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

A = 2000²⁰⁰², B = 2001²⁰⁰¹, C = 2002²⁰⁰⁰

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) = clog_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 10¹⁰⁰.

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

logA > logB

if and only if A > B. Hence

logA = 2002 log2000 ≈ 2002(3.010) ≈ 6608.662

logB = 2001 log2001 ≈ 6605.795

logC = 2000 log2002 ≈ 6602.928

The approximate difference is given by : logA − logB = logA/B ≈ 3, hence A ≈ 10³B. Similarly B ≈ 10³C. Thus A > B > C.

Here is Koopa Koo's more general result.

Claim: A > B > C

Proof: A > B if and only if logA > logB.

I shall prove logA − logB > 0 i.e. 2002log2000 −2001log2001 > 0.

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

Differentiating this function,

f ' (x) = (x + 2)/x + logx − 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) = logA − logB > 0 and it follows that A > B.

The proof that B > C is similar.