jmc

A term of the binomial expansion takes the form $$a_k=\left(\genfrac{}{}{0ex}{}{31}{k}\right){\left(\frac{1}{2}\right)}^k,$$where $0\le k\le 31.$ To find which $a_k$ is the largest, we evaluate the ratio $\frac{a_{k+1}}{a_k}$: $$\frac{a_{k+1}}{a_k}=\frac{\left(\genfrac{}{}{0ex}{}{31}{k+1}\right){\left(\frac{1}{2}\right)}^{k+1}}{\left(\genfrac{}{}{0ex}{}{31}{k}\right){\left(\frac{1}{2}\right)}^k}=\frac{\frac{31!}{(k+1)!(30-k)!}{\left(\frac{1}{2}\right)}^{k+1}}{\frac{31!}{k!(31-k!)}{\left(\frac{1}{2}\right)}^k}=\frac{31-k}{2(k+1)}.$$Now, the inequality $\frac{31-k}{2(k+1)}>1$ is equivalent to $31-k>2k+2,$ or $k<\frac{29}{3},$ or $k\le 9.$ Furthermore, $\frac{31-k}{2(k+1)}<1$ when $k>\frac{29}{3},$ or $k\ge 10.$ Therefore, $a_{k+1}>a_k$ for $k\le 9$ and $a_{k+1}<a_k$ for $k\ge 10.$ It follows that $a_{10}$ is the largest term of the binomial expansion. We have $$a_{10}=\left(\genfrac{}{}{0ex}{}{31}{10}\right){\left(\frac{1}{2}\right)}^{10},$$so it suffices to find the power of $2$ in the prime factorization of $\left(\genfrac{}{}{0ex}{}{31}{10}\right).$ We have $$\left(\genfrac{}{}{0ex}{}{31}{10}\right)=\frac{31\cdot 30\cdot 29\cdot 28\cdot 27\cdot 26\cdot 25\cdot 24\cdot 23\cdot 22\cdot 21}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=\frac{A\cdot 2^8}{B\cdot 2^8}=\frac{A}{B},$$where $A$ and $B$ are odd integers. Therefore, $\left(\genfrac{}{}{0ex}{}{31}{10}\right)$ is odd, and so the answer is $2^{10}=\boxed{1024}.$