problems like these are just patterns
write out powers of 18 until a pattern forms
18^2--last 2 didgets are 24
18^3--last 2 didgets are 32
18^4--last 2 didgets are 76
18^5--last 2 didgets are 68
18^6--last 2 didgets are 24
we see that it starts repeating when last 2 didgets are 24 thus their are 4 unique last 2 didgets
18^(2-5)
18^(6-9)
18^(10-13)
18^(14-17)
18^(18-21)
18^(22-25)
18^(26-29)
18^(30-33)
18^(34-37)
as you can see 37 is the last of the set or the last 2 didgets are 68
+26367 Find the last 2 digits of 18^(37)
\(\begin{array}{|rcll|} \hline && 18^{37} \pmod {100} \\ &\equiv& \Big(18^4 \Big)^{9}\cdot 18 \pmod {100} \quad & | \quad 18^4 \equiv 76 \pmod {100} \\ &\equiv& 76^{9}\cdot 18 \pmod {100} \\ &\equiv& 76^8\cdot 76 \cdot 18 \pmod {100} \quad & | \quad 76^{2n} \equiv 76 \pmod {100} \text{ or } 76^8 \equiv 76 \pmod {100} \\ &\equiv& 76\cdot 76 \cdot 18 \pmod {100} \\ &\equiv& 76^2 \cdot 18 \pmod {100} \quad & | \quad 76^2 \equiv 76 \pmod {100} \\ &\equiv& 76 \cdot 18 \pmod {100} \\ &\equiv& 1368 \pmod {100} \\ &\mathbf{ \equiv } & \mathbf{ 68 \pmod {100} } \\ \hline \end{array}\)
The last 2 digits are 68
