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# Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers,

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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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45% (medium)

Question Stats:

68% (01:46) correct 32% (02:06) wrong based on 207 sessions

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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. $$a^{27}*b^{26}*c^{25}$$

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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
2
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)

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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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Joined: 24 Jan 2016
Posts: 18
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
1
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
2
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

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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.

Bunuel

Looking 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.

Thanks!
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Posts: 51
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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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers,  [#permalink]

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08 Jul 2018, 11:55
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Re: Given that N=a^3*b^4*c^5 where a, b and c are distinct prime numbers, &nbs [#permalink] 08 Jul 2018, 11:55
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