What is the smallest number that leaves a remainder of $4$ on division by $5$, $5$ on division by $6$, $6$ on division by $7$, and $7$ on division by $8$?

The correct option is C $839$
Let the number be $N.$
When a number is divided by $5$, a remainder of $4$ can be thought of as a remainder of $-1$.
So, $N=5a-1$ or $N+1=5a$
Similarly,
$N=6b-1$ or $N+1=6b$
$N=7c-1$ or $N+1=7c$
$N=8d-1$ or $N+1=8d$
$N+1$ can be expressed as a multiple of $(5,6,7,8)$
Now, smallest value of $N+1=$ LCM of $(5,6,7,8)$
$\Rightarrow$ Smallest value of $N=$ LCM of $(5,6,7,8)-1$
$=840-1=839$