Sum of Multinomial Coefficients!

Putting n=2020 , m=24 and all ai =1 in the multinomial theorem we get:

$(\sum_{i=1}^m a_{i} )^{n}= \sum_{\sum_{i=1}^m j_{i} =n} \frac{n!}{\prod_{i=1}^m j_{i} !}\prod_{i=1}^m (a_{i})^{j_{i}}$

$(\sum_{i=1}^{24} 1 )^{2020}= \sum_{\sum_{i=1}^{24} j_{i} =2020} \frac{2020!}{\prod_{i=1}^{24} j_{i} !}\prod_{i=1}^{24} (1)^{j_{i}}$

$24^{2020}= \sum_{\sum_{i=1}^{24} j_{i} =2020} \frac{2020!}{\prod_{i=1}^{24} j_{i} !} \times 1$ and $24^{even}$ has the last two digits as 76 which is the answer.

Hope you enjoyed the problem :)