# 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
5 years, 7 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.

- 5 years, 7 months ago

- 5 years, 7 months ago

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

- 5 years, 7 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 $$9$$.

- 5 years, 7 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.

- 5 years, 7 months ago

it goes for 81 also

- 5 years, 7 months ago

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

- 5 years, 7 months ago

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

- 5 years, 7 months ago

thanks

- 5 years, 7 months ago

right but first prove 19^93>13^99

- 5 years, 7 months ago

not needed to show the greater than sign.

- 5 years, 7 months ago

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

- 5 years, 7 months ago

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

- 5 years, 7 months ago

hmm thikie bolecho go sohomda

- 5 years, 7 months ago

But why is it required?

- 5 years, 7 months ago

yeah i got my mistake

- 5 years, 7 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 $$81$$.

- 5 years, 7 months ago