If (2^n)(3^k)(5^12) is equal to one percent of 60^10, then what

(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

(\(2^{N}\)) (\(3^{K}\)) (\(5^{8}\)) = \(10\)% 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

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:

GMAT assassins aren't born, they're made,
Rich

Hello,
please, how to get 60 exponent 8 ? iis there a formula used? Thank you
@bunnuel

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.