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# Giants

**Why do this problem?**

**Possible approach**

**Key questions**

**Possible extension**

### Possible support

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### Telescoping Series

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Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions
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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.

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.

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

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

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

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

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 $\sqrt{2}$?