Is the product of two integers A and B odd?

Question

Is the product of two integers A and B odd? -> in another phrase, is A = Odd and B = Odd?

(1) A is the number of factors of N, where N is a perfect square and B=A^3−1
1st part of the information about A is useless. We can tell from B=A^3-1 that B and A are even/odd or odd/even. This means product of A&B is even. So this is sufficient.

(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.
Got tricked on the first part. Since 2 is not excluded from prime number so A can be even or odd.
From the 2nd part of this statement, we know B+3^11+A is odd -> B+A is even -> means B & A are even/even or odd/odd. This result in different A*B product. Thus not sufficient.

Answer therefore is A.

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-NP-e1533292073470.jpg

ENGRTOMBA2018 · Answer

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.

kamni30 · Answer

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.

ENGRTOMBA2018 · Answer

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.

testcracker · Answer

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 ....

please help me ....

AV19 · Answer

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

mkarthik1 · Answer

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.

stonecold · Answer

Here we need to get  if AB is even or odd 
Statement 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^3-1 => B=O-O=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=odd-odd=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

ronakk29 · Answer

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

EgmatQuantExpert · Answer

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, e-GMAT https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-NP-e1533292073470.jpg

Shiv2016 · Answer

Hello e-gmat team This surely is a knockout question. I marked D very confidently assuming that I read somewhere in e-gmat 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. :thumbup:

Alexey1989x · Answer

(1)
Perfect squares have odd number of factors, thus A is ODD.
B=A^3-1=A-1=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=ODD-ODD
B+A=EVEN, thus A and B can be even or odd
Suff.

Answer A.

stan3544 · Answer

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.

GyMrAT · Answer

Is  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

zhikyang · Answer

Is the product of two integers A and B odd? -> in another phrase, is A = Odd and B = Odd?

(1) A is the number of factors of N, where N is a perfect square and B=A^3−1
1st part of the information about A is useless. We can tell from B=A^3-1 that B and A are even/odd or odd/even. This means product of A&B is even. So this is sufficient.
 
(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.
Got tricked on the first part. Since 2 is not excluded from prime number so A can be even or odd. 
From the 2nd part of this statement, we know B+3^11+A is odd -> B+A is even -> means B & A are even/even or odd/odd. This result in different A*B product. Thus not sufficient.

Answer therefore is A.

ShubhiChauhan · Answer

for stmt-2 it says A is a product of 2 consecutive prime numbers.

2,3 are the only consecutive prime numbers and according to that A is even and hence B is even?

Please explain this?

Thanks

CEdward · Answer

This step throws me off here:  The statement also tells us that A+B+311=A+B+311= odd, since 3 is an odd number, 311311 would also be odd and subtracting an odd from an odd number would give us an even number. So, we can rewrite

Why do we have to rearrange? Is it not evident that odd + odd + odd = odd? In which case A and B are odd. If that is so, then the product of AB is necessarily odd.

Thelegend2631 · Answer

Why are we not considering 1² = 1, 1 as a perfect square? Is 1 not generally considered as a perfect square?

Posted from my mobile device

Thelegend2631 · Answer

Why are we not considering 1² = 1, 1 as a perfect square? Is 1 not generally considered as a perfect square

Posted from my mobile device

ueh55406 · Answer

yes, A is the one.

for a product to be odd, both numbers have to be odd.

St1 uses the fact the in a perfect square the number of distinct factors is always odd (reverse is true as well) (also, for perfect squares, the sum of the distinct factors is also odd) anyway, so we know A is odd. is B odd ?

B= A^3-1  . just take any odd value for A, say it's 3. so,  3^3-1  = odd ? Nope. it's even. 26. therefore, B is even. therefore A*B is not odd. you can try more values if you like. it will render even number.

A is Suff.

St2> could be a trap if you forget 2 is also a prime number, and more importantly, it's an even prime. so, A , the product of two primes can be both even and odd. thus, A can take both values, even/ odd. not good enough. Not suff.

hit kudos ! if you like the explanation.

MBAB123 · Answer

Hey  Hemanthdasu13

1 is definitely a perfect square and indeed even if you consider 1 in the first statement, the answer would still be the same. A*B = 0, which is even.

Is the product of two integers A and B odd?

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Is the product of two integers A and B odd?

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