February 20, 2019 February 20, 2019 08:00 PM EST 09:00 PM EST Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST February 21, 2019 February 21, 2019 10:00 PM PST 11:00 PM PST Kick off your 2019 GMAT prep with a free 7day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 25 Aug 2011
Posts: 137
Location: India
WE: Operations (Insurance)

If x and y are positive integers is y odd?
[#permalink]
Show Tags
04 Mar 2012, 22:42
Question Stats:
34% (02:24) correct 66% (02:26) wrong based on 931 sessions
HideShow timer Statistics
If x and y are positive integers is y odd? (1) \(\frac{(y+2)!}{x!} = odd\) (2) \(\frac{(y+2)!}{x!}\) is greater than 2
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 53020

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
04 Mar 2012, 23:37
If x and y are positive integers is y odd?(1) (y+2)!/x! = odd. Notice that \(\frac{(y+2)!}{x!}=odd\) can happen only in two cases: A. \((y+2)!=x!\) in this case \(\frac{(y+2)!}{x!}=1=odd\). For this case \(y\) can be even: \(y=2=even\) and \(x=4\): \(\frac{(y+2)!}{x!}=\frac{24}{24}=1=odd\); B. \(y=odd\) and \(x=y+1\), in this case \(\frac{(y+2)!}{(y+1)!}=y+2=odd\). For example, \(y=1=odd\) and \(x=y+1=2\): \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\) (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...). Not sufficient. (2) (y+2)!/x! is greater than 2. Clearly insufficient: consider \(y=1=odd\) and \(x=2\) OR \(y=2=even\) and \(x=1\). (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient. Answer: C. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 11 Jul 2012
Posts: 28

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
24 Sep 2012, 22:17
very good explanation



Intern
Joined: 06 Jan 2013
Posts: 2
GPA: 3.34
WE: Analyst (Computer Software)

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
19 Aug 2013, 07:08
Bunuel wrote: If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd. Notice that \(\frac{(y+2)!}{x!}=odd\) can happen only in two cases:
A. \((y+2)!=x!\) in this case \(\frac{(y+2)!}{x!}=1=odd\). For this case \(y\) can be even: \(y=2=even\) and \(x=4\): \(\frac{(y+2)!}{x!}=\frac{24}{24}=1=odd\);
B. \(y=odd\) and \(x=y+1\), in this case \(\frac{(y+2)!}{(y+1)!}=y+2=odd\). For example, \(y=1=odd\) and \(x=y+1=2\): \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\) (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...).
Not sufficient.
(2) (y+2)!/x! is greater than 2. Clearly insufficient: consider \(y=1=odd\) and \(x=2\) OR \(y=2=even\) and \(x=1\).
(1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.
Answer: C.
Hope it's clear. *********** Hi Bunuel, **B. y=odd and x=y+1, in this case \frac{(y+2)!}{(y+1)!}=y+2=odd. For example, y=1=odd and x=y+1=2: \frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...). ** You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient. Are you considering y=1 as odd? can you explain how y=1 satisfies the solution.



Math Expert
Joined: 02 Sep 2009
Posts: 53020

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
19 Aug 2013, 07:17
abhisheksriv85 wrote: Bunuel wrote: If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd. Notice that \(\frac{(y+2)!}{x!}=odd\) can happen only in two cases:
A. \((y+2)!=x!\) in this case \(\frac{(y+2)!}{x!}=1=odd\). For this case \(y\) can be even: \(y=2=even\) and \(x=4\): \(\frac{(y+2)!}{x!}=\frac{24}{24}=1=odd\);
B. \(y=odd\) and \(x=y+1\), in this case \(\frac{(y+2)!}{(y+1)!}=y+2=odd\). For example, \(y=1=odd\) and \(x=y+1=2\): \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\) (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...).
Not sufficient.
(2) (y+2)!/x! is greater than 2. Clearly insufficient: consider \(y=1=odd\) and \(x=2\) OR \(y=2=even\) and \(x=1\).
(1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.
Answer: C.
Hope it's clear. *********** Hi Bunuel, **B. y=odd and x=y+1, in this case \frac{(y+2)!}{(y+1)!}=y+2=odd. For example, y=1=odd and x=y+1=2: \frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...). ** You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient. Are you considering y=1 as odd? can you explain how y=1 satisfies the solution. Well 1 is an odd number. If \(y=1=odd\) and \(x=y+1=2\), then \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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 DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Jan 2013
Posts: 2
GPA: 3.34
WE: Analyst (Computer Software)

