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If n is an integer between 2 and 100 and if n is also the square of an

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If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 23 Feb 2014, 07:51
2
1
31
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

58% (01:43) correct 42% (01:35) wrong based on 1029 sessions

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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 23 Feb 2014, 07:51
1
2
SOLUTION

If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n?

Given: n is a perfect square between 2 and 100 (a perfect square is an integer that can be written as the square of some other integer, for example 16=4^2, is a perfect square).

(1) n is even --> n can be any even perfect square in the given range: 4, 16, 36, ... Not sufficient.

(2) The cube root of n is an integer --> so n is also a perfect cube between 2 and 100. There are 4 perfect cubes in this range: 2^3=8, 3^3=27 and 4^3=64 but only one of them namely 64 is also a perfect square, so n=64=8^2=4^3. Sufficient.

Answer: B.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 24 Feb 2014, 00:23
1
If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n?

(1) n is even.
(2) The cube root of n is an integer.



Sol: Given n= I^2 where I some integer and 2<n<100 or 2<I^2<100

St 1: n is even and only possible squares in the given range are for 4,6,8 which are 16,36,64 respectively...So n can be 4,6 or 8.
A and D ruled out

St2: Cube root of n= a where " a " is an integer So Taking cube we get
n=a^3.So n is a cube of an integer "a" but " a^3" should be between 2 and 100 so a^3=27 or 64. Since 2 (n=3 or 4) answers are there so not possible.

Combining we get,

n is even and cube root of n is an integer so only possible value of n=64

Ans is C
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 24 Feb 2014, 03:09
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Option B
From S1:2<n<100 and n is a square.
=>n=4,9,16,25,36,49,64,81
According to S1:n=4,16,36,64.Not suff.

From S2:2<n<100
n=4,9,16,25,36,49,64,81 AND
Cube root of n=integer
Only possible for 64.Hence,Sufficient.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 24 Feb 2014, 04:28
From the statement- it states that 'n' is a square (of an integer) between 2 and 100.
Thus the possible values of n are:-
4,9,16,25,36,49,64,81.

Statement 1:- n is even
This leaves us with 4,16,36 and 64 - Four values for n hence this statement is not sufficient

Statement 2:- The cube root of n is an integer.
Of all the values of n possible above only 64 has an integer cuberoot.
Thus there is only 1 value of n. Hence it is sufficient

Thus answer is Option (B)
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 24 Feb 2014, 04:29
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WoundedTiger wrote:
If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n?

(1) n is even.
(2) The cube root of n is an integer.



Sol: Given n= I^2 where I some integer and 2<n<100 or 2<I^2<100

St 1: n is even and only possible squares in the given range are for 4,6,8 which are 16,36,64 respectively...So n can be 4,6 or 8.
A and D ruled out

St2: Cube root of n= a where " a " is an integer So Taking cube we get
n=a^3.So n is a cube of an integer "a" but " a^3" should be between 2 and 100 so a^3=27 or 64. Since 2 (n=3 or 4) answers are there so not possible.

Combining we get,

n is even and cube root of n is an integer so only possible value of n=64

Ans is C


Bro 27 is not a square of an integer hence you cannot consider that :)
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 26 Feb 2014, 07:07
n is an integer between 2 and 100, n is also a square,what is n?

1) n is even -
So n = 4,16,36,64
Not sufficient

2)Cube root of n is an integer.

n =8,27,64 out of these only 64 is a square also.

hence B is sufficient.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 27 Feb 2014, 22:02
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Best part of this question is the piece of sentence n is also the square of an integer
It narrow downs the answer to 4,9,16,25,36,49,64,81 (between 2 and 100)

A.) n is even. As seen in the above list there are more than one even no.
INSUFFICIENT
B.) cube root of n is integer. Find the cube root of all above no. Only 64 yields an integer
SUFFICIENT

