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 ?

Note by Nishant Sharma
6 years, 4 months ago

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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.

- 6 years, 4 months ago

- 6 years, 4 months ago

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

- 6 years, 4 months 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 $9$.

- 6 years, 4 months 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.

- 6 years, 4 months ago

it goes for 81 also

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago

thanks

- 6 years, 4 months ago

right but first prove 19^93>13^99

- 6 years, 4 months ago

not needed to show the greater than sign.

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago

hmm thikie bolecho go sohomda

- 6 years, 4 months ago

But why is it required?

- 6 years, 4 months ago

yeah i got my mistake

- 6 years, 4 months 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 $81$.

- 6 years, 4 months ago