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Posts: 3,651
Updated on: 13 Aug 2018, 02:10
Originally posted by EgmatQuantExpert on 08 Apr 2015, 02:50.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:10, edited 8 times in total.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:10, edited 8 times in total.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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44%
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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: e-GMAT
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(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: e-GMAT
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Posts: 3,651
Updated on: 07 Aug 2018, 00:01
Originally posted by EgmatQuantExpert on 10 Apr 2015, 04:40.
Last edited by EgmatQuantExpert on 07 Aug 2018, 00:01, edited 3 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 00:01, edited 3 times in total.
Detailed Solution
Step-I: Given Info
The question tells us about two integers \(A\) & \(B\) and asks us if their product is odd.
Step-II: 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.
Step-III: Statement-I
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 Statement-I is sufficient to get the answer.
Step-IV: Statement-II
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.
Step-V: 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 even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2.The number of factors of a perfect square would always be odd.
3.Know the properties of Even-Odd combinations to save the time spent deriving them in the test
Regards
Harsh
_________________
Step-I: Given Info
The question tells us about two integers \(A\) & \(B\) and asks us if their product is odd.
Step-II: 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.
Step-III: Statement-I
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 Statement-I is sufficient to get the answer.
Step-IV: Statement-II
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.
Step-V: 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 even-odd questions, simplify complex expressions into simpler expressions using the properties of even-odd combinations.
2.The number of factors of a perfect square would always be odd.
3.Know the properties of Even-Odd combinations to save the time spent deriving them in the test
Regards
Harsh
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Posts: 3,651
Updated on: 07 Aug 2018, 00:03
Originally posted by EgmatQuantExpert on 19 Jun 2015, 20:52.
Last edited by EgmatQuantExpert on 07 Aug 2018, 00:03, edited 2 times in total.
Last edited by EgmatQuantExpert on 07 Aug 2018, 00:03, edited 2 times in total.
riyazgilani
Dear riyazgilani
Please 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
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General Discussion
EgmatQuantExpert
1. Number of factors of a perfect square is odd. Therefore, A is odd.
B=Odd^3-1=Even
Therefore, A*B=Odd*Even=Even.
Answer : No
Sufficient
2. A could be Even if one of the prime numbers is 2.
a*b=Even
A could be Odd if the smaller prime no is not 2.
a+b+3^11=Odd
Odd+b+Odd=Odd
b=Odd
a*b=Odd
Therefore, a*b could be both even and odd depending on the value of a which could be both even and odd.--->Not Sufficient.
Answer: A
GMAT 1: 610 Q49 V25
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Statement 1: A is odd ( no. of factors of a perfect square) ; B is even (odd no.(A^3) - 1). Therefore A*B is even. Hence sufficient.
Statement 2: A can be odd(3*5) or even(2*3). Since 3^11 is odd, we need A+B=even for A+B+3^11 to be odd. So, when A is odd, B is also odd and vice versa. Therefore A*B=odd(when both are odd) or even(when both are even). Hence Insufficient.
Answer: (a)
Statement 2: A can be odd(3*5) or even(2*3). Since 3^11 is odd, we need A+B=even for A+B+3^11 to be odd. So, when A is odd, B is also odd and vice versa. Therefore A*B=odd(when both are odd) or even(when both are even). Hence Insufficient.
Answer: (a)
Tmoni26
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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
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
Question IS A X B = ODD ?
For A X B = ODD, A and B Shall be ODD.
1. First option is fairly easy to judge, if you knew No of factors of a perfect square is odd.
A= N ( No of Factors of perfect Square ) => ODD
B = A^ 3 - 1 ( We Know A = ODD, There fore A * A * A = ODD ( 3 *3*3 = 81 ODD)
B = 81 -1 = Even
Therefore : ODD X EVEN = EVEN ( Sufficient to Answer)
2: Statement
A is a product of two consecutive prime Nos, ( Prime Nos : 2 , 3, 5, 7 , 11...... )
Only Two is Even. 1 st Case 2 * 3 = 6 Even, 2nd Case 3 * 5 = 15 ODD )
So A Can be Either Even Or ODD
Lets Analyze B : B + 3^11 is added to A, the sum is an odd number
ODD = B + ODD + EVEN (We Know 3 multiplied by any no of time of 3 will be an ODD Number)
Now We have 4 Cases 1. B= ODD, 2. B = EVEN, 3. A= ODD 4, A= EVEN
Since sum has to be Odd : 1 st ODD + ODD + ODD = ODD 2nd EVEN + ODD + EVEN = ODD. B and A Can ODD or Can be EVEn
Statement 2 Inconclusive
For A X B = ODD, A and B Shall be ODD.
