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A perfect square is defined as the square of an integer and [#permalink]
22 Jan 2011, 03:49

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45% (02:34) correct
55% (01:39) wrong based on 133 sessions

A perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer. How many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube?

Re: A perfect square is defined as the square of an integer and [#permalink]
22 Jan 2011, 08:12

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amod243 wrote:

A perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer. How many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

don't know the OA

Given: positive integer \(n\) is a perfect square and a perfect cube --> \(n\) is of a form of \(n=x^6\) for some positive integer \(x\) --> \(0<x^6<10^3\) --> \(0<x^2<10\) --> \(x\) can be 1, 2 or 3 hence \(n\) can be 1^6, 2^6 or 3^6.

Answer: B.

amod243 wrote:

hi Mackieman

Agreed with your explanation. but i think we need to consider 0 also???

because 0 also satisfy the condition..

\(n\) can not be 0 as given that \(n\) is a positive integer. _________________

Re: A perfect square is defined as the square of an integer and [#permalink]
18 Jul 2014, 23:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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If the square root of any of these numbers results in an integer then they will be both perfect square and perfect cube.

By looking at the numbers we can see that only three numbers results in integer, sqrt(1) =1, sqrt(4) = 2 and sqrt(9) = 3 => the answer is 3.

(ex sqrt(8^3) = sqrt(8*8*8) = sqrt(8)*sqrt(8)*sqrt(8) = 8*sqrt(8) = not integer)

Didn't get your explanation. 5^3 = 125 , 8^3 = 512 , 6^3 = 216.

IMO , the answer should be 2 as there are only two numbers which can come as perfect square as well as cube between 1 and 1000. The numbers would be 1 and 64

If the square root of any of these numbers results in an integer then they will be both perfect square and perfect cube.

By looking at the numbers we can see that only three numbers results in integer, sqrt(1) =1, sqrt(4) = 2 and sqrt(9) = 3 => the answer is 3.

(ex sqrt(8^3) = sqrt(8*8*8) = sqrt(8)*sqrt(8)*sqrt(8) = 8*sqrt(8) = not integer)

Didn't get your explanation. 5^3 = 125 , 8^3 = 512 , 6^3 = 216.

IMO , the answer should be 2 as there are only two numbers which can come as perfect square as well as cube between 1 and 1000. The numbers would be 1 and 64

Re: A perfect square is defined as the square of an integer and [#permalink]
26 Jul 2015, 08:15

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: A perfect square is defined as the square of an integer and [#permalink]
27 Jul 2015, 02:56

amod243 wrote:

A perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer. How many positive integers n are there such that n is less than 1,000 and at the same time n is a perfect square and a perfect cube?

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

After reading the question it will be a bit scary. But there is <1 min logic for this question.

For a number to be a square and cube at the same time, it should be in the power of 6. X^6. We have to Start counting from 1

1^6 = 1 2^6 = 64 3^6 = 729 4^6 will be greater than 1000. So the Answer is 3

gmatclubot

Re: A perfect square is defined as the square of an integer and
[#permalink]
27 Jul 2015, 02:56

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