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

Note by Biswaroop Roy 4 years, 9 months ago

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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.

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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:)

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

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`\sin \theta`

`\boxed{123}`

## Comments

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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.

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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.

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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 \( 3^{n+2} \mid 10^{3^n} - 1 \).

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

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

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