# Giants

## Problem

Which is bigger: $9^{10}$ or $10^{9}$?

Now find a way to use your calculator to compare $99^{100}$ and $100^{99}$.

Work out which is bigger out of $999^{1000}$ and $1000^{999}$.

This challenge calls for some experimentation with numbers and for some ingenuity in finding a solution. There are several possible methods of solution.

Optional extension challenge: If you solved the above, you might wish to consider which is bigger for the same sort of problem with a billion 9s and a billion 0s.

## Getting Started

Some possible methods:

1) Can you split up the powers into smaller pieces, or consider another quantity which your calculator can handle from which you can deduce the answer?

or 2) Can you use inequalities and logic such as 'if X > Y and Y > Z then we know that X > Z'

or 3) What happens when you raise $100^{1/100}$ to the power $9900$?

## Student Solutions

Antoine from Lycée St Genes submitted the following solution:

I checked with the calculator and saw that $9^{10}>10^9$.

Obviously we can't do that for the next questions.

I decided to deal with logarithms:

$log\, 99^{100}=100\, log\, 99=199.56$

$log\, 100^{99}=99\, log\, 100=198$

$log\, 99^{100}>log\, 100^{99}\,$

Therefore, $99^{100}>100^{99}$.

Same for $999^{1000}$ and $1000^{999}$:

$log\, 999^{1000}=1000\, log\, 999=2999.56$

$log\, 1000^{999}=999\, log\, 1000=2997$

$log\, 999^{1000}>log\, 1000^{999}$

Therefore, $999^{1000} > 1000^{999}$.

Michael sent in this more general approach:

My approach not only works for large numbers, but can also be applied to smaller numbers too.

$$x^y>y^x \iff ln\, x^y > ln\, y^x \iff y\, ln\, x > x\, ln\, y \iff \frac{ln\, x}{x} > \frac{ln\, y}{y}$$

Let $f(n) = \frac{ln\, n}{n}$. Hence $f(x) > f(y) \iff x^y > y^x$.

Differentiating $f(n)$ with respect to $n$ using the Quotient Rule,

$$f'(n) = \frac{n \times \frac{1}{n} - ln\, n \times 1}{n^2} = \frac{1-ln\, n}{n^2}$$

Hence the function has a stationary point at $ln\, n=1$, i.e. at $n = e$, $f(n) = e^{-1}$, and no others.

More importantly, it shows that $f(n)$ is a decreasing function for $n>e$, since this is where $f'(n) = \frac{1-ln\, n}{n^2}$ is negative.

Hence $$e \leq x < y \Rightarrow f(x) > f(y)$$

And so $$e \leq x < y \Rightarrow x^y > y^x$$

So $$9^{10} > 10^9,$$

$$99^{100} > 100^{99},$$

$$999^{1000} > 1000^{999},$$

$$...,$$

$$(a\, billion\, 9s)^{(1\, and\, a\, billion\, 0s)} > (1\, and\, a\, billion\, 0s)^{(a\, billion\, 9s)},$$

and so on.

Here is another neat way of thinking about the problem:

It can be shown that $e$ is the limit as $n \to \infty$ of the increasing sequence $(1+\frac{1}{n})^n$. So we have $$e>(1+\frac{1}{n})^n.$$

Then if $n>e$, $$n>(1+\frac{1}{n})^n$$

So $$n>\frac{(n+1)^n}{n^n}$$

and $$n^{n+1}>(n+1)^n$$

So for all natural numbers $n$ greater than $e$, we can see straight away that $n^{n+1}>(n+1)^n$.

## Teachers' Resources

**Why do this problem?**

This problem challenges students to solve a problem that is too big to fit on their calculator, so they will need to draw together what they know about indices.

**Possible approach**

Start by posing the question "Which is bigger, $10^9$ or $9^{10}?$" Students will probably reach for their calculators. Then pose the question "Which is bigger, $100^{99}$ or $99^{100}$? Your calculator probably can't cope with numbers that big, so you'll have to apply some of your knowledge about powers to help you..."

Give students some time to explore, circulating the classroom to look for good ideas to share with the rest of the group. Once students have come up with some methods, invite them up to the board to share what they did, focussing on the way they manipulate indices.

**Key questions**

What happens when you raise $100^{1/100}$ to the power 9900?

Consider $(\frac{99}{100})^{99}$. How could this help?

**Possible extension**

Students who have met logarithms could be offered a range of "Which is bigger" questions to tackle.

### Possible support

Tens provides some gentle revision of indices in the form of a simpler challenge.