1. First option is fairly easy to judge, if you knew No of factors of a perfect square is odd.
A= N ( No of Factors of perfect Square ) => ODD
B = A^ 3 - 1 ( We Know A = ODD, There fore A * A * A = ODD ( 3 *3*3 = 81 ODD)
B = 81 -1 = Even
Therefore : ODD X EVEN = EVEN ( Sufficient to Answer)
2: Statement
A is a product of two consecutive prime Nos, ( Prime Nos : 2 , 3, 5, 7 , 11...... )
Only Two is Even. 1 st Case 2 * 3 = 6 Even, 2nd Case 3 * 5 = 15 ODD )
So A Can be Either Even Or ODD
Lets Analyze B : B + 3^11 is added to A, the sum is an odd number
ODD = B + ODD + EVEN (We Know 3 multiplied by any no of time of 3 will be an ODD Number)
Now We have 4 Cases 1. B= ODD, 2. B = EVEN, 3. A= ODD 4, A= EVEN
Since sum has to be Odd : 1 st ODD + ODD + ODD = ODD 2nd EVEN + ODD + EVEN = ODD. B and A Can ODD or Can be EVEn
Statement 2 Inconclusive
Statement 2 mentions that A is a product of two consecutive prime numbers.
I do not see how 3 and 5 are consecutive!
I do not see how 3 and 5 are consecutive!
nassib787
You should read more carefully: A is a product of two consecutive prime numbers. 3 and 5 ARE consecutive primes.
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riyazgilani
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EgmatQuantExpert
E-Gmat 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..?
Statement 2 suggests that A is a product of two prime numbers. The only two consecutive prime numbers are 2 and 3 which infers that A = 6 which is an even number. Therefore if the sum of A+ (B+3^11) = odd, (B+3^11) is odd which also implies that B has to be even in order for (B+3^11) to be odd. Dosen't this make statement 2 also sufficient to answer the question?
Kjaffer
The highlighted part shows the mistake you have made. Consecutive Prime Numbers mean two prime numbers which have no other prime number falling between the two of them
2 and 3 are the only numbers which are consecutive Integers and Consecutive Prime Numbers as well
(2,3) or (3,5) or (5,7) or (7,11) or (11,13) or (13,17) etc. are all representing pairs of two consecutive Primes
I hope it clears your doubt!
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Kjaffer
Hi Kjaffer,
Please refer to the post here which answers your query.
Let me know in case you still have any trouble in the explanation
Regards
Harsh
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For the first statement, I only analysed the second part:
B= A^3-1
B= A-1 (since power won't change the odd nature of A)
So from equation we can see that A & B will have different sign ie if A is even, B is odd or A is odd, B is even. In either case the product will be even. Hence statement 1 is sufficient.
B= A^3-1
B= A-1 (since power won't change the odd nature of A)
So from equation we can see that A & B will have different sign ie if A is even, B is odd or A is odd, B is even. In either case the product will be even. Hence statement 1 is sufficient.
[quote="EgmatQuantExpert"]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.
First thing first this is a mind blowing question that pushes you to bring almost all of the number property rules on the surface of your brain.
Here's my take from this question.
A*B= odd then A and B must both be odd.
1)A=number of factor of N,and N is perfect square (1st half of the statement)
infer - N^even -that leads to ---n^(even + 1) = odd , thus A must be ODD.
B=A^3-1 (other half of the statement)
so far we know A is ODD --thus-- odd^3-1--- 3^3-1=27-1= 26(even) & 5^3-1= 125-1 = 124 (even)
so know we know A=odd & B= Even.
we can conclude - odd+ even= must be even ------Sufficient !!
2)A is product of two consecutive prime number. (1st half of the statement)
A= 2*3= 6 (even) or 3*5 = 15 ( odd).
&
A+B +3^11= ODD
here comes the cyclist rule of unit digit for 3^11, the unit digit be 11/4 = (remainder)= 3---3^1*3^2=3*9=7(unit digit)
Those who don't understand this part must check rule of cyclist rules.
we know now 3^11= odd.
A (could be odd/even) + b+odd= odd ------Insufficient !!
Answer= B.
(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.
First thing first this is a mind blowing question that pushes you to bring almost all of the number property rules on the surface of your brain.
Here's my take from this question.
A*B= odd then A and B must both be odd.
