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Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

(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 x-8 or x+2 and x-2.. 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 (x-m)?

(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 (m-8)(m-2)=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 (x-m): 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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Please kudos if you found this post helpful. I am trying to unlock the tests

Given that x is a positive integer, what is the greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

(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 x-m is a positive iteger)

Question : Greatest common divisor (GCD) of the two positive integers, (x+m) and (x-m)?

Statement 1:m^2 – 10m + 16 = 0 m^2 – 8m - 2m + 16 = 0 i.e. m(m – 8) - 2(m - 8) = 0 i.e. (m-8) (m-2) = 0 i.e. m = 2 or 8 @x=9 and m=2, (x-m) = 7 and (x+m) = 11 i.e. GCD = 1 @x=8 and m=2, (x-m) = 6 and (x+m) = 10 i.e. GCD = 2

NOT SUFFICIENT

Statement 2:x + 26 is a prime number. But m is unknown so NOT SUFFICIENT e.g.@x+26=29(Prime), i.e. x=3, and m=2, (x-m) = 1 and (x+m) = 5 i.e. GCD = 1 and @x+26=31(Prime), i.e. x=5, and m=3, (x-m) = 2 and (x+m) = 8 i.e. GCD = 3

Combining the Two statements If x+26=(Prime) then x MUST be and ODD number odd+m(2or8) = ODD odd-m(2or8) = ODD

i.e. Diffience between (x+m) and (x-m) will be = 2m = 4 or 16 two odd number at the difference of 4 or 16 will always have GCD = 1 i.e. SUFFICIENT

Answer: 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

1

This post received KUDOS

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) & (x-2) 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