Using Fermat's Little Theorem: 18^{802}(mod29) Calculation

[bendaddy]

Homework Statement

Use Fermat's Little Theorem to calculate 18^{802}(mod29)

Homework Equations

Fermat's Little Theorem: a^{p-1}\equiv1(modp)
where in this case, a=18 and p=29

The Attempt at a Solution

By FLT, I found that 18^{28}\equiv1(mod29)
So, 18^{802}\equiv(18^{28})^{28.5}*18^{4}(mod29)\equiv18^{4}(mod29)\equiv25(mod29)

So my solution is 25(mod29).
However, the solution my professor posted is 4(mod29) (**NOT -4(mod29)**). Pretty confused here... Is there something wrong in my logic?

 

[tiny-tim]

hi bendaddy!

By FLT, I found that 18^{28}\equiv1(mod29)
So, 18^{802}\equiv(18^{28})^{28.5}*18^{4}(mod29)\equiv18^{4}(mod29)\equiv25(mod29)

isn't 18¹⁴ minus 1 ?