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

Question

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.

ENGRTOMBA2018 · Accepted Answer

Given x >0 and is an integer. GCD (x+m, x-m)=?

1. Statement 1 gives m=8 or 2.

This is not sufficient as follows:

Assume x is 16 .

Then we get x+8=24, x-8=8 : GCD =8

But GCD (16+2,16-2)=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,x-8)=1, take x=11, GCD(19,3)=1, similarly, x=15, GCD (23,7)=1 etc.

and GCD(x+2,x-2)=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.

chetan2u · Answer

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

reto · Answer

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.

ENGRTOMBA2018 · Answer

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.

reto · Answer

Yes you are right. I need to focus more

MBAHopeful99 · Answer

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.

GMATinsight · Answer

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

HappyQuakka · Answer

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

aserghe1 · Answer

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

tyildirim92 · Answer

Here is the official Veritasprep explanation. https://gmatclub.com/forum/divisibility ... l#p1589618

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

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