Adding Two Large Products?

So we know that

$\begin{aligned} 27! &=& 1\times 2\times3\times\cdots \times 27 \\ 28! &=& 1\times 2\times3\times\cdots \times 27\times28 \end{aligned}$

And also, $28! = 28 \times 27!$ , so adding these numbers gives

$\begin{aligned} 27! + 28! &=& 27! + 28 \times27! \\ &=& 27! \times (1 + 28) \\ &=& 27! \times 29 \end{aligned}$

This tells us that we want to find the maximum prime factor of the product, $27! \times 29$ . Since all the prime factors of $27!$ must be less than or equals to 27, and 29 is already a prime number, this tells us that $\boxed{29}$ is indeed the largest prime factor.

$27!+28! = 27!+28*27!$

then it's equal to , $27!(1+28) = 27!*29$

Thus, 29 is the $largest prime factor$ .

If n+1 = m, n+2 = j, and j is a prime, so finding the largest prime divisor for n! x m! is j