Just an easy simplification

3 147 7 3 1 3 + 7 1 3 = a b c \large 3 \sqrt{147} - \frac{7}{3} \sqrt{ \frac{1}{3} } + 7 \sqrt{ \frac{1}{3} } = \frac{a}{b} \sqrt{c}

where a a and b b are positive co-primes and c c is not a perfect square .

Find a + b + c \large a + b + c .

This problem is a part of the sets - 1's & 2's & " N " for number theory .


The answer is 215.

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1 solution

3 147 7 3 1 3 + 7 1 3 = 3 7 3 7 3 3 3 + 7 3 3 = 3 9 ( 21 9 7 + 7 3 ) = 203 9 3 = a b c a + b + c = 203 + 9 + 3 = 215 3\sqrt { 147 } -\frac { 7 }{ 3 } \sqrt { \frac { 1 }{ 3 } } +7\sqrt { \frac { 1 }{ 3 } } =3·7\sqrt { 3 } -\frac { 7 }{ 3 } ·\frac { \sqrt { 3 } }{ 3 } +7·\frac { \sqrt { 3 } }{ 3 } =\frac { \sqrt { 3 } }{ 9 } \left( 21·9-7+7·3 \right) =\frac { 203 }{ 9 } \sqrt { 3 } =\frac { a }{ b } \sqrt { c } \Rightarrow a+b+c=203+9+3=215

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