Let m, n in N$ and gcd (2, n)=1. If


Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If

$30\left(\begin{array}{c}30 \\ 0\end{array}\right)+29\left(\begin{array}{c}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)=\mathrm{n} \cdot 2^{\mathrm{m}}$

then $\mathrm{n}+\mathrm{m}$ is equal to


Let $S=\sum_{r=0}^{30}(30-r)^{30} C_{r}$

$=30 \sum_{r=0}^{30}{ }^{30} C_{r}-\sum_{r=0}^{30} r^{30} C_{r}$

$=20 \times 2^{30}-\sum_{r=1}^{30} r \cdot \frac{30}{4} \cdot{ }^{29} \mathrm{C}_{r-1}$

$=30 \times 2^{30}-30 \cdot 2^{29}$

$=(30 \times 2-30) \cdot 2^{29}=30 \cdot 2^{29} \Rightarrow 15 \cdot 2^{30}$

$=n=15, \quad m=30$


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