What is the largest integer n such that 3^n is a factor of 1*3*5*...*97*99?
\(\begin{array}{|rcll|} \hline 99!! &=& 1\cdot 3 \cdot 5\cdot 7\cdot \ldots \cdot 95\cdot 97\cdot 99 \\\\ &&\text{Formula:}~\boxed{(2n-1)!! = \frac{(2n)!}{2^n\cdot n!}} \\\\ && (2\cdot 50 - 1)!! = \dfrac{(2\cdot 50)!} {2^{50}\cdot 50! }\\\\ 99!! &=& \dfrac{100!} {2^{50}\cdot 50! } \\ \hline \end{array}\)
100!
The factor \(3^i\)
\(\begin{array}{|rcll|} \hline \frac{100}{3} &=& [33].\overline{3} \\ \frac{100}{3^2} &=& [11].\overline{1} \\ \frac{100}{3^3} &=& [3].\overline{703} \\ \frac{100}{3^4} &=& [1].\overline{234567901} \\ \frac{100}{3^5} &=& 0 \\ \hline 33+11+3+1 &=& 48 \\ i &=& 48 \\ \hline \end{array} \)
50!
The factor \(3^j\)
\(\begin{array}{|rcll|} \hline \frac{50}{3} &=& [16].\overline{6} \\ \frac{50}{3^2} &=& [5].\overline{5} \\ \frac{50}{3^3} &=& [1].\overline{851} \\ \frac{50}{3^4} &=& 0 \\ \hline 16+5+1 &=& 22 \\ j &=& 22 \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline && \dfrac{3^{48}}{3^{22}} \\\\ &=& 3^{48-22} \\ &=& 3^{26} \\ \hline \end{array} \)
\(\text{The largest integer $n$ such that $3^n$ is a factor of $1*3*5*...*97*99$ is $\mathbf{26}$ }\)
