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Prove 6^n - 1 is divisible by 5 by induction

Step by step, with the reason for every step · Induction · Discrete Math

Step by step

The question

Question: prove by induction that 6ⁿ − 1 is divisible by 5, for every whole number n.

6n − 1 is divisible by 5

Like dominoes: show it for n = 1, then show that k passes it on to k + 1.

1 Check n = 1

For n = 1: 5, and that is 5 × 1.

Divisible by 5: the first domino falls.

61 − 1 = 5 = 5 × 1

2 Assume it works for k

Suppose it works for k: 6ᵏ − 1 is a multiple of 5.

So it is 5 times some whole number.

Call it m.

6k − 1 = 5m
m is a whole number

Move the rest to the other side: 6ᵏ equals 5m + 1.

We will need this.

6k = 5m + 1

3 What we must show

We must show the same for k + 1: every n becomes k + 1.

= 6k + 1 − 1 is divisible by 5

Careful!

One more power means one more times the number.

It is 6 times as big, not 1 more.

Careful: not 6k + 1 = 6k + 1 ✗

4 Use the assumption

One more power: 6 times the power with k.

6k + 1 − 1 = 6 × 6k − 1
6k + 1 = 6 × 6k

Now swap in 5m + 1 for 6ᵏ.

= 6(5m + 1) − 1

Open the bracket.

= 30m + 6 − 1

Like terms together.

= 30m + 5

5 Take out 5

Every term has a 5 in it.

Take out 5: 5(6m + 1).

= 5(6m + 1)
6m + 1 is a whole number

So it works for k + 1 too.

True for n = 1, and passed on each time: true for every whole number n.

The proof is done.

divisible by 5 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 215.

n = 3: 63 − 1 = 215

215 is 5 times 43.

Divisible by 5.

Right!

215 = 5 × 43

Let's look back.

First: check n = 1.

Then: assume it works for k.

Then: what we must show.

Then: use the assumption.

Then: take out 5.

Last: check.

Answer
6m + 1 is a whole number

Your turn

Try this one yourself, then check:

23n − 1 is divisible by 7
Show the answer
7(8m + 1)

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