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# If a, b and c are integers greater than one, and 6^15 = a*b^c, what is

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Joined: 02 Sep 2009
Posts: 60605
If a, b and c are integers greater than one, and 6^15 = a*b^c, what is  [#permalink]

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19 Nov 2019, 01:23
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38% (01:49) correct 62% (01:51) wrong based on 60 sessions

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If a, b and c are integers greater than one, and $$6^{15}=a*b^c$$, what is the value of c?

(1) a is not divisible by 3.
(2) b is prime.

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Joined: 16 Feb 2015
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Concentration: Finance, Operations
If a, b and c are integers greater than one, and 6^15 = a*b^c, what is  [#permalink]

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Updated on: 19 Nov 2019, 02:23
2
1
Bunuel wrote:
If a, b and c are integers greater than one, and $$6^{15}=a*b^c$$, what is the value of c?

(1) a is not divisible by 3.
(2) b is prime.

Are You Up For the Challenge: 700 Level Questions

Given, 6^15=a*b^C

6^15=2^15*3^15

1. a can be 2^15 or 3^15, but given that a is not divisible by 3, hence a is 2^15. So C is 15. Sufficient alone.

2. Again, b can be 2 or 3 which is prime numbers. So not sufficient

Pls give kudos, if you find my explanation good enough

Originally posted by rajatchopra1994 on 19 Nov 2019, 02:03.
Last edited by rajatchopra1994 on 19 Nov 2019, 02:23, edited 1 time in total.
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Joined: 18 Jul 2019
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Re: If a, b and c are integers greater than one, and 6^15 = a*b^c, what is  [#permalink]

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19 Nov 2019, 02:17
1
IMO A

6^15 = (3*2)^15 = 3^15 * 2^15 = a * b^c

1) a is not divisible be 3, which implies that b^c = 3^15 ==> c= 15

2) bis a prime number, but in the equation we can deduce from 3^15 * 2^15 = a * b^c that so is a and hence not enough

VP
Joined: 19 Oct 2018
Posts: 1294
Location: India
Re: If a, b and c are integers greater than one, and 6^15 = a*b^c, what is  [#permalink]

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19 Nov 2019, 03:47
3
Statement 1-

Case 1-

$$6^{15}= 2^{15}*3^{15}$$

where $$a= 2^{15}$$, $$b=3$$ and $$c=15$$

Case 2-

$$6^{15}= 2^3*2^{12}*3^{15}= 2^3 * (2^4*3^5)^3$$

where $$a=2^3$$, $$b=2^4*3^5$$ and $$c=3$$

Insufficient

Statement 2-

b is a prime number

Case 1-

$$6^{15}= 2^{15}*3^{15}$$

where $$a= 2^{15}$$, $$b=3$$ and $$c=15$$

Case 2-

$$6^{15}= 2^3*2^{12}*3^{15}= (2^4*3^5)^3 *2^3$$

where $$b=2$$, $$a=(2^4*3^5)^3$$ and $$c=3$$

Insufficient

Combining both equations
a is not divisible by 3, and b is prime

hence, a must be $$2^{15}$$ and b =3 and c=15

Sufficient

Bunuel wrote:
If a, b and c are integers greater than one, and $$6^{15}=a*b^c$$, what is the value of c?

(1) a is not divisible by 3.
(2) b is prime.

Are You Up For the Challenge: 700 Level Questions
VP
Joined: 24 Nov 2016
Posts: 1094
Location: United States
If a, b and c are integers greater than one, and 6^15 = a*b^c, what is  [#permalink]

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20 Nov 2019, 09:40
1
Bunuel wrote:
If a, b and c are integers greater than one, and $$6^{15}=a*b^c$$, what is the value of c?

(1) a is not divisible by 3.
(2) b is prime.

$$6^{15}=a*b^c…2^{15}3^{15}=ab^c$$

(1) a is not divisible by 3. insufic

$$2^{15}3^{15}=ab^c…a=2^{something}…b=(2•3)^{something}…b=3^{something}$$
$$a=2^3…b^c=2^{12}3^15=(2^43^5)^3$$
$$a=2^5…b^c=2^{10}3^15=(2^23^3)^5$$
$$a=2^{15}…b^c=3^{15}$$

(2) b is prime. insufic

$$b^c=3^{something}…b^c=2^{something}$$

(1&2) sufic

$$a=2^{15}…b^c=3^{15}…c=15$$

Ans (C)
If a, b and c are integers greater than one, and 6^15 = a*b^c, what is   [#permalink] 20 Nov 2019, 09:40
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