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.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

What I had done is that I wrote it as $(18+1)^{93}-(12+1)^{99}$ and now it is easy to see that a very few terms are left, after the binomial expansion, that are not divisible by $162$,i.e., $3^4*2$. And then it is very easy to show that the remaining terms add up to be a multiple of $162$ as well.

We will have to prove that its divisible by $81$ and $2$. But its obviously divisible by $2$ and if we consider both numbers mod 9, it turns out that its divisible by $9$.

@Superman Son
–
inequality korchhis bole sob jaegae inequality?
positivity or negativity is not at all needed as the problem is asking to prove divisibility.

A way to reduce the bulk of this problem is to use Euler's totient function.
Also, it is inevitably evident that the given number is divisible by $2$, so we have to worry about $81$ only.

After applying Euler's totient function, we only need to show that:
$19^{39} \equiv c \pmod{81}$ and $13^{45} \equiv c \pmod{81}$ i.e $19^{39}$ and $13^{45}$ leave the same remainder namely $"c"$ on division by $81$.

Easy Math Editor

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:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWhat I had done is that I wrote it as $(18+1)^{93}-(12+1)^{99}$ and now it is easy to see that a very few terms are left, after the binomial expansion, that are not divisible by $162$,i.e., $3^4*2$. And then it is very easy to show that the remaining terms add up to be a multiple of $162$ as well.

Log in to reply

latex not loading proprly pls type it

Log in to reply

Another method is to note that $13^2 \equiv 7 \pmod {162}$.

Log in to reply

We will have to prove that its divisible by $81$ and $2$. But its obviously divisible by $2$ and if we consider both numbers mod 9, it turns out that its divisible by $9$.

Log in to reply

Got that it is divisible by 2 and 9. But we need to prove that it is divisible by 162. Anyways, I could not think of approaching it this way.

Log in to reply

it goes for 81 also

Log in to reply

hey vikram bhaiya please mujhe yei problem latex chor kar likh do mere computer par load nehi ho raha latex

Log in to reply

show that 19^93 - 13^99 is a positive integer divisible by 162

Log in to reply

Log in to reply

right but first prove 19^93>13^99

Log in to reply

not needed to show the greater than sign.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

A way to reduce the bulk of this problem is to use Euler's totient function. Also, it is inevitably evident that the given number is divisible by $2$, so we have to worry about $81$ only.

After applying Euler's totient function, we only need to show that: $19^{39} \equiv c \pmod{81}$ and $13^{45} \equiv c \pmod{81}$ i.e $19^{39}$ and $13^{45}$ leave the same remainder namely $"c"$ on division by $81$.

Log in to reply