Short Question, Long Number

here is a solution, that is fundamentally not different from what is provided by @Kaizen Cyrus . However, there are estimations involved, which may influence the accuracy of the final solution.

If we define $LCM(1,2,\dots,n)=f(n)$ , then there is a relation

$e^{\Lambda(n)}=\frac{f(n)}{f(n-1)}$

$\Lambda(n)$ is Mangoldt function.

$e^{\Lambda(n)}\cdot e^{\Lambda(n-1)} \cdot \dots \cdot e^{\Lambda(2)} =\frac{f(n)}{f(n-1)} \cdot \frac{f(n-1)}{f(n-2)} \cdot \dots \cdot \frac{f(2)}{f(1)}=\frac{f(n)}{f(1)}=\frac{f(n)}{1}$

For $n=100$

$e^{\sum_{n=2}^{100} \Lambda(n)}=e^{\sum_{n=1}^{100} \Lambda(n)} =f(100)$

It has been proven that

$\lim _{x \rightarrow \infty }\frac{\sum_{n=1}^{x} \Lambda(n)}{x}=1$

hence $\sum_{n=1}^{x} \Lambda(n)$ can be approximated by $x$ . Therefore, the answer to our question can be approximated to be

$f(100)= e^{\sum_{n=2}^{100} \Lambda(n)} \approx e^{100}$

the number of digits is calculated to be $\log_{10}e^{100}=43$ , which is 2 units away from the correct answer.

To find the LCM of the said sequence of numbers, we must find all largest $p^{n}$ , where $p$ is a prime number and that $p^{n}<100$ , and multiply them.

The $p^{n}$ are:

$\begin{array}{cccccccccc} 2^{6} &&&&&&&&& 43 \\ 3^{4} &&&&&&&&& 47 \\ 5^{2} &&&&&&&&& 53 \\ 7^{2} &&&&&&&&& 59 \\ 11 &&&&&&&&& 61 \\ 13 &&&&&&&&& 67 \\ 17 &&&&&&&&& 71 \\ 19 &&&&&&&&& 73 \\ 23 &&&&&&&&& 79 \\ 29 &&&&&&&&& 83 \\ 31 &&&&&&&&& 89 \\ 37 &&&&&&&&& 97 \\ 41 &&&&&&&&& \end{array}$

The product of the numbers above is $\approx 6.972 × 10^{40}$ , which has $\boxed{41}$ digits.