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# Is the product of integers M and N even?

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Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 05:04
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Is the product of integers M and N even?

(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.

(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.

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Last edited by EgmatQuantExpert on 13 Dec 2016, 05:45, edited 3 times in total.

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Re: Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 09:26
1) For N, we can surely say that it is odd. Since 2 & 3 are the only consecutive prime numbers, N = $$3^2 - 2^2$$ = 5 which is odd.
Form M, we know that M = P*Q where Q is odd. But we don't know anything about P. So nature of M can't be found.
So, insufficient.

2) For N, we can surely say that it is even because difference of squares of any two prime numbers which differ by 2 shall always be even.This is because N = $$(X+2)^2-X^2 = X^2 + 4 + 2X - X^2$$ = 2X + 4 which is even for any number X.
Since N is always even here so product of M & N will be even no matter what M is.
Sufficient.

So B.

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Re: Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 10:05
Wow.. really a thriller question. Took me somewhere around 3 mins.

Solution

The only central idea is to know if either M or N or Both are even numbers.

From Option, B it is quite clear that N= even. [Exactly at that point, I stopped further calculation].

EgmatQuantExpert wrote:
Is the product of integers M and N even?

(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.

(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.

We will provide the OA in some time. Till then Happy Solving

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Re: Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 10:36
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Hi tapas.shyam

By 2 consecutive prime numbers it need not neccesarily mean that it would be 2 and 3.According to me, it means that out of primes numbers like 2,3,5,7... the numbers would be consecutive.
Also, in condition 2, M would also be even.
Consider multiples of 4- 4,8,12
So, Z would be 3,7,11
You can check the sum of these numbers using n(n+1)/2. It will be even always.

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Re: Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 17:52
Naina1 wrote:
Hi tapas.shyam

By 2 consecutive prime numbers it need not neccesarily mean that it would be 2 and 3.According to me, it means that out of primes numbers like 2,3,5,7... the numbers would be consecutive.
Also, in condition 2, M would also be even.
Consider multiples of 4- 4,8,12
So, Z would be 3,7,11
You can check the sum of these numbers using n(n+1)/2. It will be even always.

Hi Naina

Thanks for your reply and opinion. However, I think my reasoning is correct as the problem says - 2 consecutive prime numbers atleast one of which is odd. Now, except 2 & 3 there is no other case where we can have an even prime number - so emphasising 'atleast one is odd' in the problem then would seem redundant. In every other case of consecutive prime numbers, we shall always have two odd prime numbers no matter what.

Secondly, in second condition, we don't even need to look into M as once we have ascertained that N is even, we can conclude that M*N shall be even no matter what value of M.

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Re: Is the product of integers M and N even? [#permalink]

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09 Apr 2015, 18:21
answer is B. Second one says the two prime numbers is at a distance of two hence "2" is ruled out since the next prime 3 is one away from it and 5 is 3 away from 2. Then the difference between squares of two off terms is even. N is even the value of M doesnt matter.

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Re: Is the product of integers M and N even? [#permalink]

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10 Apr 2015, 06:53
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Detailed Solution

Step-I: Given Info

We are given two integers M and N and are asked to find if their product is even.

Step-II: Interpreting the Question Statement

The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.

Step-III: Statement-I

The statement tells us that M is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:

• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd

• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.

From the above two cases, we can’t say with certainty whether N is odd or even.

The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.

Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.

So, N would be even.

Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.

Hence, Statement-II is 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 B

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature

tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.

Regards
Harsh
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Re: Is the product of integers M and N even? [#permalink]

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12 Apr 2015, 13:10
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given two integers M and N and are asked to find if their product is even.

Step-II: Interpreting the Question Statement

The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.

Step-III: Statement-I

The statement tells us that M is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:

• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd

• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.

From the above two cases, we can’t say with certainty whether N is odd or even.

The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.

Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.

So, N would be even.

Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.

Hence, Statement-II is 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 B

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature

tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.

Regards
Harsh

Harsh, thanks for the detailed solution. I have a doubt in statement #1 (I am definitely missing something here!). If Q= 2P+1, with both Q and being NATURAL NUMBERS, we can clearly see that Q is Odd. This leads to show that P will then be an even number. Thus, with one of P/Q determined to be an even number >> M=even. So MN = even irrespective of what N is. Thus the OA in my opinion should be D (both are sufficient!).

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Re: Is the product of integers M and N even? [#permalink]

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12 Apr 2015, 21:14
Engr2012 wrote:
EgmatQuantExpert wrote:
Detailed Solution

Step-I: Given Info

We are given two integers M and N and are asked to find if their product is even.

Step-II: Interpreting the Question Statement

The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.

