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Consider the sequence \(a_1 = 101, a_2 = 10101, a_3 = 1010101 \) and so on. Prove that \(a_k\) is composite iff \(k\geq 2\).

Note by D K 2 years ago

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Aha I see it. Any even one a sub 2k can be written in the form (a sub k)^2-100^k which neatly factors to (a sub k+10^k)(a sub k-10^k) for all evens. Sorry for my latex. And the odd case is trivial.

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I have another unanswered discussion question. Can you take a look at it?

Nice observation. Kudos.

That's interesting. I can show that if \(k\) is composite, then \(a_k\) is composite and has a factor of 101...101.

you mean if k is odd ? then ak will have 101 as a factor. but I also need proof for when k is even.

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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

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TopNewestAha I see it. Any even one a sub 2k can be written in the form (a sub k)^2-100^k which neatly factors to (a sub k+10^k)(a sub k-10^k) for all evens. Sorry for my latex. And the odd case is trivial.

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I have another unanswered discussion question. Can you take a look at it?

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Nice observation. Kudos.

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That's interesting. I can show that if \(k\) is composite, then \(a_k\) is composite and has a factor of 101...101.

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you mean if k is odd ? then ak will have 101 as a factor. but I also need proof for when k is even.

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