2022 CSAT — Q9

Quant Number theory 2.5 marks Medium

Which number amongst $2^{40}$ , $3^{21}$ , $4^{18}$ and $8^{12}$ is the smallest?

A $2^{40}$
B $3^{21}$ Answer
C $4^{18}$
D $8^{12}$

Worked rationale

Reduce everything to base $2$ where possible:

4^{18} = (2^2)^{18} = 2^{36},\qquad 8^{12} = (2^3)^{12} = 2^{36}.

So $4^{18} = 8^{12} = 2^{36}$ , and $2^{40} > 2^{36}$ . Three of the four are powers of $2$ ; the smallest among them is $2^{36}$ .

Now compare $3^{21}$ against $2^{36}$ . Bring to a common exponent — use exponent $3$ :

3^{21} = (3^7)^3 = 2187^3,\qquad 2^{36} = (2^{12})^3 = 4096^3.

Since $2187 < 4096$ , we get $3^{21} < 2^{36} = 4^{18} = 8^{12} < 2^{40}$ .

Answer: (b) $3^{21}$ .

Why the other options miss

A

solved the wrong question: picks the largest when asked for the smallest.
C

stopped comparing too early: stops after reducing $4^{18}=8^{12}=2^{36}$ and forgets to beat it against $3^{21}$ .
D

stopped comparing too early: same as (c); treats $2^{36}$ as the floor without testing the lone base- $3$ term.

Specialist insight

Two moves crack power-comparison items: first collapse the common base ( $4^{18}=8^{12}=2^{36}$ removes two options at a stroke), then equalise exponents to compare the odd one out ( $3^{21}$ vs $2^{36}$ via the cube $2187^3$ vs $4096^3$ ). Never estimate magnitudes blindly — a shared exponent makes the comparison a single inequality between bases.

The trap, in one line

$4^{18}=8^{12}=2^{36}$ ; $3^{21}=2187^3 < 4096^3 = 2^{36}$ , so $3^{21}$ is least $\Rightarrow$ (b).

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