Given that $N=a^3b^4c^5$ where $a, b$ and $c$ are distinct prime numbers, what is the smallest number with which

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

by VJesus12 » Fri Jul 24, 2020 7:09 am

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Given that $N=a^3b^4c^5$ 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. $a^3b^4c^5$
B. $a^5b^4c^3$
C. $a^2b^3c^5$
D. $a^7b^6c^5$
E. $a^{27}b^{26}c^{25}$

[spoiler]OA=E[/spoiler]

Source: Veritas Prep

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Source: Beat The GMAT — Problem Solving | Fri Jul 24, 2020 7:09 am

sahilaro23: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Jul 24, 2020 11:07 am; Location: INDIA

Re: Given that $N=a^3b^4c^5$ where $a, b$ and $c$ are distinct prime numbers, what is the smallest number with whi

by sahilaro23 » Fri Jul 24, 2020 11:27 am

multiple of square, cube and 5th power; simply take the LCM of power required = 2x3x5 = 30.
We need 30 in each power to make it follow what is asked.
It'll be a^30 b^30 c^30
Hence, C.

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Srishtim23: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Sat Jul 25, 2020 6:13 am

Re: Given that $N=a^3b^4c^5$ where $a, b$ and $c$ are distinct prime numbers, what is the smallest number with whi

by Srishtim23 » Sat Jul 25, 2020 6:54 am

To be a perfect square, a perfect cube as well as a perfect fifth power, such number should be in power of this expression which is divisible by 2,3 and 5 , In simpler terms, we are required to find LCM of 2, 3 & 5 which is 30... as it should be perfect square AND cube AND Fifth.

So, 30 should be the power of each expression a, b, & c

OA: E (30-3) (30-4) (30-5)

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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

Re: Given that $N=a^3b^4c^5$ where $a, b$ and $c$ are distinct prime numbers, what is the smallest number with whi

by deloitte247 » Sun Jul 26, 2020 2:52 am

The power exponents 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.
Therefore, we need to find the LCM of 2, 3, and 5.
2 = 1 * 2
3 = 1 * 3
5 = 1 * 5
LCM = 1 * 2 * 3 * 5
$$Therefore,\ the\ smallest\ integer\ N=a^{30}b^{30}c^{30}$$
$$N=a^{\left(30-3\right)}b^{\left(30-4\right)}c^{\left(30-5\right)}$$
$$N=a^{27}b^{26}c^{25}\ \ \ \ \ \ \ \ \ Answer\ =\ option\ E$$

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Nik13: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sun Jul 26, 2020 4:57 am

Re: Given that $N=a^3b^4c^5$ where $a, b$ and $c$ are distinct prime numbers, what is the smallest number with whi

by Nik13 » Sun Jul 26, 2020 5:10 am

Thank you for the solution. That was really meaningful.

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