What are the last 2 terms in the equation 13^1234?

This is equivalent to computing 13^1234 (mod 100).

Since φ(100) = φ(2^2) φ(5^2) = (4 - 2)(25 - 5) = 40, Euler's Theorem implies that
13^φ(100) = 13^40 = 1 (mod 100).

Hence, 13^1234 = (13^40)^(300) * 13^34 = 1 * 13^34 = 13^34 (mod 100).

However, 13^3 = 97 = -3 (mod 100).
==> 13^34 = (13^3)^11 * 13 = (-3)^11 * 13 = 53 * 13 = 89 (mod 100).

So, the last two digits are 89.

I hope this helps!