Use the EulerFermat theorem to find the remainder when 2549

Use the Euler-Fermat theorem to find the remainder when 2549 is divided by 29.

Solution

I AM SORRY THERE IS AN URGENT MATTER TO BE ATTENDED ..

I SHALL POST THE ANSWER LATER BY EDITING ..

THE ANSWER IS WELL KNOWN BY NORMAL DIVISION ...

2549 = 29*87+ 6

HENCE REMAINDER = 6 ...OR .....2549 = 6 [MOD 29]

BUT TO PROVE IT USING EULER-FERMAT THEOREM IS WHAT YOU WANT ...WHICH I HAVE TO WORK UPON...

WELL ....I AM BACK & THOUGHT QUITE A BIT ....

FERMATS THEOREM TELLS US THAT IF P IS A PRIME & [A,P] = 1, THEN ..

A^(P-1) = 1 [MOD P]

SO IT IS USED TO FIND DIVISIBILITY & MODULII ON DIVISION BY LARGE POWRS OF NUMBERS AND CONGRUENCE RELATIONS THERE ON ....BUT OUR QUESTION IS A VERY SIMPLE CONGRUENCE OF ..

2549=X[MOD 29] ....OR........X = 2549[MOD 29]

THOUGH WE DO HAVE 29 & 2549 AS A PRIME NUMBERS & [2549 , 29]=1 ..., IT WILL ONLY HELP US TO PROVE ..SAY ..

2^28=1[MOD 29] ........ETC WHICH HELP US TO SOLVE FOR LARGE NUMBERS AS POINTED OUT ...& NOT 2549 ...WHICH IS IS VERY SMALL NUMBER INDEED..

EVEN WITH A VERY SNMALL BASE OF 2 OR 3 & ALL WE GO WAY BEYOND THE REQUIRED NUMBER ..

SO I REGRET TO POINT OUT THAT , I AM UNABLE TO OFFER YOU ANY WAY TO USE FERMATS THEOREM TO SOLVE THIS SEEMINGLY VERY SIMPLE PROBLEM ...WHERE A STRAIGHT FORWARD DIRECT CALCULATIO GIVES ...2549 = 20 [MOD 29] WITHOUT ANY SWEATING ..

IN THE BEGINING