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, $k \mid 99$ .

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Md Zuhair - 3 years, 7 months ago

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

Calvin Lin Staff - 3 years, 7 months ago

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

Md Zuhair - 3 years, 7 months ago

can anyone please help me out with this problem

Kalpa Roy - 3 years, 8 months ago

What have you tried? Where are you stuck?

Calvin Lin Staff - 3 years, 8 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

Md Zuhair - 3 years, 7 months ago

not correct

Michael Fitzgerald - 3 years, 7 months ago

@Michael Fitzgerald – Ya... Calvin sir told it before also... I know its wrong

Md Zuhair - 3 years, 7 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.

Sathvik Acharya - 3 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$