$12^\text{th}$ Root

Algebra Level 3

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

This problem was created by José Carlos Pereira

8 18 4 7

2 solutions

Joao Miguel Coelho
Feb 20, 2016

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

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

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

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

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

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

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

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

$A + B = 12 + 6 = 18$

nice question and good solution buddy...

Shreyansh Choudhary - 5 years, 3 months ago

Thanks mate

Joao Miguel Coelho - 5 years, 3 months ago

Hung Woei Neoh
Jul 10, 2016

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

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

1 2 th 12^\text{th} 1 2 th Root

2 solutions

0 pending reports

$12^\text{th}$ Root