w, x, y, and z are integers. If w >x>y>z>0, is y a common : GMAT Data Sufficiency (DS)
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# w, x, y, and z are integers. If w >x>y>z>0, is y a common

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26 Jan 2012, 21:19
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w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

[Reveal] Spoiler:
Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Sep 2012, 02:15, edited 3 times in total.
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26 Jan 2012, 22:33
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enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.
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10 Sep 2012, 22:26
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Initially I went for B but after carefully examining bunuel's solution, though it was a bit tough one, I arrived on D.
Moreover I went a little further and deduced the values of w,x,y,z as 4,3,2,1.
Please let me know if these values are correct.
Thank you.
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10 Jan 2013, 09:27
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Bunuel wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.

Bunuel,
How to start in this type of question? In both the statement i arrived up to w=x+1 and w-y=2 , but after that I lost my way...
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26 May 2013, 20:45
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enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

1 ) w/x = 1/z + 1/x
since W is greater than x , w/x should be greater than 1
hence 1/z +1/x should also be greater than 1. since x and z are integers grater than 0 , and z<x , the only way , 1/z +1/x can be >1 is if z=1
hence equation becomes - w/x= 1/1 + 1/x ======> w-1=x ..
hence x and w are consecutive integers and since y is not 1 and is >1 ( since z is 1) , the answer is NO
Sufficient

2) w^2-wy-2w=0
w(w-y-2) =0
since w != 0 w=y+2 .. w, x and y are consecutive integers . Same as statement 1 , the answer is NO.
Suff

-Jyothi
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26 May 2013, 23:21
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

From F.S 1, we know that $$\frac{w}{x}= z^{-1}+x^{-1}$$ --> $$w = \frac{x}{z}+1$$

or $$w-1 = \frac{x}{z}$$.

On the number line,we can have this arrangement ---x ----- x/z ---- w ; where $$\frac{x}{z}$$ and w are consecutive integers.

Note that we can not have ---x/z---x---w as because there cannot be an integer(x) between 2 consecutive integers.

Thus, we have $$\frac{x}{z}>x$$ --> z<1. However, this is not possible as z is at-least 1. Thus, the only solution is when $$\frac{x}{z}$$ and x coincide i.e. $$\frac{x}{z}$$ = x --> z=1.

As w and x are consecutive integers, y(which is not equal to 1) can never be a divisor for both x and w. Sufficient.

From F. S 2, we know that as$$w\neq{0}$$ w = y+2. Thus, on the number line,

--y ---(y+1)----w. Thus, there is only one integer between y and w and as y<x<w, we have x = y+1. Just as above, y can't be a divisor for both w and x. Sufficient.

D.
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22 Jul 2013, 05:29
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enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

[Reveal] Spoiler:
Folks - started to solving it like this, but got stuck after a while. Can someone please help?

Basically this question is asking whether y is factor of both w & x?

Considering Statement 1

w/x=1/z+1/x. Since w>x , the quotient w/x must be greater than 1. For example 17/6 = 2 5/6

Since w/x is greater than 1 then 1/z + 1/x must be greater than 1 as well. For 1/z+1/x to be greater than 1 both z and x has to be 1. x cannot be equal to 1 because of the condition w > x > y > z > 0, therefore, z must be equal to 1. We can now re write the equation as w/x = 1+1/x which will give us w = x+1. If w= x+1 then mean x and w are consecutive integers and therefore no number other than 1 can be a factor of both.

I got stuck after this. But from my guess work I think this statement looks sufficient. I know not the right way of doing MATHS, therefore can someone please help?

Considering Statement 2

w^2-wy-2w = 0

If we factor above into, that will give w(w-y-2) = 0. This means that either w = 0 or w-y-2 = 0. w cannot be zero because w > x > y > z > 0, therefore w-y-2 = 0 or w = y+2. Since x falls between y and w we can tell that x, y and w are consecutive integers and therefore no number other than 1 can be a factor of both. Since y >1 it cannot be a factor of both w & x. Therefore this statement is sufficient to answer the question.

Awesome sum, here's my small bit

$$\frac{w}{x} = \frac{1}{z} +\frac{1}{x}$$ , Given

$$\frac{w}{x} - \frac{1}{x}= \frac{1}{z}$$, bringing terms with same denominator to LHS

$$\frac{w-1}{x}=\frac{1}{z}$$, simplification

now if z>1 then RHS is a fraction so LHS is a fraction too.

Now if LHS is a fraction then x>w-1

lets look at this statement x>w-1, since w and x are positive and $$x \neq w$$ hence x>w-1 is never possible.
Lets take a few positive integers such that w>x and test this. e.g. x = 6 and w =7 , 6 > 6 not possible, x= 6 w=12 , 6>11 not possible , hence x> w-1 not possible if w>x and positive.So LHS cannot be a fraction and hence RHS also cannot be a fraction so Z cannot be > 1

if Z>1 not possible then Z=1 and as $$0 < z \not> 1$$ ( z is greater than 0 but not greater than 1 and z is an integer so z can be 1 only)

so if z = 1 then from $$\frac{w-1}{x}=\frac{1}{z}$$ we have w-1 = x , which tells us that x is one less than w or x and w are consecutive integers. Hence they are co prime . Therefore they cannot have a common factor other than 1, but we know z is 1 so y cannot be 1, because y>z , Hence y is not a common factor for w and x as y is not 1 but >1.
Sufficient.
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Last edited by stne on 08 Dec 2016, 06:21, edited 2 times in total.
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18 Nov 2013, 05:28
This was a really tough question. Totally stumped me : (
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16 May 2014, 13:51
Another easy approach is to introduce y in the equation by inserting...w=py and x=qy
The equation formed after putting in the values in I is not valid & hence w & x do not have a common factor x ...as the equation cannot be untrue
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16 May 2014, 23:49
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Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.
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enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

(1) w = x/z + 1 -> x = kz and w = k + 1 ( with k is an positive integer)
since w > x -> k+1> kz -> k(z-1) < 1
because of k and z are positive integers, z must equal 1 -> x = k , and w = k + 1 -> y is not a common divisor of w and x.

