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# Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...

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EMPOWERgmat Instructor
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Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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Updated on: 24 Nov 2015, 01:09
00:00

Difficulty:

5% (low)

Question Stats:

89% (02:06) correct 11% (02:09) wrong based on 97 sessions

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QUANT 4-PACK SERIES Problem Solving Pack 4 Question 4 If (2^n)(3^k)(5^8)...

If ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) is equal to one percent of $$60^{10}$$ , then what is the value of N – K?

A) 7
B) 8
C) 12
D) 14
E) 15

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This question is part of the Quant 4-Pack series

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Originally posted by EMPOWERgmatRichC on 19 Nov 2015, 17:51.
Last edited by EMPOWERgmatRichC on 24 Nov 2015, 01:09, edited 1 time in total.
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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19 Nov 2015, 18:19
1
If ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) is equal to one percent of $$60^{10}$$ , then what is the value of N – K?

A) 7
B) 8
C) 12
D) 14
E) 15

60=(2^2)*3*5

($$1/100$$)*($$(2^2)*3*5$$)^100= ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$)
($$2^{20}$$)*($$3^{10}$$)*($$5^{10}$$)= ($$2^{N+2}$$) ($$3^{K}$$) ($$5^{10}$$)

N+2=20
N=18

K=10

18-10=8

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Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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23 Nov 2015, 09:28
1
EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Problem Solving Pack 4 Question 4 If (2^n)(3^k)(5^12)...

If ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) is equal to one percent of $$60^{10}$$ , then what is the value of N – K?

A) 7
B) 8
C) 12
D) 14
E) 15

48 Hour Window Answer & Explanation Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation

Given, ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = 1% of $$60^{10}$$= $$60^8$$

---> ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = $$60^8$$ =$$[(2^2)*(3)*5]^8$$ = $$[(2^{16})*(3^8)*5^8]$$---> n=16 and m=8 , giving n-k=8. B is the correct answer.
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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23 Nov 2015, 09:45
1
(2^N) * (3^K)*(5^8) = (10^-2)*(60^10)
RHS = (2*5)^-2 * (2^2*3*5)^10 = (2^-2)*(5^-2)*(2^20)*(3^10)*(5^10) = (2^18)*(3^10)*(5^8)
N = 18, K = 10; N-K = 8
=>B
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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23 Nov 2015, 12:05
1
EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Problem Solving Pack 4 Question 4 If (2^n)(3^k)(5^12)...

If ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) is equal to one percent of $$60^{10}$$ , then what is the value of N – K?

A) 7
B) 8
C) 12
D) 14
E) 15

($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = $$10$$% of $$60^{10}$$

($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = $$60^{8}$$

To make the highlighted part a factor of 60, we must have -

($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = $$60^{8}$$

($$2^{16}$$) ($$3^{8}$$) ($$5^{8}$$) = $$60^{8}$$ since - Since 60 = $$2^{2}$$ ( $$3$$)( $$5$$)

Now we get N = 16 and K = 8

So, N - K = 8
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EMPOWERgmat Instructor
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^8)...  [#permalink]

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24 Nov 2015, 01:26
1
EMPOWERgmatRichC wrote:
QUANT 4-PACK SERIES Problem Solving Pack 4 Question 4 If (2^n)(3^k)(5^8)...

If ($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) is equal to one percent of $$60^{10}$$ , then what is the value of N – K?

A) 7
B) 8
C) 12
D) 14
E) 15

Hi All,

This question is based on the concept of Prime Factorization (the idea that every positive integer greater than 1 is either a prime number OR the product of prime numbers).

To start, we're told that ($$2^{N}$$) ($$3^{K}$$) ($$5)^{8}$$) = $$60^{10}$$ / 100

The 'right side' of the equation can be rewritten as...

$$(2x2x3x5)^{10}$$/ (2x2x5x5) =
$$(2)^{18}$$ $$(3)^{10}$$ $$(5^{8}$$

When this calculation is substituted into the larger equation, we have...

($$2^{N}$$) ($$3^{K}$$) ($$5^{8}$$) = $$(2)^{18}$$ $$(3)^{10}$$ $$(5)^{8}$$

'Canceling out' the common terms leaves us....

N = 18 and K = 10

Thus, N - K = 18 - 10 = 8

GMAT assassins aren't born, they're made,
Rich
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...  [#permalink]

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26 Oct 2018, 02:27
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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^12)...   [#permalink] 26 Oct 2018, 02:27
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