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

[tonit]

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.

Homework Equations

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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

 

[I like Serena]

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.

Homework Equations

-

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?

 

[tonit]

that would be $3^k * 3 = 3$^k+1
thanks btw for the reply.

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

 

[I like Serena]

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.

 

[tonit]

ok thanks :D