Question:
Prove that LCM $\{6 !, 7 !, 8 !\}=8 !$
Solution:
To Prove : LCM $\{6 !, 7 !, 8 !\}=8 !$
Formula: $n !=n \times(n-1) !$
LCM is the smallest possible number that is a multiple of two or more numbers.
Here, we observe that (8!) is the first number which is a multiple of all three given numbers i.e. $6 !, 7 !$ and $8 !$.
$1 \times(8 !)=8 !$
$8 \times(7 !)=8 !$
$8 \times 7 \times(6 !)=8 !$
Therefore, $8 !$ is the LCM of $\{6 !, 7 !, 8 !\}$
Conclusion : Hence proved