Hint : Induction. :)
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Zi Song Yeoh
·
3 years, 8 months ago

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@Zi Song Yeoh
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As Zi Song has pointed out, direct induction is one of the easiest ways to go. You can also recognize that \[a_{n} = a_{n-1} \times \left(10^{2 \cdot 3^n} + 10^{3^n} + 1 \right) = a_{n-1} \times \left( \left(10^{3^n} \right)^2 + 10^{3^n} + 1\right)\]
It should be easy to show that \(3\) divides \(\left( \left(10^{3^n} \right)^2 + 10^{3^n} + 1\right)\). Hence, you can now conclude that \(3a_{n-1}\) divides \(a_n\).
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Marvis Narasakibma
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3 years, 8 months ago

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I came to the same solution as that of Marvis N. Question's not difficult, just presence of mind is needed!
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Siddharth Kumar
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3 years, 8 months ago

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this type of questions i often come across
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Sayan Chaudhuri
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3 years, 8 months ago

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but i think solution may be somehow critical
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Sayan Chaudhuri
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3 years, 8 months ago

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I DONT NOTHING ABOUT HOW TO SOLVE THIS
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Sayan Chaudhuri
·
3 years, 8 months ago

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TopNewestHint : Induction. :) – Zi Song Yeoh · 3 years, 8 months ago

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– Marvis Narasakibma · 3 years, 8 months ago

As Zi Song has pointed out, direct induction is one of the easiest ways to go. You can also recognize that \[a_{n} = a_{n-1} \times \left(10^{2 \cdot 3^n} + 10^{3^n} + 1 \right) = a_{n-1} \times \left( \left(10^{3^n} \right)^2 + 10^{3^n} + 1\right)\] It should be easy to show that \(3\) divides \(\left( \left(10^{3^n} \right)^2 + 10^{3^n} + 1\right)\). Hence, you can now conclude that \(3a_{n-1}\) divides \(a_n\).Log in to reply

I came to the same solution as that of Marvis N. Question's not difficult, just presence of mind is needed! – Siddharth Kumar · 3 years, 8 months ago

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this type of questions i often come across – Sayan Chaudhuri · 3 years, 8 months ago

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but i think solution may be somehow critical – Sayan Chaudhuri · 3 years, 8 months ago

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I DONT NOTHING ABOUT HOW TO SOLVE THIS – Sayan Chaudhuri · 3 years, 8 months ago

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