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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]  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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Founder
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Re: The integer v is greater than 1. If v is the square of an i [#permalink]  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]  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

Cheers,
Brent
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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
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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.

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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 $$A and B$$

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]  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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