Exponents and Division

Algebra Level 2

When 1 2 18 12^{18} is divided by 1 8 12 18^{12} , the result is ( m n ) 3 \left(\dfrac{m}{n}\right)^3 . What is m n m-n , where m m and n n are coprime positive integers?

245 247 243 240 244

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Marco Brezzi
Aug 11, 2017

1 2 18 1 8 12 = ( 2 2 3 ) 18 ( 2 3 2 ) 12 = 2 36 3 18 2 12 3 24 = 2 24 3 6 = ( 2 8 3 2 ) 3 = ( 256 9 ) 3 = ( m n ) 3 \dfrac{12^{18}}{18^{12}}=\dfrac{(2^2\cdot 3)^{18}}{(2\cdot 3^2)^{12}}=\dfrac{2^{36}\cdot 3^{18}}{2^{12}\cdot 3^{24}}=\dfrac{2^{24}}{3^6}=\left(\dfrac{2^8}{3^2}\right)^3=\left(\dfrac{256}{9}\right)^3=\left(\dfrac{m}{n}\right)^3

m n = 256 9 = 247 \Longrightarrow m-n=256-9=\boxed{247}

Thank you for sharing your solution.

Hana Wehbi - 3 years, 10 months ago
Hana Wehbi
Aug 11, 2017

1 2 18 1 8 12 = 4 18 × 3 18 2 12 × 9 12 = 2 36 3 18 2 12 3 24 = 2 24 3 6 = ( 2 8 3 2 ) 3 = ( 256 9 ) 3 256 9 = 247 \Large\frac{12^{18}}{18^{12}} = \frac{4^{18}\times3^{18}}{2^{12}\times 9^{12}}= \frac {2^{36}3^{18}}{2^{12}3^{24}} = \frac{2^{24}}{3^6} = (\frac{2^8}{3^2})^3= (\frac{256}{9})^3\implies 256 - 9=247

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...