Number Theory Problem

Show that $19^{93}$ - $13^{99}$ is a positive integer divisible by 162.

NOTE:Actually I could not understand how to proceed. Can anyone show me the path ?

#NumberTheory #HelpMe!

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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.

Shourya Pandey - 8 years ago

latex not loading proprly pls type it

superman son - 8 years ago

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

Shourya Pandey - 8 years ago

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 $9$ .

Vikram Waradpande - 8 years ago

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.

Nishant Sharma - 8 years ago

it goes for 81 also

superman son - 8 years ago

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

superman son - 8 years ago

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

Anoopam Mishra - 8 years ago

@Anoopam Mishra – thanks

superman son - 8 years ago

right but first prove 19^93>13^99

superman son - 8 years ago

not needed to show the greater than sign.

Shourya Pandey - 8 years ago

@Shourya Pandey – yes if we do bionomial then not but otherwise Show that 19^93 - 13^99 is a positive integer

superman son - 8 years ago

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

Soham Chanda - 8 years ago

@Soham Chanda – hmm thikie bolecho go sohomda

superman son - 8 years ago

@Superman Son – But why is it required?

Shourya Pandey - 8 years ago

@Shourya Pandey – yeah i got my mistake

superman son - 8 years ago

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 $81$ .

Aditya Parson - 8 years 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}`	$\boxed{123}$