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x^n = x^(n+2) for any integer n. Is it true that x > 0?
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21 Dec 2014, 20:41
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x^n = x^(n+2) for any integer n. Is it true that x > 0? (1) x = x^2  2 (2) 2x < x^5
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x^n = x^(n+2) for any integer n. Is it true that x > 0?
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21 Jan 2015, 02:59
x^n = x^(n+2) for any integer n. Is it true that x > 0? For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or 1. (1) x = x^2  2 > x^2  x  2 = 0 > (x  2)(x + 1) = 0 > x = 2 or x = 1. Since we know that x can only be 0, 1, or 1, then x = 1 < 0. Sufficient. (2) 2x < x^5 > from the possible values of x (0, 1, 1), only 1 satisfies 2x < x^5, thus x = 1 < 0. Sufficient. Answer: D.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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22 Dec 2014, 02:42
viktorija wrote: x^n=x^(n+2) for any integer n. Is it true that x>0?
(1) x=x^22 (2)2x<x^5 \(x^n = x^{n+2}\) st.1 x=x^22 now x^22x=0 x^22x+x2=0 x(x2)+1(x2)=0 (x2)(x+1)=0 x=2 or x=1 out of these values of x, only x=1 will satisfy the equation \(x^n = x^{n+2}\). hence we will reject x=2. thus only possible value of x=1. which answer the original question as false. Thus st.1 alone is sufficient. st.2 x^5>2x x(x^42)>0 this equation is satisfied by x=1, x=2,3,4... now out of these values of x, only x=1, satisfy the equation \(x^n = x^{n+2}\). hence x=1 . which answer the original question as false. thus st.2 alone is sufficient. hence answer must be D



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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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22 Dec 2014, 15:40
Hi viktorija, manpreetsingh86's explanation is spot on, so I won't rehash that here. The information given in the prompt involves a really specific Number Property. Knowing this Number Property would make solving the rest of the question considerably easier: We're told that x^n=x^(n+2) for ANY integer n. This equation offers a SERIOUS restriction; for it to hold true for ANY integer value of n, the value of X can only be one of 3 things: 1, 0 or 1. Taking THAT knowledge into Fact 1 and Fact 2 would allow you to work through them quickly and avoid any excess math. GMAT assassins aren't born, they're made, Rich
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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21 Jan 2015, 06:00
Here x can only be 0 OR 1 OR 1(if n=even) Statement 1 tells that x=2 or x=1. So x=1. Sufficient.
Statement 2 can holds only when x=1 and not when x=0 OR x=1. So x=1. Sufficient.
Answer is D.



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x^n = x^(n+2) for any integer n. Is it true that x > 0?
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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10 Jan 2016, 19:16
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. x^n = x^(n+2) for any integer n. Is it true that x > 0? (1) x = x^2 2 (2) 2x < x^5 When you modify the original condition and the question, it becomes x=1,0,1. There is 1 variable(x), which should match with the number equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely make D the answer. For 1), from x^2x2=0, (x2)(x+1)=0, x=2,1, only x=1 is possible, which is no and sufficien.t For 2), what satisfies 2x<x^5 is only x=1, which is no and sufficient. Therefore, the answer is D. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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06 Nov 2017, 21:27
Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or 1.
(1) x = x^2  2 > x^2  x  2 = 0 > (x  2)(x + 1) = 0 > x = 2 or x = 1. Since we know that x can only be 0, 1, or 1, then x = 1 < 0. Sufficient.
(2) 2x < x^5 > from the possible values of x (0, 1, 1), only 1 satisfies 2x < x^5, thus x = 1 < 0. Sufficient.
Answer: D. Hi Bunuel I have one question for x must be 0,1,1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/whatisthe ... 51809.htmlwhere 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or 1 Please help Thanks



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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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06 Nov 2017, 21:53
hellosanthosh2k2 wrote: Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or 1.
(1) x = x^2  2 > x^2  x  2 = 0 > (x  2)(x + 1) = 0 > x = 2 or x = 1. Since we know that x can only be 0, 1, or 1, then x = 1 < 0. Sufficient.
(2) 2x < x^5 > from the possible values of x (0, 1, 1), only 1 satisfies 2x < x^5, thus x = 1 < 0. Sufficient.
Answer: D. Hi Bunuel I have one question for x must be 0,1,1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/whatisthe ... 51809.htmlwhere 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or 1 Please help Thanks 0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part. Apart from these exception x = 0 is perfectly OK.
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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 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.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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06 Nov 2017, 23:04
Bunuel wrote: hellosanthosh2k2 wrote: Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or 1.
(1) x = x^2  2 > x^2  x  2 = 0 > (x  2)(x + 1) = 0 > x = 2 or x = 1. Since we know that x can only be 0, 1, or 1, then x = 1 < 0. Sufficient.
(2) 2x < x^5 > from the possible values of x (0, 1, 1), only 1 satisfies 2x < x^5, thus x = 1 < 0. Sufficient.
Answer: D. Hi Bunuel I have one question for x must be 0,1,1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat? I came across similiar question https://gmatclub.com/forum/whatisthe ... 51809.htmlwhere 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or 1 Please help Thanks 0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part. Apart from these exception x = 0 is perfectly OK. Thanks Bunuel for explanation



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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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09 Nov 2017, 08:58
viktorija wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
(1) x = x^2  2 (2) 2x < x^5 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. \(x^n = x^{n+2}\) \(x^{n+2}  x^n = 0\) \(x^n(x^21) = 0\) \(x^n(x+1)(x1) = 0\) \(x 1\) or \(x = 0\) or \(x = 1\) The original condition states \(x 1\) or \(x = 0\) or \(x = 1\). Condition 1) \(x = x^2  2\) \(x^2  x  2 = 0\) \((x+1)(x2) = 0\) \(x = 1\) or \(x = 2\) From the original condition, we have x = 1. Then answer is "No" This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t Condition 2) \(2x < x^5\) \(x^5  2x > 0\) \(x(x^4  2 ) > 0\) Only \(x = 1\) from the original condition satisfies this condition. Then answer is "No" This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t The answer is D. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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07 Mar 2018, 05:01
viktorija wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
(1) x = x^2  2 (2) 2x < x^5 hard problem from x^n<x^(n+2) we see that x =0 or 1 or 1 put these number into condition 1 and 2. we see that for both, only x=1 is good. so, each condition is ok D.



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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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02 Dec 2018, 15:27
This came up in a GMAT Focus test for me and I selected (B).
Pluggingin numbers and being cognizant of limitations given would have helped me solve this. I think other people fall into this trap as well judging by the difficulty.



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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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05 Dec 2018, 12:54
I do not know why no body has solved in this way \(x^n\)=\(x^{(n+2)}\) \(x^n\)=\(x^n\)*\(x^2\) so \(x^2\)=1 x=1 or +1 Statement 1: sufficient put 1 and +1 in the expression only 1 satisfy the equation so x=1 Statement 1: sufficient put 1 and +1 in the expression only 1 satisfy the equation so x=1 so both the statements gives the solution hence answer is D
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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29 Dec 2018, 05:04
Bunuel wrote: x^n = x^(n+2) for any integer n. Is it true that x > 0?
For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or 1. I know this is a silly doubt, but this is how I solved it. x^n = x^(n+2) So, x^n / x^(n+2) =1 x^(n  n  2) = 1 x^2 = 1 1/x^2 = 1 x^2 = 1 So, x = 1 or x = 1. With this approach, I am not getting x = 0.




Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?
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