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Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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19 Feb 2016, 02:45
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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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19 Feb 2016, 05:37
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The powers of a, b and c have to be divisible by 2, 3 and 5 for N to be a perfect square, perfect cube and perfect 5th power. LCM of 2, 3 and 5 is 30 The smallest integer N = (a^30)(b^30)(c^30)
Answer: E



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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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23 Feb 2016, 03:01
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Bunuel wrote: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?
A. a^3*b^4*c^5 B. a^5*b^4*c^3 C. a^2*b^3*c^5 D. a^7*b^6*c^5 E. a^27*b^26*c^25
Kudos for correct solution. Given: N=a^3*b^4*c^5 We need to make this a perfect square, a perfect cube as well as a perfect fifth power. LCM of 2, 3 and 5 = 30 Therefore we need to have the powers of a, b and c as 30 Option E gives us the term that on multiplication will make N as (a^30)(b^30)(c^30) Correct Option : E



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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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09 Nov 2016, 12:27
Answer:E N1=a^3*b^4*c^5 should be perfect square, a perfect cube as well as a perfect fifth power Perfect square: ^2 Perfect cube: ^3 Perfect fifth: ^5 to combine all, we need a common factor 2x3x5 = 30 then N3=a^30*b^30*c^30 is the perfect of the three numbers. N1*N2=N3 N2 will be a^27*b^26*c^25



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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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13 Jan 2017, 02:35
Great Question this one. Perfect square => Powers of all primes=> Multiple of 2. Perfect cube=>Powers of all the primes => Multiple of 3. Perfect fifth power => Powers of all the primes => Multiple of 5
Hence powers of all the primes must be a multiple of all>2,3,5 => LCM=30 Smallest multiplication => \(a^{27}*b^{26}*c^{25}\) Hence E.
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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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13 Apr 2017, 06:27
Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power? A. a^3*b^4*c^5 B. a^5*b^4*c^3 C. a^2*b^3*c^5 D. a^7*b^6*c^5 E. a^27*b^26*c^25 The catch in this question is 'smallest integer' but if don't catch it, can this be a way to solve this question: Option E: (a^27*a^3)(b^26*b^4)(c^25*c^5) That gives us: a^30*b^30*c^30 Can anyone please answer this query?
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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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04 May 2017, 06:28
Bunuel wrote: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?
A. a^3*b^4*c^5 B. a^5*b^4*c^3 C. a^2*b^3*c^5 D. a^7*b^6*c^5 E. a^27*b^26*c^25
Kudos for correct solution. BunuelLooking at a question, how can we determine that we have to find the LCM? This is little confusing and I am having a little trouble understanding it. Really appreciate your help. Thanks!



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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, [#permalink]
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04 May 2017, 12:08
lcm of 2,3,5=30 a^30/a^3=a^27 b^30/b^4=b^26 c^30/c^5=c^25 ans: E




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