1)A=number of factor of N,and N is perfect square (1st half of the statement)
infer - N^even -that leads to ---n^(even + 1) = odd , thus A must be ODD.
B=A^3-1 (other half of the statement)
so far we know A is ODD --thus-- odd^3-1--- 3^3-1=27-1= 26(even) & 5^3-1= 125-1 = 124 (even)
so know we know A=odd & B= Even.
we can conclude - odd+ even= must be even ------Sufficient !!
2)A is product of two consecutive prime number. (1st half of the statement)
A= 2*3= 6 (even) or 3*5 = 15 ( odd).
&
A+B +3^11= ODD
here comes the cyclist rule of unit digit for 3^11, the unit digit be 11/4 = (remainder)= 3---3^1*3^2=3*9=7(unit digit)
Those who don't understand this part must check rule of cyclist rules.
we know now 3^11= odd.
A (could be odd/even) + b+odd= odd ------Insufficient !!
Answer= B.
ggurface
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Is it necessary to know that # of factors in perfect square is always odd?
I found that the answer is A without knowing that theory by doing this:
Is AB odd? Only way for AB to be odd is if A & B are both odd.
(1) Using just B = A^3 -1
=> This says that only one of A and B are odd. Try it for yourself, odd^anything - 1 = even. Even^anything - 1 = odd.
Sufficient.
(2) A is the product of 2 consecutive primes.
=> Doesnt say anything about A, because this could be 2*3 which is even, or all the other consecutive primes which would be odd.
When B + 3^11 is added to A, the sum is odd
=> odd ^ anything = odd so
=> B + odd + A = odd
=> odd + x = odd if x is even
=> A + B = even.
A&B could be both even, or both odd, still insufficient.
I found that the answer is A without knowing that theory by doing this:
Is AB odd? Only way for AB to be odd is if A & B are both odd.
(1) Using just B = A^3 -1
=> This says that only one of A and B are odd. Try it for yourself, odd^anything - 1 = even. Even^anything - 1 = odd.
Sufficient.
(2) A is the product of 2 consecutive primes.
=> Doesnt say anything about A, because this could be 2*3 which is even, or all the other consecutive primes which would be odd.
When B + 3^11 is added to A, the sum is odd
=> odd ^ anything = odd so
=> B + odd + A = odd
=> odd + x = odd if x is even
=> A + B = even.
A&B could be both even, or both odd, still insufficient.
Hi ggurface ,
Is it necessary to know that # of factors in perfect square is always odd?
let's say we have a number 84700, and we need to know the number of factors it has,
Then - 84700 = 2^2 * 5^2* 7^1* 11^2
so in order for us to determine number of factors 84700 has we would add 1 to each power and then would add them all together.
(2+1)+(2+1)+(1+1)+(2+1)= 3+3+2+3=11 factors, now we know 84700 has 11 factors in it.
Coming to your question- statement 1 says " A is the number of factors of N", where N is a perfect square"
It clearly means we need to know whether A is even/odd.
N^even +1 = odd=A
I figured out in above mentioned method.
I don't know if there's any other method exists by which the problem could be solved. There might be.Please comment GMAT Experts if i'm wrong in my explanation.
Is it necessary to know that # of factors in perfect square is always odd?
let's say we have a number 84700, and we need to know the number of factors it has,
Then - 84700 = 2^2 * 5^2* 7^1* 11^2
so in order for us to determine number of factors 84700 has we would add 1 to each power and then would add them all together.
(2+1)+(2+1)+(1+1)+(2+1)= 3+3+2+3=11 factors, now we know 84700 has 11 factors in it.
Coming to your question- statement 1 says " A is the number of factors of N", where N is a perfect square"
It clearly means we need to know whether A is even/odd.
N^even +1 = odd=A
I figured out in above mentioned method.
I don't know if there's any other method exists by which the problem could be solved. There might be.Please comment GMAT Experts if i'm wrong in my explanation.
ggurface
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jaspreets
but the only way for AB to be odd is if A & B are BOTH ODD, and if B = A^3 - 1 then doesnt that mean they always differ in even/odd?
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ggurface
Yes, ggurface, you are right that \(B = A^3-1\) in itself is sufficient and you do not need to know any other information. As \(B = A -1\) (simplified version as the nature of A will same as that for A^n, n\(\in\) positive integer).
Thus , from B = A-1 , we know that if A = even, B = odd and if A = odd, B = even. Thus in either scenario, AB = even and we get a unique answer for all the cases.
Posts: 6
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
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
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