When we calculate the HCF of two numbers by the long division method, the last divisor is 2 and the successive quotients from first to last are 2, 4, and 2 respectively. Find these two numbers.

Working backwards using the long division steps:
Let quotients be $q_1=2$, $q_2=4$, $q_3=2$, and the last divisor $d_3=2$.

* Step 3 (Last Step):
$$\text{Remainder}=0$$

$${\text{Dividend}}_3={\text{Divisor}}_3\times q_3+0=2\times 2=4$$

* Step 2: The dividend of step 3 becomes the divisor of step 2.

$${\text{Divisor}}_2=4,{\text{Remainder}}_2={\text{Divisor}}_3=2$$

$${\text{Dividend}}_2={\text{Divisor}}_2\times q_2+{\text{Remainder}}_2=4\times 4+2=18$$

* Step 1: The dividend of step 2 becomes the divisor of step 1.

$${\text{Divisor}}_1=18,{\text{Remainder}}_1={\text{Divisor}}_2=4$$

$${\text{Dividend}}_1={\text{Divisor}}_1\times q_1+{\text{Remainder}}_1=18\times 2+4=40$$

The two numbers are the first divisor (18) and the first dividend (40).