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The integer v is greater than 1. If v is the square of an i

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The integer v is greater than 1. If v is the square of an i [#permalink] New post 19 Jan 2016, 08:04
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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) \(4v^2 + 4 \sqrt{v} + 1\)

Practice Questions
Question: 16
Page: 342
Difficulty: medium/hard
[Reveal] Spoiler: OA

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Re: The integer v is greater than 1. If v is the square of an i [#permalink] New post 19 Jan 2016, 08:22
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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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Re: The integer v is greater than 1. If v is the square of an i [#permalink] New post 03 Mar 2018, 07:20
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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


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
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Re: The integer v is greater than 1. If v is the square of an i [#permalink] New post 03 Sep 2018, 14:36
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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
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Re: The integer v is greater than 1. If v is the square of an i [#permalink] New post 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 \(A and B\)


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

Practice Questions
Question: 16
Page: 342
Difficulty: medium/hard


[Reveal] Spoiler: 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] 14 Dec 2018, 18:46
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