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
19 Aug 2013, 07:20
Bunuel wrote: abhisheksriv85 wrote: Bunuel wrote: If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd. Notice that \(\frac{(y+2)!}{x!}=odd\) can happen only in two cases:
A. \((y+2)!=x!\) in this case \(\frac{(y+2)!}{x!}=1=odd\). For this case \(y\) can be even: \(y=2=even\) and \(x=4\): \(\frac{(y+2)!}{x!}=\frac{24}{24}=1=odd\);
B. \(y=odd\) and \(x=y+1\), in this case \(\frac{(y+2)!}{(y+1)!}=y+2=odd\). For example, \(y=1=odd\) and \(x=y+1=2\): \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\) (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...).
Not sufficient.
(2) (y+2)!/x! is greater than 2. Clearly insufficient: consider \(y=1=odd\) and \(x=2\) OR \(y=2=even\) and \(x=1\).
(1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.
Answer: C.
Hope it's clear. *********** Hi Bunuel, **B. y=odd and x=y+1, in this case \frac{(y+2)!}{(y+1)!}=y+2=odd. For example, y=1=odd and x=y+1=2: \frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... > y+2=odd, 3, 5, 7, ... > y=odd, 1, 3, 5, ...). ** You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient. Are you considering y=1 as odd? can you explain how y=1 satisfies the solution. Well 1 is an odd number. If \(y=1=odd\) and \(x=y+1=2\), then \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\). Thanks Bunuel... got it!!



Intern
Joined: 13 Jan 2013
Posts: 4

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
10 Sep 2013, 11:38
devinawilliam83 wrote: If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd (2) (y+2)!/x! is greater than 2 From (1): assume: x = y + 2, so (y + 2)!/x! = 1 = odd y can odd or even, so (1) Insufficient. From (2): assume x = y + 1, so (y + 2)!/x! = y + 2 > 2 because y is a positive integer y can odd or even, so (2) Insufficient. From (1) + (2): (y + 2)!/x! > 2 and (y + 2)!/x! = odd Assume: (y + 2)!/x! = 3, so (y + 2)! =x!*(2k + 1) with k >=1 If x = y + 2, so 1 = 2k + 1 (wrong because k >=1) If x = y + 1, so y + 2 = 2k + 1, therefore y = odd If x = y, so (y + 1)(y + 2) = 2k + 1 (wrong) because the product of 2 consecutive integers is even. If x = y  1, so y(y + 1)(y + 2) = 2k + 1 (wrong) because the product is always even. Only one solution is right: If x = y + 1, so y + 2 = 2k + 1, therefore y = odd Answer is C.



Intern
Joined: 04 Sep 2013
Posts: 12

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
16 Sep 2013, 06:17
[quote="Bunuel"]If x and y are positive integers is y odd?
I am sorry i don't want to sound dumb, i just want to know how you assumed the cases above?? also what did you mean by:"(1)+(2) From (2) we cannot have case A, hence we have case B, which means . Sufficient."
Will appreciate your help.



Manager
Joined: 04 Oct 2013
Posts: 167
Concentration: Finance, Leadership
GMAT 1: 590 Q40 V30 GMAT 2: 730 Q49 V40
WE: Project Management (Entertainment and Sports)

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
23 Dec 2013, 03:30
skamran wrote: also what did you mean by:"(1)+(2) From (2) we cannot have case A, hence we have case B, which means . Sufficient."
Will appreciate your help.
Let's see if I can help you out. This question involves the knowledge of few concepts. 1) The division of two factorial expressions has to be odd. When does it happen? since a factorial say y! is the product amongst y, (y1), (y2) and so on, this sequence will always contain a multiple of two unless y is 1 or 0 (remember 1! =1 and 0! =1). In our case x and y are positive integers thus the lowest value y can assume is 1. If y=1 then (y+2)!=3!. We will not face the zero factorial case thus we can safely get rid of it. Back to our question: if we want to obtain an odd integer from \(\frac{(y+2)!}{x!}\) either 1. the two numbers have to be equal (entailing 1 as a quotient) EG y=1 > (y+2)!=3! x!=3!  result is 1 = odd integer. In this case y is odd y=2 (y+2)!=4! x!=4!  result is 1 = odd integer. In this case y is even. 2. or x! has to be one less than (y+2)! with y as an odd number. EG y=3 (y+2)!=5! x!=4!  5! upon 4! yields 5 and y must be ODD. Statement one is thus insufficient by itself.2) this statement doesn't tell us anything interesting besides that x! must be different from (y+2)! y can assume both an even and an odd value. Not sufficient1+2) Statement 2 conveys us that x! must be different from (y+2)! This said we can get rid of case one in our statement 1 analysis. Now we are sure that Y is an odd integer. Sufficient
_________________
learn the rules of the game, then play better than anyone else.



