Out of four numbers, the average of the first three is $15$ and that of the last three is $16$. If the last number is $19$, then the first number is:

Let the four numbers be $a,b,c,$ and $d$.
According to the first condition, the average of the first three numbers is $15$:

$$\frac{a+b+c}{3}=15⟹a+b+c=45\text{— (Equation 1)}$$

According to the second condition, the average of the last three numbers is $16$:

$$\frac{b+c+d}{3}=16⟹b+c+d=48\text{— (Equation 2)}$$

We are given that the last number $d=19$. Substitute $d=19$ into Equation 2:

$$b+c+19=48$$

$$b+c=48-19=29$$

Now substitute the value of $(b+c)=29$ into Equation 1:

$$a+29=45$$

$$a=45-29=16$$

Thus, the first number is $16$.