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# M16-18

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:59
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Difficulty:

95% (hard)

Question Stats:

34% (01:07) correct 66% (01:29) wrong based on 122 sessions

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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

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Joined: 02 Sep 2009
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15 Sep 2014, 23:59
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.

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18 Mar 2016, 02:37
Hi,

Hope zero can be considered rite?

If its taken, then the answer is E. Kindly debrief?
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18 Mar 2016, 02:57
HARRY113 wrote:
Hi,

Hope zero can be considered rite?

If its taken, then the answer is E. Kindly debrief?

Zero is an even integer, so the answer is still C.
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27 Aug 2016, 04:43
I think this is a high-quality question and I agree with explanation.
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Joined: 12 Aug 2015
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Schools: Boston U '20 (M)
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21 Nov 2016, 07:22
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
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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07 Jul 2017, 02:31
amazing questions
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21 Nov 2017, 12:39
(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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21 Nov 2017, 20:08
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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07 Jan 2018, 23:28
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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07 Jan 2018, 23:36
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:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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23 Jul 2018, 00:33
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!
Re: M16-18 &nbs [#permalink] 23 Jul 2018, 00:33
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# M16-18

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