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

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- May 15th 2010, 02:42 PMhmmmmfermat's little theorem
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

- May 15th 2010, 02:59 PMDinkydoe
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 - May 15th 2010, 03:00 PMundefined
- May 15th 2010, 03:08 PMDinkydoeQuote:

$\displaystyle 36^{50} + 30^{19} \equiv 2^{50} + 13^{19} \equiv 2^{3\cdot16+2}+13^{16+3} \equiv 2^2+13^3\equiv4+2197\equiv8\ \text{(mod }17\text{)}$

The book answer is 11? I did a direct calculation and also got 8.

Yeah me too, 11 is definately wrong! - May 15th 2010, 03:46 PMhmmmm
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 - May 15th 2010, 03:56 PMundefined
- May 15th 2010, 04:06 PMhmmmm
oh dear sorry its 31 (Headbang)

- May 15th 2010, 04:12 PMundefined
- May 15th 2010, 04:20 PMhmmmm
ah ok thanks very much sorry about that