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Given that integer x>1 and x is not a perfect square,
[#permalink]
18 Sep 2019, 20:53

Question Stats:

Given that integer \(x>1\) and x is not a perfect square, which of the following could be a perfect square?

Indicate all possible choices.

a)\(4x\)

b)\(x^2−1\)

c)\(x^3+1\)

d)\(x^2−2x+1\)

e)\(√27x\)

f)\(17x\)

Indicate all possible choices.

a)\(4x\)

b)\(x^2−1\)

c)\(x^3+1\)

d)\(x^2−2x+1\)

e)\(√27x\)

f)\(17x\)

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
18 Sep 2019, 20:55

My question, the explanation in the practice tests are suggesting too long answer according to the limited time.

So, is there any shortcuts for this question?

My way was that I chose 3 as a number and substitute but couldn't find all the answers.

So, is there any shortcuts for this question?

My way was that I chose 3 as a number and substitute but couldn't find all the answers.

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
18 Sep 2019, 21:00

I couldn't understand this explanation

Factorize (x^3+1)=(x+1)(x^2−x+1) . For this to be a perfect square the two factors has to be equal, x+1=(x^2−x+1), which leads to x^2=2x . Therefore, if x is 2 the expression can be a perfect square, which is 9.

Also, why did they exclude A and include F. Since we can take 4 for X?

Factorize (x^3+1)=(x+1)(x^2−x+1) . For this to be a perfect square the two factors has to be equal, x+1=(x^2−x+1), which leads to x^2=2x . Therefore, if x is 2 the expression can be a perfect square, which is 9.

Also, why did they exclude A and include F. Since we can take 4 for X?

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
18 Sep 2019, 22:21

Given that integer x>1 and x is not a perfect square, which of the following could be a perfect square?

Can someone clear this up please? Does it mean that it's not necessary in all cases that the number should form a perfect square, but if it does in one instance, it has to be considered?

Can someone clear this up please? Does it mean that it's not necessary in all cases that the number should form a perfect square, but if it does in one instance, it has to be considered?

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
18 Sep 2019, 22:29

Could anyone help with the quickest solution here?

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
20 Sep 2019, 02:22

1

Expert Reply

First and foremost an important thing to notice here is that the stem says the word could be , which means that the possible answer can be a perfect square or NOT.

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
23 Oct 2019, 11:01

Carcass wrote:

First and foremost an important thing to notice here is that the stem says the word could be , which means that the possible answer can be a perfect square or NOT.

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

While plugging in value for A), why have you excluded 4*4 ?

GRE Prep Club Tests Editor

Joined: **13 May 2019 **

Affiliations: **Partner at MyGuru LLC.**

Posts: **186**

Location: **United States**

GMAT 1: **770 Q51 V44**

WE:**Education (Education)**

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
23 Oct 2019, 11:09

Expert Reply

3152gs wrote:

Carcass wrote:

First and foremost an important thing to notice here is that the stem says the word could be , which means that the possible answer can be a perfect square or NOT.

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

While plugging in value for A), why have you excluded 4*4 ?

Because 4 itself is precluded from being a possible value of x, since x cannot be a perfect square and of course 2 x 2 = 4, making 4 a perfect square.

Re: Given that integer x>1 and x is not a perfect square,
[#permalink]
23 Oct 2019, 11:15

MyGuruStefan wrote:

3152gs wrote:

Carcass wrote:

The wrong answer will never be a perfect square.

Moreover, the students sometimes pick the wrong and difficult way to solve it. Your goal is to pick the right answer, then the way you pick it does not matter. Of course, all this in a reasonable amount of time.

Pick number as the stem tells you.

A) 4*2=8; 4*3=12;4*5=20.......Never a perfect square

B) 2^2-1=3; 3^2-1=8;5^2-1=24.......Never.

C) 2^3+1=9 perfect square; 3^3+1=28 not ......could be. This is a good one.

Repeat for the rest and you will get the answer.

For instance, it is also a good practice to go a bit so far and in depth to find the solution

F) 17*2=NOT; 17*3=not.....it seems is not a good one. But what if \(17*17= \sqrt{289} = 17\) ....A good one

I took no more than a minute to do this process. On average is 1.30 seconds per quant question.

Take away: you must be flexible. Find a path, if it is no good switch to an alternative approach. Sometimes the answer is not straight. Sometimes there is a short cut, sometimes does not and you need to go through trial and errors.....thei sis the GRE.

NOT a straight highway.

Regards

While plugging in value for A), why have you excluded 4*4 ?

Because 4 itself is precluded from being a possible value of x, since x cannot be a perfect square and of course 2 x 2 = 4, making 4 a perfect square.

Ahh. Damn. Missed that little information. Thanks for the response.

gmatclubot

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