Compute the following:
a) $2^{208} \pmod{53}$
b) $2^{288} \pmod{73}$
c) $7^{19} \pmod{28}$
Here is what I did so far:
a) ∅ (53) = 52
$(2^{52})^4 => (1)^4 => 1$
b) ∅ (73) = 72
$(2^{72})^4 => (1)^4 => 1$
c) ∅ (28) = 12
$7^{12} * 7^7 \pmod{28}$
$1 * 7^7 \pmod{28} => 7$
Just wondering if what I've worked out is correct?