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

Statement 1: (m-2)(m-8)=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

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.

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

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

Hi chetan2u,
For statement 1, how did you determine that the two numbers will be coprime if the two numbers are odd?

Here is the official Veritasprep explanation.

https://gmatclub.com/forum/divisibility ... l#p1589618

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