# 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
1 year 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.

- 11 months, 4 weeks ago

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

Staff - 11 months, 4 weeks ago

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

- 11 months, 4 weeks ago

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

- 11 months, 3 weeks ago

Comment deleted 11 months ago

No, the statement holds for all $$m$$, not just the $$m$$ that are palindromic.

For example, $$9 \mid 63$$ and $$9 \mid 36$$.

Staff - 11 months, 4 weeks ago

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

- 11 months, 4 weeks ago

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

Staff - 11 months, 4 weeks ago

- 1 year 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

- 11 months, 4 weeks ago

not correct

- 11 months, 1 week ago

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

- 11 months, 1 week ago