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A perfect square is defined as the square of an integer and

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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
[Reveal] Spoiler: OA

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New post 22 Jan 2011, 05:39
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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New post 22 Jan 2011, 08:25
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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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.
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New post 22 Jan 2011, 09:19
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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Re: A perfect square is defined as the square of an integer and [#permalink]

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New post 22 Jan 2011, 09:25
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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New post 18 Feb 2011, 14:59
cool question. thanks.
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New post 19 Jul 2014, 17:21
Bunuel - how did you come up with x^6?

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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 :-D

Anyways, Thanks to Bunuel for his awesome explanation !!!! :-D
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New post 21 Jul 2014, 06:12
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

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New post 21 Jul 2014, 06:33
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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New post 27 Jul 2015, 03:56
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

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New post 18 Aug 2017, 09:07
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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New post 19 Aug 2017, 02:02
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] 19 Aug 2017, 02:02
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