# Reversed number

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, $k \mid 99$. Note by Kalpa Roy
3 years, 9 months ago

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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.

- 3 years, 9 months ago

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

Staff - 3 years, 9 months ago

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

- 3 years, 9 months ago

- 3 years, 9 months ago

Staff - 3 years, 9 months ago

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

- 3 years, 9 months ago

not correct

- 3 years, 8 months ago

Ya... Calvin sir told it before also... I know its wrong

- 3 years, 8 months ago

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

- 3 years, 9 months ago