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Re: Is product 2*x*5*y an even integer? [#permalink]

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22 Feb 2012, 08:04

I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.

Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.

Now if we combine A & B:

Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.

There is something wrong with the question. Do you have a source for this one?
_________________

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I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.

Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.

Now if we combine A & B:

Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.

There is something wrong with the question. Do you have a source for this one?

Is product 2*x*5*y an even integer?

Notice that we are not told that x and y are integers.

Question: \(2*x*5*y=even\). As there is 2 as a multiple, then this expression will be even if \(5xy=integer\). Basically we are asked is \(5xy=integer\) true?

Note that \(x\) and \(y\) may not be integers for \(2*x*5*y\) to be even (example \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\)) BUT if they are integers then \(2*x*5*y\) is even.

(1) \(2+x+5+y=even\) --> \(7+x+y=even\) --> \(x+y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)

(2) \(x-y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)

(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\) --> \(2x=even\) --> \(x=integer\) --> \(y=integer\) --> Both \(x\) and \(y\) are integers. Hence sufficient.

Re: Is product 2*x*5*y an even integer? [#permalink]

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22 Feb 2012, 08:48

Is product 2*x*5*y an even integer?

1. 2 + x + 5 + y is an even integer 2. x - y is an odd integer

1. 7 + x + y = even (7+1+2 satisfy the statement, 7+2.5+0.5 satisfy the statement as well --> not sufficient) 2. x - y = odd (5-2 = odd and 5.5-2.5 = odd --> not sufficient)

Together: (x+y)+(x-y) = odd + odd --> 2x = even 2x can only be even if x is an integer, thus y must be an integer too

Re: Is product 2*x*5*y an even integer? [#permalink]

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24 Feb 2012, 11:18

Bunuel wrote:

omerrauf wrote:

I dont get it. \(2\) is already being multiplied to the original number. so unless \(x\) and \(y\) is a fraction, let's say \(\frac{1}{2}\) then it could be odd, but as long as \(x\) and \(y\) are integers, there is no way this could be an odd number since it is being multiplied by \(2\). Now if we look at Statement A, it still does not tell us that \(x\) and \(y\) are integers or not. Since \(x\) could be \(\frac{1}{2}\) and \(y\) again could be \(\frac{1}{2}\) so A is obviously Insufficient. We cannot even establish if \(2*x*5*y\) is an integer, let alone it is even or not. And the presence of \(2\) is extremely misleading.

Now B says that \((x - y)\) is an odd integer. Let's suppose \(x=\frac{4}{3}\) and \(y=\frac{1}{3}\), then \((x-y)=1\) which is an odd integer as it is supposed to be but that does not make \(2*x*5*y\) an even integer an integer at all. On the other hand let x=4 and y=3 than \((x-y)=1\) which is again an odd integer so yes \(2*x*5*y\) is an even integer. Two different answers, Hence Insufficient.

Now if we combine A & B:

Statement A: \(2+x+5+y\)is \(even\) so \((x+y)+7\) is \(even\) so \((x+y)\) has to be odd Statement B: \(x-y=odd\) which is basically just restating Statement A.

There is something wrong with the question. Do you have a source for this one?

Is product 2*x*5*y an even integer?

Notice that we are not told that x and y are integers.

Question: \(2*x*5*y=even\). As there is 2 as a multiple, then this expression will be even if \(5xy=integer\). Basically we are asked is \(5xy=integer\) true?

Note that \(x\) and \(y\) may not be integers for \(2*x*5*y\) to be even (example \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\)) BUT if they are integers then \(2*x*5*y\) is even.

(1) \(2+x+5+y=even\) --> \(7+x+y=even\) --> \(x+y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)

(2) \(x-y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)

(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\) --> \(2x=even\) --> \(x=integer\) --> \(y=integer\) --> Both \(x\) and \(y\) are integers. Hence sufficient.

Answer: C.

Hope it's clear.

A real tricky question and an awesome explanation. Thanks

Re: Is product 2*x*5*y an even integer? [#permalink]

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27 Feb 2013, 00:18

i knw its a silly question to ask but can anybody pls explain: 2x can only be even if x is an integer,?? what if x is a value like .7.. then 2x is 1.4.. which is even i assume.. or is it not?? any help would be highly appreciated!!!

i knw its a silly question to ask but can anybody pls explain: 2x can only be even if x is an integer,?? what if x is a value like .7.. then 2x is 1.4.. which is even i assume.. or is it not?? any help would be highly appreciated!!!

No, 0.7 is not even. Only integers can be even or odd.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. An even number is an integer of the form \(n=2k\), where \(k\) is an integer.

An odd number is an integer that is not evenly divisible by 2. An odd number is an integer of the form \(n=2k+1\), where \(k\) is an integer.

Re: Is product 2*x*5*y an even integer? [#permalink]

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17 Mar 2013, 00:05

No matter what the value of x and y is (as long as it is an integer) the product will always be an Even number , therefore what this question is asking essentially is whether x and y are integers..

Statement (1) : Simplified we get, x+y = Odd Integer... This is satisfied by x being 2.5 and y being 0.5 , therefore we are not certain whether x and y are integers .. Not Suff.

Statement (2) : x - y = odd integer, this again is satisfied with 3.5 (x) - 0.5 (y) .. therefore is also insuff.

Combining 1 and 2, and lining them up we get, 2x = Odd int + Odd Int , or 2x = even integer (odd Integer + Odd integer is always an even integer) . Solving further we now know that X is an integer (Even integer divided by 2 is always an integer) .. Similarly substituting this information in any one of the 2 equations we can verify that y is also an integer.

Therefore the Answer is C...

Hope this helps..
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"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: Is product 2*x*5*y an even integer? [#permalink]

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18 Mar 2013, 13:10

vomhorizon wrote:

No matter what the value of x and y is (as long as it is an integer) the product will always be an Even number , therefore what this question is asking essentially is whether x and y are integers..

Statement (1) : Simplified we get, x+y = Odd Integer... This is satisfied by x being 2.5 and y being 0.5 , therefore we are not certain whether x and y are integers .. Not Suff.

Statement (2) : x - y = odd integer, this again is satisfied with 3.5 (x) - 0.5 (y) .. therefore is also insuff.

Combining 1 and 2, and lining them up we get, 2x = Odd int + Odd Int , or 2x = even integer (odd Integer + Odd integer is always an even integer) . Solving further we now know that X is an integer (Even integer divided by 2 is always an integer) .. Similarly substituting this information in any one of the 2 equations we can verify that y is also an integer.

Re: Is product 2*x*5*y an even integer? [#permalink]

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30 May 2015, 08:26

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