Step-III: Statement-I

The statement tells us that N is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:

• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd

• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.

From the above two cases, we can’t say with certainty whether N is odd or even.

The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.

Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.

Step-IV: Statement-II

Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.

So, N would be even.

Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.

Hence, Statement-II is 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 B

Key Takeaways

1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature

tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.

Regards
Harsh

Harsh, thanks for the detailed solution. I have a doubt in statement #1 (I am definitely missing something here!). If Q= 2P+1, with both Q and being NATURAL NUMBERS, we can clearly see that Q is Odd. This leads to show that P will then be an even number. Thus, with one of P/Q determined to be an even number >> M=even. So MN = even irrespective of what N is. Thus the OA in my opinion should be D (both are sufficient!).

Hi Engr2012, You are absolutely right when you say Q is an odd number as it is in the form of 2P + 1, but if you observe you would see that Q will be an odd number irrespective of the even/odd nature of P as 2P would always be even irrespective of whether P is odd or even.

For example: Consider P = 10, Q= 21, also if P= 11, Q =23. So in both the cases, Q would be odd even when P is even or odd.

Since, we don't know for sure the even/odd nature of P in statement-I , we would not be able to comment on even/odd nature of N.

Hope its's clear

Regards
Harsh

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Re: Is the product of integers M and N even? [#permalink]

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13 Apr 2015, 01:56
Thanks for the reply. I understand your point but let me present to you another way of looking at the given information. If Q=2P+1, we know for sure that Q will be ODD. Also we can write P= (Q-1)/2 and as Q is odd, P has to be even.

Let me know where I'm going wrong with my thinking!

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Re: Is the product of integers M and N even? [#permalink]

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13 Apr 2015, 03:13
Engr2012 Assume Q as 7, so Q-1= 6 which is even. If I divide (Q-1) by 2 I would get 3, which is odd.
Similarly, assume Q=9, so Q-1 = 8. If I divide (Q-1) by 2 I would get 4, which is even. Hence, Q can be even or odd.

An even number can be represented in the form of 2n. When this number is divided by 2, the resultant would be n. Now, n can be even or odd. So, we can say that a number when divided by 2 would still be even, if the original number was a multiple of at least $$2^2$$ i.e. 4. In this case, we know that (Q-1) is even but do not know if (Q-1) is a multiple of 4. Hence, $$\frac{(Q-1)}{2}$$ may be even or odd.

Regards
Harsh

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Re: Is the product of integers M and N even? [#permalink]

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13 Apr 2015, 09:44
Hi Harsh

Thanks for the reply. Yes, now I understand my mistake. I was ignoring that 6 is an odd multiple of 2, in effect making P both odd and even with different sets of values for Q. Should've been more careful with my assumption.

Thanks

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Re: Is the product of integers M and N even? [#permalink]

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08 Nov 2015, 21:46
tapas.shyam wrote:
1) For N, we can surely say that it is odd. Since 2 & 3 are the only consecutive prime numbers, N = $$3^2 - 2^2$$ = 5 which is odd.
Form M, we know that M = P*Q where Q is odd. But we don't know anything about P. So nature of M can't be found.
So, insufficient.

2) For N, we can surely say that it is even because difference of squares of any two prime numbers which differ by 2 shall always be even.This is because N = $$(X+2)^2-X^2 = X^2 + 4 + 2X - X^2$$ = 2X + 4 which is even for any number X.
Since N is always even here so product of M & N will be even no matter what M is.
Sufficient.

So B.

I think consecutive prime number can be 3,5,7,11,13,. Am i Correct. Please clarify.

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Re: Is the product of integers M and N even? [#permalink]

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11 Dec 2017, 21:38
EgmatQuantExpert wrote:
Is the product of integers M and N even?

(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.

(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.

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Hello,
Thanks for your response but I have a question.
In the first statement, "N can be expressed as a difference of squares of two consecutive prime numbers", only consecutive prime nos are 2 and 3. So doesn't become the case in hand?

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Re: Is the product of integers M and N even? [#permalink]

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11 Dec 2017, 21:45
SamriddhiPan wrote:
EgmatQuantExpert wrote:
Is the product of integers M and N even?

(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.

(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.

This is

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Hello,
Thanks for your response but I have a question.
In the first statement, "N can be expressed as a difference of squares of two consecutive prime numbers", only consecutive prime nos are 2 and 3. So doesn't become the case in hand?

The two consecutive integers, of which both are primes are indeed 2 and 3. But consecutive primes are {2, 3}, {3, 5}, {5, 7}, {7, 11}...
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Re: Is the product of integers M and N even?   [#permalink] 11 Dec 2017, 21:45
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