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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewestHint: 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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Lets 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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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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