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

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

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New post Updated on: 24 Nov 2015, 01:09
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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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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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New post 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

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

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New post 23 Nov 2015, 09:28
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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
Earn KUDOS! Post your answer and explanation.
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

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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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New post 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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New post 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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Re: Problem Solving Pack 4, Question 4 If (2^n)(3^k)(5^8)...  [#permalink]

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New post 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

Final Answer:

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

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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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