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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.

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

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23 Feb 2014, 23:23

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 sq [#permalink]

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24 Feb 2014, 03:28

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

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

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24 Feb 2014, 03:29

1

This post received KUDOS

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

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

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27 Feb 2014, 21:02

2

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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

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

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18 May 2014, 09:07

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.

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

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20 Jul 2014, 02:38

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 sq [#permalink]

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13 Aug 2014, 08:05

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!!!!!!

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.