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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x?

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dvinoth86
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   Updated on: 21 May 2018, 00:30
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If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

(1) y > x
(2) x < 0
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A

Originally posted by dvinoth86 on 23 Feb 2012, 20:27.
Last edited by Bunuel on 21 May 2018, 00:30, edited 5 times in total.
Edited the question and added the OA
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   24 Feb 2012, 00:18
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\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   24 Dec 2012, 06:32
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   16 Mar 2012, 00:04
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if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel wrote:
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   16 Mar 2012, 02:18
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devinawilliam83 wrote:
if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand

Bunuel wrote:
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.


The stem DOES NOT say that \(x=y\), it says: EITHER \(xy=0\) OR \(x=y\).

(1) says \(y > x\), which means that \(x=y\) is not possible, hence \(xy=0\).

Hope it's clear.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^-3,what is the value of x? [#permalink] New post   25 Dec 2012, 13:18
Bunuel wrote:
If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\) --> \(8xy^3 + 8x^3y=8*2x^2y^2=0\) --> reduce by 8 and re-arrange: \(xy^3+x^3y-2x^2y^2\) --> factor out xy: \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> \(xy=0\) or \(y-x=0\).

(1) y > x. Since \(y>x\), then \(y-x\neq{0}\), thus \(xy=0\). Sufficient.

(2) x < 0. Not sufficient.

Answer: A.


This solution does not answer to the question. Question asks about value xy.
if y>x then xy can't be equal 0. It can't be discussed.
Really I don't understand your solution please explain
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^-3,what is the value of x? [#permalink] New post   26 Dec 2012, 03:43
Expert Reply
akshin wrote:
Bunuel wrote:
If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?

\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\) --> \(8xy^3 + 8x^3y=8*2x^2y^2=0\) --> reduce by 8 and re-arrange: \(xy^3+x^3y-2x^2y^2\) --> factor out xy: \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> \(xy=0\) or \(y-x=0\).

(1) y > x. Since \(y>x\), then \(y-x\neq{0}\), thus \(xy=0\). Sufficient.

(2) x < 0. Not sufficient.

Answer: A.


This solution does not answer to the question. Question asks about value xy.
if y>x then xy can't be equal 0. It can't be discussed.
Really I don't understand your solution please explain


You should read a solution carefully.

From the stem we have that either \(xy=0\) or \(y-x=0\).

(1) says that y > x, so y - x > 0, which means that \(y-x\neq{0}\), thus \(xy=0\).
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   14 Nov 2014, 06:23
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Bunuel wrote:
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.


Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that.
i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end.
Why cant we cancel the x and y ? Please help.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   14 Nov 2014, 06:28
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Ankur9 wrote:
Bunuel wrote:
8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> reduce by 8: \(xy^3+x^3y=2x^2*y^2\) --> rearrange: \(xy^3+x^3y-2x^2*y^2=0\) --> factor out \(xy\): \(xy(y^2+x^2-2xy)=0\) --> \(xy(y-x)^2=0\) --> either \(xy=0\) or \(y-x=0\) (\(x=y\)).

(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.


Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that.
i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end.
Why cant we cancel the x and y ? Please help.


If you divide (reduce) 8x*y^3 + 8x^3*y = 2x^2*y^2/2^(-3), by xy you assume, with no ground for it, that xy does not equal to zero thus exclude a possible solution (notice that xy=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Check more tips on Algebra here: algebra-tips-and-hints-175003.html

Hope it helps.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   13 Aug 2015, 18:19
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I factored out the xy from both sides and that screwed up my equation. Note to self: dont factor out things when it may lead to division by zero!
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   22 Nov 2017, 09:24
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St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   22 Nov 2017, 09:34
Expert Reply
Barui wrote:
St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


You can always check your theoretical deduction by simple number plugging.

Fro (2) plug x = y = -1 or x = y = -2.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   30 Nov 2017, 06:00
Thanks

I understood now...


Bunuel wrote:
Barui wrote:
St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0.
and hence xy = 0. this sums to Answer D.

Where am I getting it wrong.

Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?

\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


You can always check your theoretical deduction by simple number plugging.

Fro (2) plug x = y = -1 or x = y = -2.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   26 Mar 2018, 00:28
This is an excellent question.
The trick is in simplyfing the question. Once you paraphrase the question into simple terms, it becomes,
xy*(x-y)^2=0.


8x∗y3+8x3∗y=2x2∗y22(−3)8x∗y3+8x3∗y=2x2∗y22(−3), What is xy?


(1) y > x ==>. Sufficient. Because, if y is greater than 1)x then x can be equal to Y.
so only one equation left. xy=0.

(2) x < 0. Clearly not sufficient.

Answer: A.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   13 Apr 2019, 04:55
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Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   28 Dec 2019, 21:01
strawberryjelly wrote:
Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?



I have the same question
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   03 Mar 2020, 07:43
strawberryjelly wrote:
Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^2-2xy)=0\); factors to \(xy(y-x)^2=0\) and not \(xy(x-y)^2=0\) I've written it out a lot and it seems to have the same product?


The bigger question is would it really matter in this case?

If it is xy(x-y)^2 =0 or it is xy(y-x)^2 =0 ???

Nope it will not matter, as eventually both the expressions will boil down to xy=0 or x=y. :) ;)

Posted from my mobile device
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   07 Feb 2023, 12:14
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Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


Hi Bunuel please advise the way we factor out 8 on both side of equation why are we not factoring out xy from both sides of equation

xy^3+x^3y=2x^2*y^2
Xy(Y^2+x^2)=2xy * xy
Y^2+x^2= 2xy
(Y-x)^2=0

I can see that we don't anything to related with xy from this but. I still curious to find out we cannot factor out xy

Bunuel

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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   07 Feb 2023, 12:27
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puneetfitness wrote:
Bunuel wrote:
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?


\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);

\(8xy^3+8x^3y=2x^2*y^2*8\);

Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);

Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);

Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);

\(xy(y-x)^2=0\);

Either \(xy=0\) or \(y-x=0\) (\(x=y\)).


(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.

(2) x < 0. Clearly not sufficient.

Answer: A.

Hope it's clear.


Hi Bunuel please advise the way we factor out 8 on both side of equation why are we not factoring out xy from both sides of equation

xy^3+x^3y=2x^2*y^2
Xy(Y^2+x^2)=2xy * xy
Y^2+x^2= 2xy
(Y-x)^2=0

I can see that we don't anything to related with xy from this but. I still curious to find out we cannot factor out xy

Bunuel

Posted from my mobile device


You can factor out xy but you cannot reduce by xy and as to why you cannot do that is explained here: https://gmatclub.com/forum/if-8xy-3-8x- ... l#p1442635

Hope it helps.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink] New post   19 Mar 2024, 23:59
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^-3, what is the value of x? [#permalink]
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