Which number amongst 2⁴⁰, 3²¹, 4¹⁸ and 8¹² is the smallest?

The correct option is 3²¹.

To find the smallest number among the given options, we can rewrite the numbers with bases that are powers of 2 so they can be easily compared. Let us write down all four numbers:
1. 2⁴⁰
2. 3²¹
3. 4¹⁸
4. 8¹²

First, let us express 4¹⁸ and 8¹² as powers of base 2:

Since 4 = 2², we have:

$$4^{18}={\left(2^2\right)}^{18}=2^{2\times 18}=2^{36}$$

Since 8 = 2³, we have:

$$8^{12}={\left(2^3\right)}^{12}=2^{3\times 12}=2^{36}$$

Now we can compare the three powers of 2: 2⁴⁰, 2³⁶, and 2³⁶. Clearly, the smallest among these is:
2³⁶ (which represents both 4¹⁸ and 8¹²).

Next, we compare 2³⁶ with the remaining number, 3²¹. To compare these two numbers, we can look for a common power or approximate them. Let us express their exponents using a common factor. Since the exponents are 36 and 21, their greatest common divisor is 3. We can rewrite them as:

$$2^{36}={\left(2^{12}\right)}^3$$

$$3^{21}={\left(3^7\right)}^3$$

Now we just need to compare the base values inside the parentheses, which are 2¹² and 3⁷:

Let us calculate their values:
2¹² = 4096
3⁷ = 3⁶ × 3 = 729 × 3 = 2187

Comparing the two results:

$$2187<4096$$

This means:

$$3^7<2^{12}$$

Raising both sides to the power of 3:

$${\left(3^7\right)}^3<{\left(2^{12}\right)}^3$$

Therefore, 3²¹ < 2³⁶.

Since 3²¹ is smaller than 2³⁶ (which equals 4¹⁸ and 8¹²) and 2³⁶ is smaller than 2⁴⁰, the smallest number among all the choices is 3²¹.