# 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 ago

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:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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

Sort by:

Top Newest

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 ago

Log in to reply

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 ago

Log in to reply

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

Staff - 3 years ago

Log in to reply

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

- 3 years ago

Log in to reply

can anyone please help me out with this problem

- 3 years ago

Log in to reply

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 ago

Log in to reply

not correct

- 2 years, 11 months ago

Log in to reply

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

- 2 years, 11 months ago

Log in to reply

Staff - 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...