(B\)  odd as well? Statement 1: Let \(N = a^x*b^y*c^z\) Since \(N\) is a perfect square, \(x,y,z\) are even & \(A = (x+1)(y+1)(z+1)\) is a product of odd integers, Hence \(A\) is \(Odd\). given, \(B = A^3  1\) = \(Odd  Odd = Even\), Hence \(B\)  \(Even\). The product \(A*B = Odd * Even\) = \(Even\) Hence \(A*B\) is not \(Odd\) Statement 1 alone is Sufficient. Statement 2: \(A = P * Q\), \(P\) & \(Q\) are consecutive prime numbers So they could be \(P = 2\) & \(Q = 3\) or could be \(P = 3\) & \(Q = 5\) or any two other consecutive primes. If \(P = 2\), we get \(A\)  \(Even\) If \(P = 3\) or any other Odd prime, then \(A\)  \(Odd\) Also given \(A + B + 3^{11}\) = \(Odd\) When \(A\)  \(Even\), the above expression will have \(B\)  \(Even\), Hence \(A * B = Even\) When \(A\)  \(Odd\), the above expression will have \(B\)  \(Odd\), Hence \(A* B = Odd\) Statement 2 alone is Not Sufficient. Answer A. Thanks, GyM




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