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Devil's Dozen!!!

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Devil's Dozen!!! [#permalink]

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The next set of tough and tricky DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. Jules and Jim both invested certain amount of money in bond M for one year, which pays for 12% simple interest annually. If no other investment were made, then Jules initial investment in bond M was how many dollars more than Jim's investment in bond M.
(1) In one year Jules earned $24 more than Jim from bond M.
(2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063846

2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!-n!
(2) p is a factor of (n+2)!/n!

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063847

3. If x and y are integers, is y an even integer?
(1) 4y^2+3x^2=x^4+y^4
(2) y=4-x^2

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063848

4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?
(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia.
(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063863

5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach?
(1) Of the astronauts who do NOT listen to Bach 56% are male.
(2) Of the astronauts who listen to Bach 70% are female.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063867

6. Is the perimeter of triangle with the sides a, b and c greater than 30?
(1) a-b=15.
(2) The area of the triangle is 50.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063871

7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order?
(1) Set A consists of 12 even consecutive integers.
(2) n=5.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063874

8. If p is a positive integer, what is the remainder when p^2 is divided by 12?
(1) p is greater than 3.
(2) p is a prime.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063884

9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers?
(1) The average (arithmetic mean) of the three numbers is 34/3.
(2) The largest number of the three distinct numbers is 24.

Solution: https://gmatclub.com/forum/devil-s-doze ... l#p1063886

10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
(1) There are total 99 snakes in Pandora's box.
(2) From any two snakes from Pandora's box at least one is a viper.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063888

11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?
(1) $15 is enough to buy 7 muffins and 11 brownies.
(2) $15 is enough to buy 10 muffins and 8 brownies.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063892

12. If x>0 and xy=z, what is the value of yz?
(1) \(x^2*y=3\).
(2) \(\sqrt{x*y^2}=3\).

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063894

13. Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?
(1) Charlie gets to the trailer in 55 minutes.
(2) Buster gets to the studio at the same time as Charlie gets to the trailer.

Solution: http://gmatclub.com/forum/devil-s-dozen ... l#p1063897
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Re: Devil's Dozen!!! [#permalink]

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New post 18 Aug 2014, 09:29
mahendru1992 wrote:
Bunuel wrote:
mahendru1992 wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:
Image
As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Answer: A.

Hi Bunuel, I don't understand how St 2 is clearly insufficient?
If we do it via venn diagram method, we know that a+b+c = 58
and we know from st 2, that a=32, so b=13, thus c=0.
But then from st 1 we get c = 13.
Where am I going wrong?


There might be a group of patients who has neither arachnophobia nor acrophobia. So, it should be a + b + c + neither = 58.

Okay but b+c=13, D can still be the answer? The question is asking us for b+c


The question asks for the number of patients who has acrophobia. Yellow box in matrix in my solution.

IF the number of patients with neither arachnophobia nor acrophobia is 0, then there will be 26 patients with acrophobia but IF the number of patients with neither arachnophobia nor acrophobia is 13, then there will be 13 patients with acrophobia.
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Re: Devil's Dozen!!! [#permalink]

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New post 13 Sep 2014, 21:57
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order?

(1) Set A consists of 12 even consecutive integers;
(2) n=5.

We should understand following two things:
1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.

2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).

Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.

Answer: B.
\\

Bunuel, pls clarify how can there be n! ways to select n out of a set which has got "k " distinct numbers?? I m stupified..
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Re: Devil's Dozen!!! [#permalink]

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New post 28 Sep 2014, 12:00
Bunuel wrote:
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia?

Tricky question.

(1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. Use double-set matrix:
Attachment:
Vertigo.png
As you can see # of patients who has acrophobia is 58-45=13. Sufficient.

(2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia. Clearly insufficient.

Answer: A.

Hi Bunuel,
I tried it with a different method but was unable to reach the answer.
58 = Arachnophobia(45) + acrophobia - Both -Neither -> Both and Neither are the same number.
13 = acrophobia - 2X
acrophobia = ?

So how do I continue from here? Or, where did I go wrong?
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Re: Devil's Dozen!!! [#permalink]

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New post 20 Oct 2014, 07:41
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


Hi,

I understand that for any 2 snakes,one would be a viper and the other would a cobra which makes 1:1 ratio.But the question is to find number of cobras as definite value right?And from option B we cannot get the number of cobras present in the box right?

Please correct me If I am wrong.

Thanks.
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New post 20 Oct 2014, 08:04
mvrravikanth wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


Hi,

I understand that for any 2 snakes,one would be a viper and the other would a cobra which makes 1:1 ratio.But the question is to find number of cobras as definite value right?And from option B we cannot get the number of cobras present in the box right?

Please correct me If I am wrong.

Thanks.


From (2) we get that there must be exactly one cobra in the box.
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Re: Devil's Dozen!!! [#permalink]

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New post 22 Oct 2014, 08:43
hey bunuel, are we assuming that y=1 in question 12? If so, why? I saw that A was not sufficient but chose C because I thought we could simply plug in different values for X (36, 81) so that y could be a number of different fractions
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New post 22 Oct 2014, 08:49
alexanthony813 wrote:
12. If x>0 and xy=z, what is the value of yz?

