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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John Ashley Capellan
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3 years, 8 months ago

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@John Ashley Capellan
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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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Yong See Foo
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3 years, 8 months ago

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@Yong See Foo
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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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Calvin Lin
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3 years, 8 months ago

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@Calvin Lin
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Yeah that's true. Or even better as you mentioned, \(v_3(10^{3^n}-1)=n+2\) or \(3^{n+2}||10^{3^n}-1\).
–
Yong See Foo
·
3 years, 8 months ago

## 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. – John Ashley Capellan · 3 years, 8 months ago

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– Yong See Foo · 3 years, 8 months ago

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

As a side note, another way of stating the question is that \( 3^{n+2} \mid 10^{3^n} - 1 \). – Calvin Lin Staff · 3 years, 8 months ago

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– Yong See Foo · 3 years, 8 months ago

Yeah that's true. Or even better as you mentioned, \(v_3(10^{3^n}-1)=n+2\) or \(3^{n+2}||10^{3^n}-1\).Log in to reply

– John Ashley Capellan · 3 years, 8 months ago

Thanks for clarifying! I also didn't notice that...Log in to reply

– Biswaroop Roy · 3 years, 8 months ago

I just solved this same problem using algebra.Would you like to try?Log in to reply

– John Ashley Capellan · 3 years, 8 months ago

Sure thing..Log in to reply

– Biswaroop Roy · 3 years, 8 months ago

Just a hint,you might find geometric progression an useful tool:)Log in to reply

– Biswaroop Roy · 3 years, 8 months ago

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