Let \(m_r\) be the reflection of \(m\). For example, \( 1234_r = 4321 \).

The positive integer k has the property,

\(\forall m\in\mathbb{N},k|m \implies k|m_r\).

Show that, \( k \mid 99 \).

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## Comments

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TopNewestLets take 4 digit \(abcd\).

So \(k|1000d+100c+10b+a\) & \(k|1000a+100b+10c+d\)

\(\implies k|999d+90c-90b-999a\)

\(\implies k|9\).

Again

\(k|1000d+100c+10b+a+1000a+100b+10c+d\)

\(\implies k|1001d+110c+110b+1001a\)

\(\implies k|11\)

Combining

\(k|99\).

YOU CAN PROVE IT USING INDUCTION.

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Not quite. Be careful of your implication signs. For example, if \( k \mid 999 \), must we have \( k \mid 1 \)?

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Oops no.. yes.. thats true... ok.. thanks sir

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Hint: Try to use the fact that even when the number is reflected it's sum of the digits still remain the same.

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Comment deleted 11 months ago

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No, the statement holds for all \(m\), not just the \(m\) that are palindromic.

For example, \( 9 \mid 63 \) and \( 9 \mid 36 \).

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Its easy. First we will prove k|9 and then we will prove k|11. So k|99

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Be careful of your implication signs. For example, we could have \( k = 9 \). However, it is not true that \( 9 \mid 11 \).

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can anyone please help me out with this problem

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Its easy I guess.. will post a soln. I didnt did it with pen and paper. Idk if I am correct becoz i did it in my mind

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not correct

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What have you tried? Where are you stuck?

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