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505-555 (Easy)|   Arithmetic|   Exponents|      
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EMPOWERgmatRichC
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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what Updated on: 03 Feb 2024, 09:40
Originally posted by EMPOWERgmatRichC on 19 Nov 2015, 17:51.
Last edited by Bunuel on 03 Feb 2024, 09:40, edited 2 times in total.
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B
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15% (low)

Question Stats:

86% (01:58) correct 14% (02:12) wrong based on 285 sessions

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

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B
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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 19 Nov 2015, 18:19
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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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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 23 Nov 2015, 09:28
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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


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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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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 23 Nov 2015, 09:45
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(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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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 23 Nov 2015, 12:05
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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
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(\(2^{N}\)) (\(3^{K}\)) (\(5^{8}\)) = \(10\)% 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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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 24 Nov 2015, 01:26
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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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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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B

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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 03 Feb 2024, 08:49
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Hello,
please, how to get 60 exponent 8 ? iis there a formula used? Thank you
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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 03 Feb 2024, 09:53
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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


Hello,
please, how to get 60 exponent 8 ? iis there a formula used? Thank you
@bunnuel
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Firstly, the question must specify that N and K are integers; otherwise, option B wouldn't necessarily be correct.

    \(60^{10}=(2^2*3*5)^{10} = 2^{20}*3^{10}*5^{10}\)

One percent of \(60^{10}\), is 1/100 of \(60^{10}\), so it's:

    \(\frac{60^{10}}{100} = \)

    \(=\frac{2^{20}*3^{10}*5^{10}}{10^2}=\)

    \(=\frac{2^{20}*3^{10}*5^{10}}{2^25^2}=\)

    \(=2^{18}*3^{10}*5^{8}\)

Hence, \((2^{N}) (3^{K})(5^{8})=2^{18}*3^{10}*5^{8}\). Since N and K must be integers, N = 18 and K = 10. Therefore, N - K = 18 - 10 = 8.

Answer: B.
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If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what 21 Apr 2026, 22:59
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