Which is the biggest and which the smallest of these numbers and how do they compare in magnitude?

$A = 2000^{2002}, B = 2001^{2001}, C = 2002^{2000}$

This solution comes from Ilham Kurnia, age 18, 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

$y = floor (\log_{a} (x)) + 1$

.

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

$\log A > \log B$

if and only if $A > B$ . Hence

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

$\log B = 2001 \log 2001 \approx 6605.795$

$\log C = 2000 \log 2002 \approx 6602.928$

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,

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

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.