Find the remainder when 51^203 is divided by 7

Question

anonymous · Answer

i don't understand your question

ganeshie8 · Answer

what have you tried so far ?

anonymous · Answer

51^203/ 7 = 2^203/7 as, dividing 51 by 7 gives 2 as a remainder! I am stuck there now

ganeshie8 · Answer

thats a very good start !

anonymous · Answer

Problem 1.13 is an example

ganeshie8 · Answer

so you have reduced $\large 51^{203}/7$  to   $\large 2^{203}/7$  is it ?

anonymous · Answer

Now what? :-/

ganeshie8 · Answer

notice that 2^3 / 7 =   8/7 = 1

anonymous · Answer

and what do we do with 2^200

ganeshie8 · Answer

$\large 2^{203} =  2^{3\times 67 + 2} =   2^2*2^{3\times 67} = 4*8^{67}$

anonymous · Answer

Well, I did notice that! I didnt get it right

ganeshie8 · Answer

whats the remainder when u divide 8^67  by 7 ?

anonymous · Answer

1 ?

ganeshie8 · Answer

yes, because  8/7 leaves a remainder 1

ganeshie8 · Answer

$\large 4*8^{67}/7 =   4*1 = 4$

anonymous · Answer

Answer is 4

ganeshie8 · Answer

yep ! but there is a better notation for remainders. heard of congruence/mod before ?

ganeshie8 · Answer

you textbook notation is bit misleading  - you should switch to below notation which is used by the rest of the world 

$$\large 2^{203} \equiv  2^{3\times 67 + 2} \equiv   2^2*2^{3\times 67} \equiv 4*8^{67} \equiv 4*1 \equiv  4 \mod 7$$

anonymous · Answer

Whoaa! Thank you so much :)

ganeshie8 · Answer

yw :)

anonymous · Answer

Nope! What is it? Its not the text book! Maybe because I am using my cellphone app to type the queation!