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M16-18

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M16-18  [#permalink]

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New post 16 Sep 2014, 00:59
4
13
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

33% (01:06) correct 67% (01:29) wrong based on 117 sessions

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Re M16-18  [#permalink]

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New post 16 Sep 2014, 00:59
Official Solution:


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

Notice 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) \(2 + x + 5 + y\) is an even integer. Given: \(2+x+5+y=even\). So, \(7+x+y=even\), which means that \(x+y=odd\). Not sufficient. (Consider \(x=1\) and \(y=2\) for an YES answer and \(x=1.3\) and \(y=1.7\) for a NO answer).

(2) \(x-y\) is an odd integer. Given: \(x-y=odd\). Not sufficient. (Consider \(x=1\) and \(y=2\) for an YES answer and \(x=1.3\) and \(y=0.3\) for a NO answer).

(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\), so \(2x=even\). This implies that \(x=integer\) and therefore \(y=integer\). Hence sufficient.


Answer: C
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Re: M16-18  [#permalink]

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New post 18 Mar 2016, 03:37
Hi,

Hope zero can be considered rite?

If its taken, then the answer is E. Kindly debrief?
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Re: M16-18  [#permalink]

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New post 18 Mar 2016, 03:57
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Re M16-18  [#permalink]

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New post 27 Aug 2016, 05:43
I think this is a high-quality question and I agree with explanation.
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Re: M16-18  [#permalink]

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New post 21 Nov 2016, 08:22
1
Great Question
Here we need to see if 10xy is even or not
Let P=10xy
Note => If x and y are integers => 10xy will be always even irrespective of the value of x and y
Lets look at statements
Statement 1
x+y+7=even
hence x+y=odd
hmm
x=1,y=2 => P = even
x=1.2,y=1.8 => P = 18*12/10 => not an integer(we don't have to perform the division here, as 5 is in the denominator and no 5 is in the numerator => This wont be an integer)
Hence not sufficient
Statement 2
x-y=odd
hmm
x=10,y=1=> P will be even
if x=10.1,y=1.1 => p will not be an integer.
Hence not sufficient
Combining the two statements

x+y=odd
x-y=odd
adding them up
2x=odd+odd = even
hence x => integer
now y=odd-x=> integer too
Hence x and y are both integers
hence P will always be even

Hence C
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Re: M16-18  [#permalink]

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New post 07 Jul 2017, 03:31
amazing questions
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Re: M16-18  [#permalink]

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New post 21 Nov 2017, 13:39
(1)+(2) Sum (1) and (2) (x+y)+(x−y)=odd1+odd2(x+y)+(x−y)=odd1+odd2, so 2x=even2x=even. This implies that x=integerx=integer and therefore y=integery=integer. Hence sufficient.

Can someone explain how 2x= even implies , X is an integer
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Re: M16-18  [#permalink]

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New post 21 Nov 2017, 21:08
HariharanIyeer0 wrote:
(1)+(2) Sum (1) and (2) (x+y)+(x−y)=odd1+odd2(x+y)+(x−y)=odd1+odd2, so 2x=even2x=even. This implies that x=integerx=integer and therefore y=integery=integer. Hence sufficient.

Can someone explain how 2x= even implies , X is an integer


2x = even;

x = even/2 = integer.
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Re: M16-18  [#permalink]

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New post 08 Jan 2018, 00:28
Hey Bunuel,

Can you explain how "2x= even" implies , X is an integer

When x = 1.6, 2*1.6 = 3.2 which is an even number.
When x = 2, 2*2 = 4 which is an even number.

Therefore x may or may not be an integer. Hence, the answer should be E - BOTH STATEMENTS ARE INSUFFICIENT
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Re: M16-18  [#permalink]

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New post 08 Jan 2018, 00:36
rogueassasin wrote:
Hey Bunuel,

Can you explain how "2x= even" implies , X is an integer

When x = 1.6, 2*1.6 = 3.2 which is an even number.
When x = 2, 2*2 = 4 which is an even number.

Therefore x may or may not be an integer. Hence, the answer should be E - BOTH STATEMENTS ARE INSUFFICIENT


3.2 is NOT an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an integer that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.

For more check here:
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Hope it helps.
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Re: M16-18  [#permalink]

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New post 23 Jul 2018, 01:33
Bunuel

Can you please provide some methods of determining x is an integer/y is an integer in such questions? While I'm pretty confident about the even/odd bit - the number and integer type of question (not the concept, but the proving/establishing) is something I've come across for the first time. Thank you in advance!
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Re: M16-18 &nbs [#permalink] 23 Jul 2018, 01:33
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