New user? Sign up

Existing user? Log in

Show that the number 111...11 with 3^n digits is divisible by 3^n.

Note by Biswaroop Roy 4 years, 5 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

We prove this by induction:

Base case:

When n = 0, it is obviously true that 1 is divisible by 1.

When n = 1:

The number is 111 which is divisible by 3.

Inductive Case: Suppose the number 1111.....111 (which has 3^n digits) is divisible by 3^n.

For 3^(n+1) digits of 1, The sum of digits for this case is always a power of 3. Hence, the assertion is true.

Numerical Proof: 3^(n+1) divides 111....111 (which has 3^(n+1) digits)

(3^n)(3) divides 111....111 (which has (3^n)(3) = 3^n + 3^n + 3^n digits)

Remembering the case for n: the number 1111.....111 (which has 3^n digits) is divisible by 3^n and the base case, the proof is done.

Log in to reply

Just to clarify your induction, as it is a bit unclear, \(111\cdots 1(3^{n+1}\) digits \()=111\cdots 1(3^n\) digits \()\times (10^{2n}+10^n+1)\). Since \((10^{2n}+10^n+1)\) is divisible by 3, by the inductive hypothesis we finish the induction.

Nicely expressed. This also shows that \(3^n\) is the largest power of 3 that divides \(111\ldots 1\).

As a side note, another way of stating the question is that \( 3^{n+2} \mid 10^{3^n} - 1 \).

@Calvin Lin – Yeah that's true. Or even better as you mentioned, \(v_3(10^{3^n}-1)=n+2\) or \(3^{n+2}||10^{3^n}-1\).

Thanks for clarifying! I also didn't notice that...

@John Ashley Capellan – I just solved this same problem using algebra.Would you like to try?

@Biswaroop Roy – Sure thing..

@John Ashley Capellan – Just a hint,you might find geometric progression an useful tool:)

Thank you very much,both of you. I should have solved myself. Thanks again:)

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe prove this by induction:

Base case:

When n = 0, it is obviously true that 1 is divisible by 1.

When n = 1:

The number is 111 which is divisible by 3.

Inductive Case: Suppose the number 1111.....111 (which has 3^n digits) is divisible by 3^n.

For 3^(n+1) digits of 1, The sum of digits for this case is always a power of 3. Hence, the assertion is true.

Numerical Proof: 3^(n+1) divides 111....111 (which has 3^(n+1) digits)

(3^n)(3) divides 111....111 (which has (3^n)(3) = 3^n + 3^n + 3^n digits)

Remembering the case for n: the number 1111.....111 (which has 3^n digits) is divisible by 3^n and the base case, the proof is done.

Log in to reply

Just to clarify your induction, as it is a bit unclear, \(111\cdots 1(3^{n+1}\) digits \()=111\cdots 1(3^n\) digits \()\times (10^{2n}+10^n+1)\). Since \((10^{2n}+10^n+1)\) is divisible by 3, by the inductive hypothesis we finish the induction.

Log in to reply

Nicely expressed. This also shows that \(3^n\) is the largest power of 3 that divides \(111\ldots 1\).

As a side note, another way of stating the question is that \( 3^{n+2} \mid 10^{3^n} - 1 \).

Log in to reply

Log in to reply

Thanks for clarifying! I also didn't notice that...

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Thank you very much,both of you. I should have solved myself. Thanks again:)

Log in to reply