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A perfect square is defined as the square of an integer and [#permalink]
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22 Jan 2011, 04:49
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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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Re: A perfect square is defined as the square of an integer and [#permalink]
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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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Re: A perfect square is defined as the square of an integer and [#permalink]
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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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Re: A perfect square is defined as the square of an integer and [#permalink]
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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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19 Jul 2014, 17:21
Bunuel  how did you come up with x^6?
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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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Re: A perfect square is defined as the square of an integer and [#permalink]
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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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Re: A perfect square is defined as the square of an integer and [#permalink]
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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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