Manager
Joined: 25 Oct 2013
Posts: 146

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
03 Feb 2014, 04:54
(1): \(\frac{(y+2)!}{x!}\) is odd. Clearly \(x<=(y+2)\). Let x=y+2 then \(\frac{(y+2)!}{x!}\) =1 y can be even or odd. so (1) is insufficient(2): \(\frac{(y+2)!}{x!} > 2\) so x has to be less than y+2 Let x=y+1 then \(\frac{(y+2)!}{x!} > 2\) is \(y+2>2\) From this y can be any +ve integer. insufficient.(1)+(2) \(\frac{(y+2)!}{x!} > 2\) and also odd. Continuing the analysis from above, if y+2>2 and y+2 is odd, then y has to be odd. Next Let x=y then \(\frac{(y+2)!}{x!} > 2\) is (y+2)(y+1) > 2. Notice (y+2) & (y+1) are consecutive integers. their product will always be even. This contradicts (1)+(2) hence x cannot be equal to y. similarly x cannot be equal to y1 and so on. x has to be equal to y+1. and y must be odd. We have a concrete answer, y is odd. (1)+(2) is sufficient.
_________________
Click on Kudos if you liked the post!
Practice makes Perfect.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8888
Location: Pune, India

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
13 Feb 2014, 03:40
devinawilliam83 wrote: If x and y are positive integers is y odd?
(1) (y+2)!/x! = odd (2) (y+2)!/x! is greater than 2 It's pretty easy to see the answer is (C) if you understand the concept of factorials well. Note that a! = 1*2*3*4...(a1)*a So a! is the product of alternate odd and even numbers. a!/b! will be an integer only if a >= b (1) (y+2)!/x! = odd So y+2 >= x Two Cases: Case (i) If y+2 = x, (y+2)!/x! = 1 (odd). y could be odd or even. Case (ii) If y+2 > x, y+2 must be only one more than x. If y+2 is 2 more than x, you wil have two extra terms in the numerator and one of them will be even. So to ensure that (y+2)!/x! is odd, y+2 must be only 1 more than x and must be odd so all we are left in the numerator is (y+2) which must be odd. y must be odd. y could be odd or even so this statement alone is not sufficient. (2) (y+2)!/x! is greater than 2 This means y+2 > x. But we have no idea about whether y is odd or even. Not sufficient. Using both statements together, we know that only Case (ii) above is possible and hence y must be odd. Answer (C)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 18 Dec 2014
Posts: 2

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
19 Dec 2014, 17:20
i igore the possibility that (y+2)! is equal to x! so i choose A... a little difficult~~



Intern
Joined: 05 Dec 2014
Posts: 3

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
25 Dec 2014, 01:35
Case 1: (Y+2)!/X! = Odd If Y=3, then the above equation can only be odd when X= (Y+2)1 which is 4. So you get 5!/4! =5. If Y=2, and X=4, 4!/4! = 1 which is also odd. Hence, Insufficient.
Case 2:(Y+2)!/X! is greater than 2 This holds for multiple values of Y and X. Hence clearly, Insufficient.
Together, if it is greater than 2 then it means Y has to be odd. Hence C!
_________________ If my post is helpful/correct, consider giving Kudos



Manager
Joined: 22 Aug 2014
Posts: 149

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
09 Mar 2015, 08:55
Hey Bunuel, I solved the problem by plugging in values
combining statement 1 and 2,
10/2=5 and 9/3=3
in this case answer will be E. Please explain where I did wrong!



