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The integer v is greater than 1. If v is the square of an i
[#permalink]
19 Jan 2016, 08:04

2

1

Expert Reply

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Question Stats:

The integer \(v\) is greater than 1. If v is the square of an integer, which of the following numbers must also be the square of an integer?

Indicate all such numbers.

A) \(81v\)

B) \(25v + 10 \sqrt{v} + 1\)

C) \(4v^2 + 4 \sqrt{v} + 1\)

_________________

Indicate all such numbers.

A) \(81v\)

B) \(25v + 10 \sqrt{v} + 1\)

C) \(4v^2 + 4 \sqrt{v} + 1\)

Practice Questions

Question: 16

Page: 342

Difficulty: medium/hard

Question: 16

Page: 342

Difficulty: medium/hard

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Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
03 Mar 2018, 07:20

3

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Bookmarks

Carcass wrote:

The integer v is greater than 1. If v is the square of an integer, which of the following numbers must also be the square of an integer?

Indicate all such numbers.

A) 81v

B) 25v + 10√v + 1

C) 4v² + 4√v + 1

Indicate all such numbers.

A) 81v

B) 25v + 10√v + 1

C) 4v² + 4√v + 1

We're told that v is the square of an integer

Let's say v = k², where k is a positive integer

This means that √v = k

Now let's examine the answer choice....

A) 81v = (81)(k²)

= (9)(9)(k)(k)

= (9k)(9k)

= (9k)²

Since k is an integer, we can be certain that 9k is an integer

So, 81v IS the square of an integer

B) 25v + 10√v + 1 = 25k² + 10k + 1

= (5k + 1)(5k + 1)

= (5k + 1)²

Since k is an integer, we can be certain that 5k is an integer, and if 5k is an integer, we can be certain that 5k + 1

So, 25v + 10√v + 1 IS the square of an integer

C) 4v² + 4√v + 1 = 4(k²)² + 4k + 1

= 4k⁴ + 4k + 1

This expression cannot be factored into the form (some integer)²

To verify this, let's replace k with some integer and see what we get.

Let k = 2

If k = 2, then 4k⁴ + 4k + 1 = 4(2)⁴ + 4(2) + 1

= 64 + 8 + 1

= 73

Since 73 is NOT the square of an integer, we CANNOT conclude that 25v + 10√v + 1 is the square of an integer

Answer: A and B

Cheers,

Brent

_________________

Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
14 Aug 2021, 06:14

2

Picking numbers isn't a bad idea here. Let's pick an easy square, like 4.

A: If 81v is a perfect square, we should be able to take its square root and get a whole number.

Sqrt(81*4) =

(81 * 4)^0.5 =

81^0.5 * 4^0.5 =

9 * 2 =

18.

Yup.

B: 25*4 + 10*2 + 1 = 100 + 20 + 1 = 121.

Sqrt(121) = 11.

Yup.

C: 4*16 + 4*2 + 1 = 64 + 8 + 1 = 73

Sqrt(73) = not a whole number

We can tell this because the nearest perfect squares are 64 [8^2] and 81 [9^2] and this guy is between the two.

Nope!

Hence, AB.

A: If 81v is a perfect square, we should be able to take its square root and get a whole number.

Sqrt(81*4) =

(81 * 4)^0.5 =

81^0.5 * 4^0.5 =

9 * 2 =

18.

Yup.

B: 25*4 + 10*2 + 1 = 100 + 20 + 1 = 121.

Sqrt(121) = 11.

Yup.

C: 4*16 + 4*2 + 1 = 64 + 8 + 1 = 73

Sqrt(73) = not a whole number

We can tell this because the nearest perfect squares are 64 [8^2] and 81 [9^2] and this guy is between the two.

Nope!

Hence, AB.

Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
19 Jan 2016, 08:22

3

Expert Reply

Solution

Here is important to notice that \(V\) is the square of an integer. Which means that, for instance, \(2^2=4\), si if V is 4, then \(\sqrt{4}\) is 2

Using this fact we can evaluate each answer choice in order to get whether they are the square of an integer as well

A) 81v= \(\sqrt{81*4}\) = 9*2 =18 So A is true

B) It is in the form of \((a + b)^2\); as such we do have 5 \(\sqrt{v}\) + 1. So B is the square of an integer as well

C) Pick \(\sqrt{v}\)\(=4=2\) that is the square of an integer. Substitution: \(64+8+1=73\) which is NOT the square of an integer. So C is not true

The answer is \(A and B\)

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Here is important to notice that \(V\) is the square of an integer. Which means that, for instance, \(2^2=4\), si if V is 4, then \(\sqrt{4}\) is 2

Using this fact we can evaluate each answer choice in order to get whether they are the square of an integer as well

A) 81v= \(\sqrt{81*4}\) = 9*2 =18 So A is true

B) It is in the form of \((a + b)^2\); as such we do have 5 \(\sqrt{v}\) + 1. So B is the square of an integer as well

C) Pick \(\sqrt{v}\)\(=4=2\) that is the square of an integer. Substitution: \(64+8+1=73\) which is NOT the square of an integer. So C is not true

The answer is \(A and B\)

_________________

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GRE: All you do need to know about the GRE Test | GRE Prep Club for the GRE Exam - The Complete FAQ

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Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
03 Sep 2018, 14:36

1

Through trial and error we can solve the problem.

