If $100!=2^m.I$ (Where $"I"$ is multiplication of odd numbers) then find the value of $"m"$
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Forgetting about $I, 2^m = 2\cdot 4\cdot 6.... \cdot 100 = 2\cdot 50!$. Proceed... – true blue anil Oct 25 '17 at 03:57
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@trueblueanil, I didn't get your approach. – User Oct 25 '17 at 04:29
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In general when you want to find the exponent of a prime $p$ in $n!$, you just need top calculate $$\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\cdots + \left[\frac{n}{p^k}\right]$$ where $k$ is the largest positive integer such that $p^k\le n\le p^k+1.$
Note: $[.]$ denotes gretest integer function.
In your case $n=100$ and $p=2$, so you need to evaluate $$\left[\frac{100}{2}\right]+ \left[\frac{100}{2^2}\right] + \left[\frac{100}{2^3}\right]+ \left[\frac{100}{2^4}\right]+\left[\frac{100}{2^5}\right]+\left[\frac{100}{2^6}\right]$$
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