# Prove that there are 3^n n-digit numbers formed with 4, 5 ,6

## Homework Statement

Prove that there are 3^n n-digit numbers formed with the numbers 4, 5, 6.
The digits can be repeated, e.g.: 444.

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## The Attempt at a Solution

What I thought was to prove it by induction. So $P$(1) is true, because we have only 4, 5, 6.

But it seems to me unreasonable to say $P$(k)$=> P$(k+1)

any help is appreciated

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## Homework Statement

Prove that there are 3^n n-digit numbers formed with the numbers 4, 5, 6.
The digits can be repeated, e.g.: 444.

-

## The Attempt at a Solution

What I thought was to prove it by induction. So $P$(1) is true, because we have only 4, 5, 6.

But it seems to me unreasonable to say $P$(k)$=> P$(k+1)

any help is appreciated
Hi tonit!

So suppose there are ##3^k## numbers with k digits.
The numbers with (k+1) digits would be one digit longer.
That is, they would have one more digit to the left.

How many numbers of (k+1) digits could you make, given that the last k digits are fixed?

that would be $3^k * 3 = 3$k+1

so is all this enough, or does the problem need other explanations. ?

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A proof by full induction involves 2 steps.

The boundary condition that ##P_{(1)}## is true.

The induction hypothesis that ##P_{(k)}## is true, followed by the proof that it implies that ##P_{(k+1)}## is true as well.

You just showed both.
So yes, that is enough.

ok thanks :D