Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# Giants

## You may also like

### Telescoping Series

### Climbing Powers

Links to the University of Cambridge website
Links to the NRICH website Home page

Nurturing young mathematicians: teacher webinars

30 April (Primary), 1 May (Secondary)

30 April (Primary), 1 May (Secondary)

Or search by topic

Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

Note that you don't need to work out the numbers exactly to decide which is bigger!

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 $9900$?

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 $9900$?

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?