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If k is an integer, what is the smallest possible value of k [#permalink]
20 Nov 2017, 10:45
Question Stats:
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If k is an integer, what is the smallest possible value of k such that 1040k is the square of an integer? A. 2 B. 5 C. 10 D. 15 E. 65 Kudos for correct solution.




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Re: If k is an integer, what is the smallest possible value of k [#permalink]
01 Dec 2017, 22:27
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Bunuel wrote: If k is an integer, what is the smallest possible value of k such that 1040k is the square of an integer?
A. 2 B. 5 C. 10 D. 15 E. 65
Kudos for correct solution. Here let 1060 can be written as = \(2^4 X 65^1\) to make it perfect square k should be equal = 65 i.e \(2^4 X 65^2\) Hence option E
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Re: If k is an integer, what is the smallest possible value of k [#permalink]
10 May 2018, 10:44
I multipled each option to 1040 and took a square root by the calculator and I got the answer in 2.03 min .. E



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Re: If k is an integer, what is the smallest possible value of k [#permalink]
11 May 2018, 04:23
65 as that is what is left post factoring. All the factors should have 2 values atleast



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Re: If k is an integer, what is the smallest possible value of k [#permalink]
14 May 2018, 00:47
One should also remember that the exponents of prime factors are even.
or we can solve the problem by taking the prime factorization of 1040 , when we do so we notice the factors of 13 and 5 are not even rather they are only 1 each..
hence we choose the option he because when we multiply 1040 by 65 13 and 5 factors make the exponents of 1040*65 even
hence option is E is correct.



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Re: If k is an integer, what is the smallest possible value of k [#permalink]
29 Dec 2018, 17:50
IshanGre wrote: One should also remember that the exponents of prime factors are even.
Please explain.



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Re: If k is an integer, what is the smallest possible value of k [#permalink]
29 Dec 2018, 19:39
AE wrote: IshanGre wrote: One should also remember that the exponents of prime factors are even.
Please explain. Since we are looking for square, the exponent has to be even.. Similarly if we are looking at a cube, the exponent should be divisible by 3.. .1040k=2*520k=\(2^2*260k=2^2*2*130k=2^4*65=2^4*5*13\) So 2 has a power of 4, and we require one more of 5 and 13 to make the entire term as square
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Some useful Theory. 1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressionsarithmeticgeometricandharmonic11574.html#p27048 2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effectsofarithmeticoperationsonfractions11573.html?sid=d570445335a783891cd4d48a17db9825 3. Remainders : https://greprepclub.com/forum/remainderswhatyoushouldknow11524.html 4. Number properties : https://greprepclub.com/forum/numberpropertyallyourequire11518.html 5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolutemodulusabetterunderstanding11281.html



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Re: If k is a positive integer, what is the smallest possible va [#permalink]
17 Apr 2020, 20:03
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workout wrote: If k is a positive integer, what is the smallest possible value of k such that 1040k is the square of an integer?
A) 2
B) 5
C) 10
D) 15
E) 65 Questions like this can often be solved by figuring out prime factors. We know that 1040 * k is a perfect square, so find the prime factorization of 1040 first: 104*10 2*52* 5*2 2*2*26*5*2 2*2*2*13*5*2 So the prime factorization of 1040 = \(2^4*5*13\) For a number to be a perfect square, each of its prime factors need to be paired with a matching prime factor. \(2^4\) is 16, a perfect square. Each factor of 2 is paired with another factor of 2. But 5 and 13 don't have matching factors, so we need another 5 * 13 to make a perfect square. That product is k. k = 5*13 = 65 Answer: E
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Re: If k is a positive integer, what is the smallest possible va [#permalink]
18 Apr 2020, 05:00
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Prime factorization of 1040 = \(2^4*5*13\)
Since \(\sqrt{2^4}\) = 4
Since \(2^4\) is already a perfect square the other factors not having a perfect square are 5 and 13
Hence the lowest number required to be multiplied to 1040 to make it a square = 5 * 13 = 65
Hence answer is E




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