1 2 th 12^\text{th} Root

Algebra Level 3

If 1 181398528 12 \sqrt [ 12 ]{ \dfrac { 1 }{ 181398528 } } can be written as A 12 B \dfrac { \sqrt [ 12 ]{ A } }{ B } , where A A and B B are positive integer, find the minimum value of A + B A+B .

This problem was created by José Carlos Pereira

8 18 4 7

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

1 181398528 12 = 1 12 181398528 12 = \sqrt [ 12 ]{ \frac { 1 }{ 181398528 } } =\frac { \sqrt [ 12 ]{ 1 } }{ \sqrt [ 12 ]{ 181398528 } } =

1 2 10 × 3 11 12 = 1 2 5 / 6 × 3 11 / 12 = \frac { 1 }{ \sqrt [ 12 ]{ { 2 }^{ 10 }\times { 3 }^{ 11 } } } =\frac { 1 }{ { 2 }^{ 5/6 }\times { 3 }^{ 11/12 } } =

1 2 5 / 6 × 3 11 / 12 × 2 6 2 6 = \frac { 1 }{ { 2 }^{ 5/6 }\times { 3 }^{ 11/12 } } \times \frac { \sqrt [ 6 ]{ 2 } }{ \sqrt [ 6 ]{ 2 } } =

2 6 2 × 3 11 / 12 = 2 6 2 × 3 11 / 12 × 3 12 3 12 = \frac { \sqrt [ 6 ]{ 2 } }{ 2\times { 3 }^{ 11/12 } } =\frac { \sqrt [ 6 ]{ 2 } }{ 2\times { 3 }^{ 11/12 } } \times \frac { \sqrt [ 12 ]{ 3 } }{ \sqrt [ 12 ]{ 3 } } =

2 6 × 3 12 2 × 3 = 2 6 × 3 12 6 = \frac { \sqrt [ 6 ]{ 2 } \times \sqrt [ 12 ]{ 3 } }{ 2\times 3 } =\frac { \sqrt [ 6 ]{ 2 } \times \sqrt [ 12 ]{ 3 } }{ 6 } =

2 1 / 6 × 3 12 6 = 2 2 / 12 × 3 12 6 = \frac { { 2 }^{ 1/6 }\times \sqrt [ 12 ]{ 3 } }{ 6 } =\frac { { 2 }^{ 2/12 }\times \sqrt [ 12 ]{ 3 } }{ 6 } =

2 2 12 × 3 12 6 = 2 2 × 3 12 6 = \frac { \sqrt [ 12 ]{ { 2 }^{ 2 } } \times \sqrt [ 12 ]{ 3 } }{ 6 } =\frac { \sqrt [ 12 ]{ { 2 }^{ 2 }\times 3 } }{ 6 } =

12 12 6 \frac { \sqrt [ 12 ]{ 12 } }{ 6 }

A + B = 12 + 6 = 18 A + B = 12 + 6 = 18

nice question and good solution buddy...

Shreyansh Choudhary - 5 years, 3 months ago

Log in to reply

Thanks mate

Joao Miguel Coelho - 5 years, 3 months ago
Hung Woei Neoh
Jul 10, 2016

1 181398528 12 = 1 2 10 × 3 11 12 = 2 2 × 3 2 12 × 3 12 12 = 4 × 3 12 6 12 12 = 12 12 6 \sqrt[12]{\dfrac{1}{181398528}}\\ =\sqrt[12]{\dfrac{1}{2^{10}\times3^{11}}}\\ =\sqrt[12]{\dfrac{2^2 \times 3}{2^{12}\times3^{12}}}\\ =\dfrac{\sqrt[12]{4\times 3}}{\sqrt[12]{6^{12}}}\\ =\dfrac{\sqrt[12]{12}}{6}

A = 12 , B = 6 , A + B = 12 + 6 = 18 A=12,\;B=6,\;A+B=12+6=\boxed{18}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...