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Let $m, n \in \mathbb{N}$ and $\gcd(2, n) = 1$. If $30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$, then $n + m$ is equal to
$S = 30 \cdot 2^{30} - 30 \cdot 2^{29} = 15 \cdot 2^{30}$. So $n = 15, m = 30$, $n + m = 45$.
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