${\displaystyle A={2000}^{2002},\hspace{0.5em}B={2001}^{2001},\hspace{0.5em}C={2002}^{2000}}$

This solution comes from Ilham, St. Patrick's College, Wellington, well done and thank you Ilham.

First let's define the function floor( $x$), where $x$ is a real number, such that floor( $x$) = the integer part of $x$.

Let

${\displaystyle y=\mathrm{floor}({\log }_a(x))+1}$

.

As a general rule, y will be the number of digits of x in base a. If we reverse this, we can say that $x$ is somewhere between $a^y$ and $a^{y+1}$.

Another basic rule is ${\log }_a(b^c)=c{\log }_a(b)$. If we don't use this rule, the calculation cannot be handled using any standard scientific calculators, as they can't handle calculation with numbers greater than ${10}^{100}$.

If we use these two rules to A, B and C in base 10, it will show that A has 6609 digits, B has 6606 digits, and C has 6603 digits in base 10.

Therefore, A is bigger than B which in turn is bigger than C. A is the biggest, and C is the smallest.

A similar solution uses the fact that the logarithm function is an increasing function so it follows that

${\displaystyle \log A>\log B}$

if and only if $A>B$. Hence

${\displaystyle \log A=2002\log 2000\approx 2002(3.010)\approx 6608.662}$

${\displaystyle \log B=2001\log 2001\approx 6605.795}$

${\displaystyle \log C=2000\log 2002\approx 6602.928}$

The approximate difference is given by : $\log A-\log B=\log A/B\approx 3$, hence $A\approx {10}^{3}B$. Similarly $B\approx {10}^{3}C$. Thus $A>B>C.$

Here is Koopa Koo's more general result.

Claim: $A>B>C$

Proof: $A>B$ if and only if $\log A>\log B.$

I shall prove $\log A-\log B>0$ i.e. $2002\log 2000-2001\log 2001>0.$

Let $f(x)=(x+2)\log x-(x+1)\log (x+1)$ so that for example f(2) = 4log2 - 3log3.

Differentiating this function,

${\displaystyle f\text{'}(x)=(x+2)/x+\log x-1-\log (x+1)=2/x-\log [(x+1)/x].}$

This derivative is positive if and only if $e^{2/x}>(x+1)/x.$

Using $e^y>1+y$ for all $y$, let $y=2/x$.

We have $e^{2/x}>1+2/x=(x+2)/x>(x+1)/x$.

So the function f is increasing, in particular, $f(2000)=\log A-\log B>0$ and it follows that $A>B$.

The proof that $B>C$ is similar.