(2) w = y + 2 -> x = y + 1 ..... -> y is not a common divisor of w and x.

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03 Sep 2014, 13:54
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

@PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z"
@Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?
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03 Sep 2014, 17:35
ronr34 wrote:
PiyushK wrote:
Another concept, two consecutive numbers are co-prime thus they have only 1 as a common factor.

1> After reducing option 1 we have Z = X/(W-1) for Z to be an integer (1) and X<W, W and Z must be consecutive numbers.
And two consecutive numbers have only one common factor that is 1, bcz Y>Z>0 and Z is also an integer then Y can not be 1. Therefore, Y can not be a common divisor for both.

2> After reducing option 2.
W=Y+2 it means W,X,Y all are consecutive integers and co prime to each other. thus, again Y can not be common divisor for W and X and neither it is equal to one as restricted by given data.

Therefore, Ans D.

@PiyushK - I think you meant "W and X must be consecutive numbers"... not "W and Z"
@Bunuel - In your solution, for st. 1, you wrote
Bunuel wrote:
since w>x>z then z=1 (if z>1 then x>w which contradicts given condition)
. How did you reach this conclusion?

w > x > y > z > 0, is y a common divisor of w and x?

We have that z(w-1)=x and w>x>z>0. If z>1, say if z=2, then 2(w-1)=x. Now tell me can in this case x be less than w?
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19 Dec 2014, 01:59
Indeed a fantastic question !! Esp. was totally stumped out by statement A. Keep throwing some more like this
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08 Mar 2015, 06:03
Bunuel wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.

Can someone pls explain how Z =1 in case 1?

Thanks.
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08 Mar 2015, 06:45
samichange wrote:
Bunuel wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.

Can someone pls explain how Z =1 in case 1?

Thanks.

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08 Mar 2015, 07:35
I did some more thinking and tried to make sense using the following argument.

After solving ( 1) , we get w-1 = x/z

w and w-1 are consecutive integers, so we can have

w> w-1 or x/z >= x > y > z ( since w-1 = x/z = integer ) i.e w-1 = x/z > x or w-1 =x/z = x

but w-1 = x/z > x is not possible because x divided by z will be less than or equal to x.

so w-1 =x/z = x ----> z=1

so the only way it can be true is when z =1

hence w-1 =x and z=1

Is the above approach correct???
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03 Jun 2015, 21:35
Bunuel wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.

Hi bunuel :
my logic to option "A" is this :

reducing equ we get X/ (w-1) = Z ( integer) --> means w-1 is a factor of X.
Now W>X and w-1 to be factor of X then W& X are consecutive integers. then only this is possible else W-1 will never be factor of X.

Is this logic right. I am thinking it's same logic you explained above but the explanation language is different language.
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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26 Jul 2015, 11:52
Bunuel wrote:
enigma123 wrote:
w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z^-1+x^-1

(2) w^2-wy-2w=0

w, x, y, and z are integers. If w > x > y > z > 0, is y a common divisor of w and x?

(1) w/x= z ^−1 + x^−1 -> $$\frac{w}{x}=\frac{1}{z}+\frac{1}{x}$$ --> multiply both sides by $$xz$$ --> $$wz=x+z$$ --> $$z(w-1)=x$$ --> since $$w>x>z$$ then $$z=1$$ (if $$z>1$$ then $$x>w$$ which contradicts given condition) --> $$w=x+1$$ --> $$w$$ and $$x$$ are consecutive integers. Now, two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. (For example 20 and 21 are consecutive integers, thus only common factor they share is 1). Thus $$w$$ and $$x$$ don not share any common factor more than 1 and as $$y>1$$ (from $$y>z>0$$) then $$y$$ is not a common factor of $$w$$ and $$x$$. Sufficient.

(2) w^2 – wy – 2w = 0 --> $$w(w-y-2)=0$$, since $$w>0$$ then: $$w-y-2=0$$ --> $$w-2=y$$ -->s $$w>x>w-2$$ (substituting $$y$$ in the given inequality) --> $$x=w-1$$. The same as above. Sufficient.

Hope it helps.

Hello Bunel,
For statement 1
z(w-1)=x
=> either w-1=x or z=x
but z cannot be equal to x as x is greater than z

so w-1 = x
is this not right ?
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common [#permalink]

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28 Jul 2015, 10:46
1
KUDOS
Assuming w=4, x=3, y=2, z=1

St1:

w/x = 1/z + 1/x
= x+z/xz
w = x (x+z)/xz
w = x+z/z
wz = x+z

Using the above values, Y is not a divisor of W & X. Sufficient.

St2 :

w^2-wy-2w=0
w^2= w(y+2)

Hence, w= y + 2

using the above values, Y is not divisible of W & X. Sufficient.

Final ans D)
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Re: w, x, y, and z are integers. If w >x>y>z>0, is y a common   [#permalink] 28 Jul 2015, 10:46

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