$\binom{2.10^n}{10^n}$ is divisible by 10 but not by 100 $\forall n\geq 2$ I know that $\binom{2n}{n}$ is divisible by $n+1$. But i am nowhere near the given problem. Any help would be appreciated. Thank you.
2026-04-07 21:22:05.1775596925
$\binom{2.10^n}{10^n}$ is divisible by 10 but not by 100
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The highest exponent of $5$ that divides $c\cdot10^n$ where integer $c\le4$
will be $$f(c)=2c(1+10+\cdots+10^{n-1}=\dfrac{2c(10^n-1)}{10-1}$$
The denominator $=f^2(1)$
What about the numerator?