INMO - 1992

I can't believe this is an INMO problem. It's actually quite easy if you do it by modulo function. 92=23x4 now, 19= -1(mod 4) 19^22=1 (mod 4) 19^22 = 1(mod 23) Therefore, 19^22=1 (mod 23x4) = 1(mod 92) so, 19^88= 1 (mod 92) therefore, 19^92=49 (mod 92)

Using Euler's generalization of fermat's theorem

$19^{\phi(92)} \equiv 1 \pmod{92}$ where $\phi(n)$ is Euler's phi function(or totient function)

$92=23 \times 2^2$

$\implies\phi(92)=92(1-\frac{1}{2})(1-\frac{1}{23})=44$

$\implies 19^{44} \equiv 1 \pmod{92}$

$\implies 19^{(44) \cdot 2} \equiv 1^2 \pmod{92}$

$\implies 19^{88} \equiv 1 \pmod{92}$

Multiplying both sides by $19^4$ yields

$\implies 19^{92}\equiv 19^4 \pmod{92}$ .. (1)

Now $19^{2}\equiv 85 \pmod{92} \equiv (-7) \pmod{92}$

$19^{4}\equiv (-7)^2 \equiv 49 \pmod{92}$ .. (2)

Using (1) and (2)

$19^{92} \equiv 49 \pmod{92}$

Hence the answer is $\boxed{49}$

We can apply euler's function