Math Expert
Joined: 02 Sep 2009
Posts: 53020

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
09 Mar 2015, 09:04



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13562
Location: United States (CA)

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
09 Mar 2015, 10:56
Hi ssriva2, You should doublecheck your math. Neither of those 2 examples 'fits' Fact 1. GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Current Student
Joined: 12 Aug 2015
Posts: 2621

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
22 Aug 2016, 05:16
Simply one of the best questions from MGMAT.!! Here we need to get if y is odd.even Statement 1 => here y can be odd or Y+2=x => insuff Statement 2 => insuff as y can be even or odd Statement one and two's combination gives us the evidence that y+2≠x hence y has to be odd. P.S =>Simply one of the best questions from MGMAT.!! CC abhimahna Go Ahead and tell me you didnt solve it in 60 sec. It took me forever to solve this
_________________
Give me a hell yeah ...!!!!!
MBA Dating: BSCHOOL with the MOST ATTRACTIVE Women MBA Recruiting: EMPLOYMENT AND SALARY STATISTICS AT TOP BSCHOOLS IN THE US! (2018) MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE!
Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs! STONECOLD's BRUTAL Mock Tests for GMATQuant(700+) AVERAGE GRE Scores At The Top Business Schools!



VP
Joined: 05 Mar 2015
Posts: 1001

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
20 Sep 2016, 18:30
sidoknowia wrote: If x and y are positive integers, is y odd?
(1) (y+2)! / x! is an odd integer. (2) (y+2)! / x! is greater than 2.
Didn't really understood OE. Can someone explain OA? (1) we can easily eliminate 1 as we can easily get 1 when we divide any unknowm variable and 1 is odd eg if y=2(EVEN) then (y+2)!=4! and we can easily choose x to be 4 for having desired odd integer for the given condition......NO similarly if y=3 then putting x=5 ...after division we get 1 (odd) but y= odd.......YES Having both scenario option is insuff.... (2) it says we get divisible value given is greater than 2... again we can assign any integer to Y and any integer for X such that x not equal y+2 then we get desired result.. Let y=3(odd) and x=2 then expression gives result>2.......Yes But if y=2(even) and x=1 then also expression gives result>2......NO so insuff... COmbining both we eliminate the possiblity of x=y+2 and as any number n!>1 always gives even number and any even number dividing any even number easily gives even number too.. but as per (1) we should have obtained a integer which is odd and it is possible in only one case when y=1 and x=2 Ans C



Board of Directors
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3631

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
20 Sep 2016, 23:53
stonecold wrote: Simply one of the best questions from MGMAT.!! Here we need to get if y is odd.even Statement 1 => here y can be odd or Y+2=x => insuff Statement 2 => insuff as y can be even or odd Statement one and two's combination gives us the evidence that y+2≠x hence y has to be odd. P.S =>Simply one of the best questions from MGMAT.!! CC abhimahna Go Ahead and tell me you didnt solve it in 60 sec. It took me forever to solve this stonecold : Just checked that you have mentioned my name here. I am not sure how but I missed this question. Now, coming back to the question. This time also I really took 58 secs to solve this question. Statement 1 : We are given (y+2)! / x ! = ODD. I took the value of ODD integer as 1,3,5. Found that I am getting y as even for y =1 and for others u as ODD. So, my mindset for the next move was ready. Statement 2 : We are given (y+2)! / x ! > 2 I know from the previous statement that for 3 and 5 it is ODD. I checked the value for 4 and 6 and found the value of y as even. Now I checked for a larger even and ODD integer, and found that for y = ODD and > 3, I will get my statements satisfied. Hence, I marked C. P.S : Going forward, please IM me such questions so that I can quickly check and respond.
_________________
My GMAT Story: From V21 to V40 My MBA Journey: My 10 years long MBA Dream My Secret Hacks: Best way to use GMATClub  Importance of an Error Log! Verbal Resources: All SC Resources at one place  All CR Resources at one place Blog: Subscribe to Question of the Day Blog GMAT Club Inbuilt Error Log Functionality  View More. New Visa Forum  Ask all your Visa Related Questions  here. New! Best Reply Functionality on GMAT Club! Find a bug in the new email templates and get rewarded with 2 weeks of GMATClub Tests for free Check our new About Us Page here.



Manager
Joined: 04 Oct 2015
Posts: 241
Location: Viet Nam
Concentration: Finance, Economics
GPA: 3.56

Re: If x and y are positive integers is y odd?
[#permalink]
Show Tags
14 Jul 2017, 00:15
In statement 1, pay attention to the SPECIAL CASE in which (y+2)!=x!
_________________
Do not pray for an easy life, pray for the strength to endure a difficult one  Bruce Lee




Re: If x and y are positive integers is y odd?
[#permalink]
14 Jul 2017, 00:15



Go to page
1 2
Next
[ 21 posts ]



