A piece of $\phi$

Number Theory Level 3

Find the last non-zero digit of $\large (a^8 \bmod 15)^{999}$ $a$ and $15$ are co-prime.

1 solution

Vilakshan Gupta
Dec 8, 2017

$\phi(15)=15\left(1-\frac 15\right)\left(1-\frac 13\right)=15\cdot \frac 45 \cdot \frac 23 = 8$

Assuming that $\gcd(a,15)=1$ , therefore $a^8 \equiv 1 \pmod {15} ~~~~~~~~\color{#3D99F6}\text{(Euler's Totient Function)}$

Therefore last digit would of $\left(a^8 \pmod {15}\right)^{999} = \boxed{1}$

It is $\large (a^8 \bmod 15)^{999}$ not $a^8 \pmod {15}^{1000}$

Sumukh Bansal - 3 years, 6 months ago

Yes. It was a typo. I have corrected it.

Vilakshan Gupta - 3 years, 6 months ago