find the remain 17der of 36^(50) + 31^(19) when divided by 17, i got 9 for this but apparently the answer is 11, thanks for any help
Have you actually used fermat's Little theorem?
It states that for any prime number $\displaystyle p$ we have $\displaystyle a^{p-1}\equiv 1$ mod $\displaystyle p$
Here we divide by $\displaystyle p=17$, thus we have $\displaystyle a^{16}\equiv 1 $mod 17 for any number $\displaystyle a$.
Use this fact and observe that $\displaystyle 36^{50}\equiv 2^{50} \equiv (*) 2^{2} \equiv 4$ mod 17
(*) here we used fermat's theorem.
Now you try $\displaystyle 30^{19}$ mod 17
ok i've checked my answer an i think i made a stupid mistake at the end,
36= 2 mod 17 and 31 = -3 mod 17 therefore
36^(50) + 30^(19) = ( 2^50 + ((-3)^19 ) mod 17
= (1 + -27) mod 17 using fermat's little theorem
= 8 mod 17
therefore the remainder is 8 (i had 9 here instead making a stupid mistake) is this correct then,
really sorry i dont know how to use latex yet but i have my exams soon so im busy revising, thanks for any help