Answer - B
Difficulty - 650
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 18 May 2014, 10:07
lets note down all the square numbers between 2 to 100. It won't be a pretty long list: 4, 9, 16, 25, 49, 64, 81, 100
1. N is even --> We have several even values in our list. So definitely not A or D. It should be either B, D or E now
2. cube root f n is an integer --> It indirectly says that the number in our list is also a cube of another integer. I can see just 64 as cube of 4 over here. Rest all numbers are not cube of any integer. Hence, B is sufficient.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 20 Jul 2014, 03:38
Statement 1) n is even --> could be 4, 16, 25 ... etc. not sufficient
Statement 2) 3 variants: 8, 27 , 64 --> cube root of 8 = 2 is not an square of an integer, same for 27; 64 is the only possible number.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 13 Aug 2014, 09:05
What does cube root mean here?? For e.g if n is 16, does it mean that "sqroot(16^3)" is an integer, then its an integer. root of 64^3 is also an integer, and how are we assuming that cube root should also be in the range 2-100. HELP!!!!!!
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 13 Aug 2014, 10:49
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sunaimshadmani wrote:
What does cube root mean here?? For e.g if n is 16, does it mean that "sqroot(16^3)" is an integer, then its an integer. root of 64^3 is also an integer, and how are we assuming that cube root should also be in the range 2-100. HELP!!!!!!


The cube root of n is an integer means \(\sqrt[3]{n}=integer\) --> raise to the third power: \(n=integer^3\), so n is the cube of some integer: ..., 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, ... Since n is between 2 and 100 and it's also the square of an integer, then it can only be 64.

Hope it's clear.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 07 Dec 2017, 02:49
n: lies between 2 and 100, inclusive & is a perfect square.

So, the power of primes that n is made up of must be divisible by 2.

St.1: n is even : Insufficient
n can be 4, 9, 16 and so on

St.2: cube root of n is an interger. This means that the power of primes that n is made up of must be divisible by 3.

So, the power of primes that n is made up of must be divisible by both 2 & 3 i.e. 6.

2 is the only prime number that has a value below 100 when raised to a power of 6.

Therefore, n= 2^6= 64.

So, Statement 2 is sufficient.

Hence, Ans B

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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 07 Dec 2017, 03:25
Option B

Already it has been given that n is the square of an integer or the square root of n is an integer.

As per the second statement, n is also the cube root of an integer.

Combining the info given in the question, we can conclude that the sixth root of n will also be an integer.

Or n = sixth power of an integer.
Only one such number , 64, exists within 2 to 100 range.

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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 18 Jun 2019, 02:06
Bunuel wrote:
SOLUTION

If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n?

Given: n is a perfect square between 2 and 100 (a perfect square is an integer that can be written as the square of some other integer, for example 16=4^2, is a perfect square).

(1) n is even --> n can be any even perfect square in the given range: 4, 16, 36, ... Not sufficient.

(2) The cube root of n is an integer --> so n is also a perfect cube between 2 and 100. There are 4 perfect cubes in this range: 2^3=8, 3^3=27 and 4^3=64 but only one of them namely 64 is also a perfect square, so n=64=8^2=4^3. Sufficient.

Answer: B.

Bunuel
between 2 and 100 should we include 2^3=8? They did not say inclusive.
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Re: If n is an integer between 2 and 100 and if n is also the square of an  [#permalink]

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New post 18 Jun 2019, 11:37
Bunuel wrote:
SOLUTION

If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n?

Given: n is a perfect square between 2 and 100 (a perfect square is an integer that can be written as the square of some other integer, for example 16=4^2, is a perfect square).

(1) n is even --> n can be any even perfect square in the given range: 4, 16, 36, ... Not sufficient.

(2) The cube root of n is an integer --> so n is also a perfect cube between 2 and 100. There are 4 perfect cubes in this range: 2^3=8, 3^3=27 and 4^3=64 but only one of them namely 64 is also a perfect square, so n=64=8^2=4^3. Sufficient.

Answer: B.

Hi Bunuel
Shouldn't it be just 3 perfect cubes in this range?
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Re: If n is an integer between 2 and 100 and if n is also the  [#permalink]

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Re: If n is an integer between 2 and 100 and if n is also the   [#permalink] 17 Oct 2019, 07:06
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