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

22
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23
21

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