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Intern  Joined: 19 Feb 2009
Posts: 46
Schools: INSEAD,Nanyang Business school, CBS,
A perfect square is defined as the square of an integer and  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 54% (02:05) correct 46% (02:19) wrong based on 210 sessions

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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

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Math Expert V
Joined: 02 Sep 2009
Posts: 57146
A perfect square is defined as the square of an integer and  [#permalink]

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6
2
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 cannot be 0 as given that n is a positive integer.
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Posts: 146
Location: Stockholm, Sweden
Re: A perfect square is defined as the square of an integer and  [#permalink]

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2
Perfect cube:
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 75
6^3 = 206
7^3 = 343
8^3 = 502
9^3 = 729
10^3 = 1000

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)
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Intern  Joined: 19 Feb 2009
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Schools: INSEAD,Nanyang Business school, CBS,
Re: A perfect square is defined as the square of an integer and  [#permalink]

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hi Mackieman

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

because 0 also satisfy the condition..
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Manager  Joined: 20 Dec 2010
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Location: Stockholm, Sweden
Re: A perfect square is defined as the square of an integer and  [#permalink]

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amod243 wrote:
hi Mackieman

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

because 0 also satisfy the condition..

Well, if we see squares as the area of polygons

#
= area 1

##
##
= area 2*2 = 4

###
###
###
= area 3*3 = 9

In my opinion 0 shouldn't be considered to be a perfect square since 0 doesn't represent the area of square.
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12/2010 GMATPrep 1 620 (Q34/V41)
01/2011 GMATPrep 2 640 (Q42/V36)
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Math Expert V
Joined: 02 Sep 2009
Posts: 57146
Re: A perfect square is defined as the square of an integer and  [#permalink]

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Mackieman wrote:
amod243 wrote:
hi Mackieman

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

because 0 also satisfy the condition..

Well, if we see squares as the area of polygons

#
= area 1

##
##
= area 2*2 = 4

###
###
###
= area 3*3 = 9

In my opinion 0 shouldn't be considered to be a perfect square since 0 doesn't represent the area of square.

Zero is both perfect square and a perfect cube but $$n$$ can not be 0 as given that $$n$$ is a positive integer.
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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cool question. thanks.
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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Bunuel - how did you come up with x^6?

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Manager  Joined: 22 Feb 2009
Posts: 158
A perfect square is defined as the square of an integer and  [#permalink]

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bankerboy30 wrote:
Bunuel - how did you come up with x^6?

Posted from my mobile device

n = a^2 = b ^3 ( with a, b, and n are positive integer)

since a^2 = b^3 --> a^2 = (x^2)^3 = x^6 ( with b = x^2, if not a^2 cannot be equal b^3)

or since b^3 = a^2 --> b^3 = x^6 ( same reason)

--> n = a^2 = b^3 = x^6

Hope it helps a little bit Anyways, Thanks to Bunuel for his awesome explanation !!!! _________________
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Manager  Joined: 28 Apr 2014
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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Mackieman wrote:
Perfect cube:
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 75
6^3 = 206
7^3 = 343
8^3 = 502
9^3 = 729
10^3 = 1000

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
Math Expert V
Joined: 02 Sep 2009
Posts: 57146
Re: A perfect square is defined as the square of an integer and  [#permalink]

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himanshujovi wrote:
Mackieman wrote:
Perfect cube:
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 75
6^3 = 206
7^3 = 343
8^3 = 502
9^3 = 729
10^3 = 1000

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

Check here: a-perfect-square-is-defined-as-the-square-of-an-integer-and-108103.html#p856696
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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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

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
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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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

We note that if an integer is a perfect square and a perfect cube at the same time, then it is the sixth power of some integer.

We need to determine how many numbers exist that when raised to the 6th power are less than 1,000.

1^6 = 1 < 1000

2^6 = 64 < 1000

3^6 = 729 < 1000

4^6 = 4,096 > 1000

Since 4^6 > 1000, we see that only 3 values exist.

Answer: B
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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To Solve this problem, we need to know that for a number to be perfect square it can be expressed in powers of 2.

1. Cubes which are less than 1000 are $$1^3,2^3,3^3,4^3,5^3,6^3,7^3,8^3,9^3$$
2. Out of the above cubes, only $$1^3, 4^3,9^3$$ can be expressed in even powers (i.e. $$1^3=1=1^2$$, $$4^3=2^6$$ and $$9^3=3^6$$)

So only 3 numbers satisfy the condition
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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Bunuel wrote:
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 cannot be 0 as given that n is a positive integer.

Bullseye, as always! Thank you Bunuel! Your help is always highly appreciated!
Intern  Joined: 02 Oct 2018
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Re: A perfect square is defined as the square of an integer and  [#permalink]

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Where is this question taken from?
Intern  B
Joined: 02 Mar 2019
Posts: 5
Re: A perfect square is defined as the square of an integer and  [#permalink]

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Easy question not 700+ and don't waste time in such question ... Approach should be...

u need to find cubes which are also expressed as square.(1^3,2^3,3^3......10^3)
Since 10^3 = 1000. Don't go beyond that.
1 to 10 we have 3 sqaures 1,4,9, so answer is 3.
Coz
4^3 = 2^(2*3) = 8^2.
9^3 = 3^(2*3) = 27^2.
Similarly 1.
Hope this helps...don't start sloving questions as soon as u read.understand the language, most of the questions doesn't require calculations

Posted from my mobile device Re: A perfect square is defined as the square of an integer and   [#permalink] 18 Jul 2019, 00:19
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