1

I have to calculate $19^{93}\equiv x\pmod {162}$.

All I can do is this,by using Euler's Theorem:-

$19^{\phi(162)}\equiv1\pmod{162}$

So,$19^{54}\equiv1\pmod{162}$

Now,I have no idea how to reach power of $93$ from $54$.Because the next number which will give the same remainder is $19^{108}$.

How do I solve it?Someone told me Ramanujan's Congruences Equations may be useful but I have no idea why he said this.

Thanks for any help!

Soham
  • 9,990

2 Answers2

2

Note that $19^{93}=(1+18)^{93}$. When we expand using the binomial theorem, we get $1+(93)(18)$ plus terms that involve $18^2$ and higher powers of $18$. These are all divisible by $162$. So all we need to do is to find the remainder when $1+(93)(18)$ is divided by $162$. But $90$ is divisible by $18$, so $x\equiv 55\pmod{162}$.

André Nicolas
  • 507,029
1

$19 = 18 + 1$ and 18 divides 162. More importantly 162 divides 18^2.

$19^2 = 18^2 + 2*18 + 1 \equiv 37 \mod 162\\ 19^n \equiv n*18+1 \mod 162\\ 19^9 \equiv 1 \mod 162$

can you get home from here?

user317176
  • 11,017