3 2 39 4 = 3 α p β {3}^2{39}^{4}={3}^\alpha \cdot p^\beta

Algebra Level 1

If 3 2 39 4 = 3 α p β , {3}^2{39}^{4}={3}^\alpha \cdot p^\beta, where p p is prime number, what is p + α + β ? p+\alpha+\beta?

22 24 23 21

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1 solution

Abhishek Singh
Mar 22, 2014

3 2 3 9 4 3^{2} 39^{4} , can be written as 3 2 ( 3 × 13 ) 4 3^{2} (3 \times 13)^{4} which gives, 3 2 + 4 × 1 3 4 3^{2+4} \times 13^{4} . Hence p = 13 , α = 6 , β = 4 p=13,\alpha =6, \beta = 4 hence p + α + β = 23 p+\alpha+\beta =\boxed{23}

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