(1) \(x^2*y=3\). If \(x=1\) then \(y=z=3\) and \(yz=9\) but if \(x=3\) then \(y=\frac{1}{3}\), \(z=1\) and \(yz=\frac{1}{3}\). Not sufficient.

(2) \(\sqrt{x*y^2}=3\) --> \(x*y^2=9\) --> \((xy)*y=9\) --> since \(xy=z\) then: \(z*y=9\). Sufficient.

Answer: B.

hey bunuel, are we assuming that y=1 in question 12? If so, why? I saw that A was not sufficient but chose C because I thought we could simply plug in different values for X (36, 81) so that y could be a number of different fractions


We are not assuming that y is 1.

From (2) we have that \((xy)*y=9\) and from the stem we know that \(xy=z\). Now, simply substitute xy with z in \((xy)*y=9\) to get \(z*y=9\).

Hope it's clear.
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Re: Devil's Dozen!!! [#permalink]

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New post 06 Nov 2014, 13:15
Dear Bunuel,

I have a question to your solution on the following problem:

2. If n is a positive integer and p is a prime number, is p a factor of n!?

you specify for statement 2 the following sentence:

(2) p is a factor of (n+2)!/n! --> \frac{(n+2)!}{n!}=(n+1)(n+2) --> if n=2 then (n+1)(n+2)=12 and for p=2 then answer will be YES but for p=3 the answer will be NO. Not sufficient.

now consider (n+2)!/n!, this can be reduced to (n+1)(n+2), for n is odd, the expression is even, further for n is even, the expression is even again.
the statement specifies that p is a factor of an even number and is prime, thus p should be even and the only even prime number is 2. How did you consider p= 3? in your solution.

Thanks in advance,
Sant
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santorasantu wrote:
Dear Bunuel,

I have a question to your solution on the following problem:

2. If n is a positive integer and p is a prime number, is p a factor of n!?

you specify for statement 2 the following sentence:

(2) p is a factor of (n+2)!/n! --> \frac{(n+2)!}{n!}=(n+1)(n+2) --> if n=2 then (n+1)(n+2)=12 and for p=2 then answer will be YES but for p=3 the answer will be NO. Not sufficient.

now consider (n+2)!/n!, this can be reduced to (n+1)(n+2), for n is odd, the expression is even, further for n is even, the expression is even again.
the statement specifies that p is a factor of an even number and is prime, thus p should be even and the only even prime number is 2. How did you consider p= 3? in your solution.

Thanks in advance,
Sant


The red part is not correct. An even number can have an odd factor. For example, 12 is even and it has an odd factor 3.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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New post 07 Nov 2014, 04:28
Thanks Bunuel!!, +1 kudos, silly mistake from me.
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New post 08 Nov 2014, 06:37
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?

tried the above problem by algebra:

from stem:
15>= 11m+7b, multiply by 3, 45>=33m+21b.

from statement 1) 15>= 7m + 11b, multiply by 3, 45 >=21m + 33b.
add the stem inequality with the inequality from statement 1, 90>= 54m+54b. divide by 2, 45 > = 27m+27b.
therefore sufficient.

from statement 2). 15>= 10m+8b. multiply by 3, 45>=30m+24b.
add the inequality with the inequalty from stem, 90>=63m+45b. as dividing by 2 splits the numbers below 27 for brownies, we cannot come to a conclusion. therefore NSF.

Answer: A
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New post 23 Dec 2014, 21:29
Bunuel wrote:
9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers?

\(a+b+ab=34\) --> \((a+1)(b+1)=35\)



I was wondering if there was any other way of getting through this- i mean, without the classical form?
Thanks in advance
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New post 17 Mar 2015, 23:58
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


Hi
plz explain
Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper.
Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!!
Now what if dr are 10 cobras and 10 viper
Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!!
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Re: Devil's Dozen!!! [#permalink]

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New post 18 Mar 2015, 00:42
dpo28 wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


Hi
plz explain
Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper.
Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!!
Now what if dr are 10 cobras and 10 viper
Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!!


hi,
the reason there is only one cobra byu statement 2....
if you pick 2 snakes atleast one is viper...
say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper..
so this statement will not be true... but we know this statement is true, so we can have only one cobra..
and when we pick up two snakes it could be two vipers or one viper and one cobra...
hope it helped
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Re: Devil's Dozen!!! [#permalink]

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New post 18 Mar 2015, 01:39
chetan2u wrote:
dpo28 wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?

Quite tricky.

(1) There are total 99 snakes in Pandora's box. Clearly insufficient.

(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.

Answer: B.


