Let m,n N and (2,n)=1.

Question

Let $m,n\in N$ and $\gcd(2,n)=1$. If $\displaystyle{30\choose0}30^{28}+{30\choose28}\cdots+2\binom{30}{28}+1\binom{30}{29}=n\cdot2^m$ (i.e. $\displaystyle\sum_{r=0}^{29}(30-r)\binom{30}{r}=n\cdot2^m$), then $n+m$ is ...

Answer: 45

Accepted Answer

$$\sum_{r=0}^{30}(30-r)\binom{30}{r} = 30\sum_{r=0}^{30}\binom{30}{r}-\sum_{r=0}^{30}r\binom{30}{r} = 30\times2^{30}-30\times2^{29}=30\times2^{29}(2-1)=30\times2^{29}$$

$$=15\times2^{30}\quad\Rightarrow\quad n=15,\ m=30\quad\Rightarrow\quad n+m=45$$

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