Find $438^{87493} \equiv ~? \pmod{11}$

Question

How to find the value of '?'
a mod m = b mod m , will this formula be used?
I am taking discrete maths course for CS. And this question is from one of its chapter

Answer 1

Observe that $\text {gcd} (11,438) = 1.$ So by Euler's theorem $438^{10} \equiv 1\ (\text {mod}\ 11).$ So $438^{87490} \equiv 1\ (\text {mod}\ 11).$ Also $438 \equiv -2\ (\text {mod}\ 11) \implies 438^3 \equiv -8 \equiv 3\ (\text {mod}\ 11).$ Therefore $$438^{87493} \equiv 3\ (\text {mod}\ 11).$$

If you don't know Euler's theorem or Fermat's little theorem then observe from my above calculation that $438^3 \equiv 3\ (\text {mod}\ 11).$ So $438^6 \equiv 9 \equiv -2\ (\text {mod}\ 11) \implies 438^{30} \equiv (-2)^5 \equiv -32 \equiv 1 (\text {mod}\ 11).$

Observe that $87493 = 87480 + 13.$ Since $30 \mid 87480$ so $438^{87480} \equiv 1\ (\text {mod}\ 11).$ Since $438^6 \equiv -2\ (\text {mod}\ 11)$ so $438^{12} \equiv 4\ (\text {mod}\ 11).$ Again $438 \equiv -2\ (\text {mod}\ 11).$ So $438^{13} \equiv 438^{12} \cdot 438 \equiv 4 \cdot (-2) \equiv -8 \equiv 3\ (\text {mod}\ 11).$ Thus we get

$(1)$ $438^{87480} \equiv 1\ (\text {mod}\ 11).$

$(2)$ $438^{13} \equiv 3\ (\text {mod}\ 11).$

Therefore what is $438^{87493} \equiv ~?\ (\text {mod}\ 11)$?

$$438^{87493} \equiv 438^{87480+13} \equiv 438^{87490} \cdot 438^{13} \equiv 1 \cdot 3 \equiv 3\ (\text {mod}\ 11).$$