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Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 02:28
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Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 04:40
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Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Given x >0 and is an integer. GCD (x+m, xm)=? 1. Statement 1 gives m=8 or 2. This is not sufficient as follows: Assume x is 16 . Then we get x+8=24, x8=8 : GCD =8 But GCD (16+2,162)=2 Thus not sufficient 2. Statement 2 tells x+26= prime Thus x+26 = 29,31,37,41,43... giving x=3,5,11,15,17... Again, this is not sufficient as m can be any value. Combining, we get m=2 or 8 and x=3,5,11,15,17... we see that GCD (x+8,x8)=1, take x=11, GCD(19,3)=1, similarly, x=15, GCD (23,7)=1 etc. and GCD(x+2,x2)=1 , take x=3, GCD(5,1)=1, similarly, x=15, GCD (17,13)=1 etc. GCD =1 is the only possible soution and thus C is the correct answer.
Last edited by ENGRTOMBA2018 on 07 Jul 2015, 10:19, edited 2 times in total.



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Re: Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 08:39
Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Statement 1: (m2)(m8)=0 >> m can be 8 or 2 > insufficient Statement 2: x+26 = prime >> x can be 1, or 3 or 5... Together: We have two statements which do not complete each other. No further information can be extracted. Answer E.
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Re: Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 08:44
reto wrote: Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Statement 1: (m2)(m8)=0 >> m can be 8 or 2 > insufficient Statement 2: x+26 = prime >> x can be 1, or 3 or 5... Together: We have two statements which do not complete each other. No further information can be extracted. Answer E. Reto, per statement 2, how can x be 1 when x+26= prime? This will give you 1+26=27 and 27 is not a prime number.



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Re: Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 08:47
Engr2012 wrote: reto wrote: Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Statement 1: (m2)(m8)=0 >> m can be 8 or 2 > insufficient Statement 2: x+26 = prime >> x can be 1, or 3 or 5... Together: We have two statements which do not complete each other. No further information can be extracted. Answer E. Reto, per statement 2, how can x be 1 when x+26= prime? This will give you 1+26=27 and 27 is not a prime number. Yes you are right. I need to focus more
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Re: Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 09:00
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Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Hi, lets look at the statements.. 1) first statement gives value of m as 8 or 2... so the two numbers become x+8 and x8 or x+2 and x2.. this tells us that either the difference in number is 16 or it is 4 and the two numbers are either both even or both odd... what does this tell us? it tells us that the numbers will have some common even GCD if x is even or the two numbers will be coprime if the two numbers are odd, giving GCD as 1... INSUFF 2) x+26 is a prime number and x is a +ive int.. x has to be odd.. in suff combined we know X is odd and as explained in stat 1 , the GCD will be 1.. suff C
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Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 09:53
Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
1) Factor this equation, and you find out that m is equal to 2 or 8. m^2 – 10m + 16 = 0 (m8)(m2)=0 m=8 m=2 This doesn't help, since you still don't know x. Insufficient.2) From this, you can see that x can be 3, 5, 11 etc. x must be an odd number. However, this doesn't say what m can be. Insufficient.Using both of these together in (x+m) and (xm): Using 3 and 2, you get 5 and 1. The GCF is 1. Using 5 and 2, you get 7 and 3. The GCF is 1. Using 3 and 8, you get 11 and 5. The GCF is 1. Using 5 and 8, you get 13 and 3. The GCF is 1. Remember to use ZONEF (Zeros, One, Negatives, Extreme, and Fractions) (only extreme applies). I used 101 as a prime, so x is 75. Using 75 and 2, you get 77 and 73. The GCF is 1. Using 75 and 8, you get 83 and 67. The GCF is 1. Answer is C.
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Re: Given that x is a positive integer, what is the greatest common diviso [#permalink]
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07 Jul 2015, 10:00
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Bunuel wrote: Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?
(1) m^2 – 10m + 16 = 0 (2) x + 26 is a prime number.
Kudos for a correct solution. Given: x is a Positive integer and m is an Integers and x>m (because xm is a positive iteger)Question : Greatest common divisor (GCD) of the two positive integers, (x+m) and (xm)?Statement 1:m^2 – 10m + 16 = 0m^2 – 8m  2m + 16 = 0 i.e. m(m – 8)  2(m  8) = 0 i.e. (m8) (m2) = 0 i.e. m = 2 or 8 @x=9 and m=2, (xm) = 7 and (x+m) = 11 i.e. GCD = 1 @x=8 and m=2, (xm) = 6 and (x+m) = 10 i.e. GCD = 2 NOT SUFFICIENTStatement 2:x + 26 is a prime number.But m is unknown so NOT SUFFICIENTe.g.@x+26=29(Prime), i.e. x=3, and m=2, (xm) = 1 and (x+m) = 5 i.e. GCD = 1 and @x+26=31(Prime), i.e. x=5, and m=3, (xm) = 2 and (x+m) = 8 i.e. GCD = 3 Combining the Two statementsIf x+26=(Prime) then x MUST be and ODD number odd+m(2or8) = ODD oddm(2or8) = ODD i.e. Diffience between (x+m) and (xm) will be = 2m = 4 or 16 two odd number at the difference of 4 or 16 will always have GCD = 1 i.e. SUFFICIENTAnswer: Option C
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Given that x is a positive integer, what is the greatest common diviso [#permalink]
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08 Sep 2017, 00:19
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Hi, I got this question wrong, albeit it took me nearly 5 minutes to do it.. As for Stmt 2, is this a trial and error thing, or is there a number property I am not aware of? Here's how I approached it: Stmt 1 M^2  10m + 16 = 0 At first, I tried to factor it and realized that I can't  which was a useless attempt  wasted some time here. Then I solved for m by trial and error, more or less. It was pretty obvious that m was 2, so not much time wasted here. m^2  10m = 16 So I came up with (x+2) & (x2) but this is insufficient since when x is 26, it's GCD ( 28, 24) = 4 but when x is 31, it's GCD (33,29) = 1 Stmt 2 x+26 is a prime number. so potential values for x are (29, 31, 37, 41, 43, 47....)  4, making it (3, 5, 11, 15, 17, 21...) Which, by itself, is insufficient because if m = 5, when x=11, GCD (16,6) = 2 when x=15, GCD (20, 10) = 10 Stmt 1 + Stmt 2 I made a (quick) table of x = 3 GCD (5, 1) = 1 x = 5 GCD (7, 3) = 1 x = 11 GCD (13, 9) = 1 x = 15 GCD (17,13) = 1 x = 17 GCD (19, 15) = 1 so I concluded that it's C. I feel like my trial and error was a little excessive, but I was terrified that I didn't want to be too quick to judge and miss a question I could've gotten wrong




Given that x is a positive integer, what is the greatest common diviso
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