As V is a square number, we can put a square integer on the options.

From 1st option we get,

a) 81V . Let's take v = 4,9

we get, 324 and 729 which are the squares of 18 and 27. So A is a answer.

Similarly from option b we will find the similar answer.

So answer option A, B

As V is a square number, we can put a square integer on the options.

From 1st option we get,

a) 81V . Let's take v = 4,9

we get, 324 and 729 which are the squares of 18 and 27. So A is a answer.

Similarly from option b we will find the similar answer.

So answer option A, B

Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
14 Dec 2018, 18:41

Carcass wrote:

Solution

Here is important to notice that \(V\) is the square of an integer. Which means that, for instance, \(2^2=4\), si if V is 4, then \(\sqrt{4}\) is 2

Using this fact we can evaluate each answer choice in order to get whether they are the square of an integer as well

A) 81v= \(\sqrt{81*4}\) = 9*2 =18 So A is true

B) It is in the form of \((a + b)^2\); as such we do have 5 \(\sqrt{v}\) + 1. So B is the square of an integer as well

C) Pick \(\sqrt{v}\)\(=4=2\) that is the square of an integer. Substitution: \(64+8+1=73\) which is NOT the square of an integer. So C is not true

The answer is \(A and B\)

Here is important to notice that \(V\) is the square of an integer. Which means that, for instance, \(2^2=4\), si if V is 4, then \(\sqrt{4}\) is 2

Using this fact we can evaluate each answer choice in order to get whether they are the square of an integer as well

A) 81v= \(\sqrt{81*4}\) = 9*2 =18 So A is true

B) It is in the form of \((a + b)^2\); as such we do have 5 \(\sqrt{v}\) + 1. So B is the square of an integer as well

C) Pick \(\sqrt{v}\)\(=4=2\) that is the square of an integer. Substitution: \(64+8+1=73\) which is NOT the square of an integer. So C is not true

The answer is \(A and B\)

The question asks what MUST be true, not what may be true. You can only use an example to eliminate choices, not to prove something MUST be true.

Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
14 Dec 2018, 18:46

1

Carcass wrote:

The integer v is greater than 1. If v is the square of an integer, which of the following numbers must also be the square of an integer?

Indicate all such numbers.

A) 81v

B) 25v + 10√v + 1

C) 4v² + 4√v + 1

Indicate all such numbers.

A) 81v

B) 25v + 10√v + 1

C) 4v² + 4√v + 1

Practice Questions

Question: 16

Page: 342

Difficulty: medium/hard

Question: 16

Page: 342

Difficulty: medium/hard

Show: :: OA

A and B

V being the square of an integer can be represented mathematically as

v = x^2

We look to see if the options can be put in the form of an integer squared.

A) 81v

81v = x^2 works, we can square root both sides and get

9sqrt(v) = x.

But v is the square of an integer so sqrt(v) gives you an integer. Thus x is an integer.

B) 25v + 10√v + 1

This can be factored into

( 5*sqrt(v) + 1 )^2

again v is the square of an integer so sqrt(v) gives you an integer. An integer times an integer plus an integer squared is of course an integer.

C) 4v² + 4√v + 1

This cannot be factored as the square of an integer. That should be a red flag that it indeed is not an integer.

Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
19 Mar 2021, 07:02

Expert Reply

Bump

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Re: The integer v is greater than 1. If v is the square of an i
[#permalink]
24 Mar 2021, 04:22

Option A is clear,

because 81 in itself is a square of an integer,

so 81v can be written as \((9x)^2\)

Option B requires a bit of thinking,

\(25v + 10\sqrt{v} +1\) can be written as \((5x+1)^2\)

Option C requires a bit of substitution, since the given terms cannot be represented in the form of \((a+b)^2\)

Only option A and B are correct

because 81 in itself is a square of an integer,

so 81v can be written as \((9x)^2\)

Option B requires a bit of thinking,

\(25v + 10\sqrt{v} +1\) can be written as \((5x+1)^2\)

Option C requires a bit of substitution, since the given terms cannot be represented in the form of \((a+b)^2\)

Only option A and B are correct

gmatclubot

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