Mathematics - Permutation & Combination Question with Solution | TestHub

$\lfloor 15\rfloor =2^{p_1}{3}^{p_2}{5}^{p_3}{7}^{p_4}11^{p_5}13^{p_6}$ replaced by $15!=2^{p_1}{3}^{p_2}{5}^{p_3}{7}^{p_4}11^{p_5}13^{p_6}$.

Then $p_1=\lfloor \frac{15}{2}\rfloor +\lfloor \frac{15}{4}\rfloor +\lfloor \frac{15}{8}\rfloor =7+3+1=11$

$p_2=\lfloor \frac{15}{3}\rfloor +\lfloor \frac{15}{9}\rfloor =5+1=6$

$p_3=\lfloor \frac{15}{5}\rfloor =3$

$p_4=\lfloor \frac{15}{7}\rfloor =2$

$p_5=\lfloor \frac{15}{11}\rfloor =1$

$p_6=\lfloor \frac{15}{13}\rfloor =1$

${\sum }_{r=1}^6{p}_r=11+6+3+2+1+1=24$

Final Answer: The final answer is $\boxed{24}$