Hi
plz explain
Question says atleast i viper and 1 cobra > This means that dr are atleat 1 cobra and 1 viper but it does not mean dr are just two of these. dr can be 10 snakes and 1 is viper and 1 cobra [according to thr question ATLEAST] or may be 6 cobra and 4 viper.
Now STATEMENT 2 states that for every i viper dr is 1 cobra..!!!
Now what if dr are 10 cobras and 10 viper
Question says atleast one, it can be more than 1 but minimum 1 cobra and i viper mus be dr..!!!


hi,
the reason there is only one cobra byu statement 2....
if you pick 2 snakes atleast one is viper...
say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper..
so this statement will not be true... but we know this statement is true, so we can have only one cobra..
and when we pick up two snakes it could be two vipers or one viper and one cobra...
hope it helped


Hi chetan2u
Thanks for the reply but am not able to get it
The questions asks the no of cobras right..!!
and for each viper dr is one cobra. Thn dr must be an equal no of both of them. But we cannot sat what's the number.
This is where am confused...!! ?
I know i sound silly but sometimes your are stuck at a point in some questions..!
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Re: Devil's Dozen!!! [#permalink]

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New post 18 Mar 2015, 02:41
dpo28 wrote:
[quote="chetan2u"

hi,
the reason there is only one cobra byu statement 2....
if you pick 2 snakes atleast one is viper...
say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper..
so this statement will not be true... but we know this statement is true, so we can have only one cobra..
and when we pick up two snakes it could be two vipers or one viper and one cobra...
hope it helped


Hi chetan2u
Thanks for the reply but am not able to get it
The questions asks the no of cobras right..!!
and for each viper dr is one cobra. Thn dr must be an equal no of both of them. But we cannot sat what's the number.
This is where am confused...!! ?
I know i sound silly but sometimes your are stuck at a point in some questions..![/quote]


hi, the question never tells us that there is one cobra for each viper....
it says there are atleast one cobra and one viper...
so it could be 1 cobra and 10 viper or 1 viper and 10 cobras or 10 viper and 100 cobras....

statement B the statement ll tells us atleast one is viper..
so for more than one cobra, atleast one viper in two snakes will not stand. it is possible to have both cobra if the cobras are more than 2 ... and this statement will not be true... but we know this statement is true, so we can have only one cobra..and when we pick up two snakes it could be two vipers or one viper and one cobra

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Re: Devil's Dozen!!! [#permalink]

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New post 18 Mar 2015, 03:02
chetan2u wrote:
dpo28 wrote:
[quote="chetan2u"

hi,
the reason there is only one cobra byu statement 2....
if you pick 2 snakes atleast one is viper...
say as u are saying there are two cobra and 8 viper.. then there can be a chance that these two cobras are picked up , but the statement ll tells us atleast one is viper..
so this statement will not be true... but we know this statement is true, so we can have only one cobra..
and when we pick up two snakes it could be two vipers or one viper and one cobra...
hope it helped


Hi chetan2u
Thanks for the reply but am not able to get it
The questions asks the no of cobras right..!!
and for each viper dr is one cobra. Thn dr must be an equal no of both of them. But we cannot sat what's the number.
This is where am confused...!! ?
I know i sound silly but sometimes your are stuck at a point in some questions..!



hi, the question never tells us that there is one cobra for each viper....
it says there are atleast one cobra and one viper...
so it could be 1 cobra and 10 viper or 1 viper and 10 cobras or 10 viper and 100 cobras....

statement B the statement ll tells us atleast one is viper..
so for more than one cobra, atleast one viper in two snakes will not stand. it is possible to have both cobra if the cobras are more than 2 ... and this statement will not be true... but we know this statement is true, so we can have only one cobra..and when we pick up two snakes it could be two vipers or one viper and one cobra
[/quote]

Hi chetan2u

Thnx for explaining
i think i get it now
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Re: Devil's Dozen!!! [#permalink]

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New post 12 Aug 2015, 01:11
Bunuel wrote:
3. If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\). Sufficient.

(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.

Answer: A.


I got the same but I considered X and Y as zero also which is a possibility. How can we ignore that. Please comment.
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Re: Devil's Dozen!!! [#permalink]

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New post 12 Aug 2015, 05:10
Awesome mind blowing questions!

Thanks Bunuel
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Re: Devil's Dozen!!! [#permalink]

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New post 16 Aug 2015, 12:18
mango1banana wrote:
Bunuel wrote:
3. If x and y are integers, is y an even integer?

(1) 4y^2+3x^2=x^4+y^4 --> rearrange: \(3x^2-x^4=y^4-4y^2\) --> \(x^2(3-x^2)=y^2(y^2-4)\). Notice that LHS is even for any value of \(x\): if \(x\) is odd then \(3-x^2=odd-odd=even\) and if \(x\) is even then the product is naturally even. So, \(y^2(y^2-4)\) is also even, but in order it to be even \(y\) must be even, since if \(y\) is odd then \(y^2(y^2-4)=odd*(odd-even)=odd*odd=odd\). Sufficient.

(2) y=4-x^2 --> if \(x=odd\) then \(y=even-odd=odd\) but if \(x=even\) then \(y=even-even=even\). Not sufficient.

Answer: A.


I got the same but I considered X and Y as zero also which is a possibility. How can we ignore that. Please comment.


0 is an even integer.
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Re: Devil's Dozen!!!   [#permalink] 16 Aug 2015, 12:18

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