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Question 27145: Find the rightmost digit of 7^(1111). How do i find the remainder when divided by 10
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! WE ARE ONLY INTERESTED IN DIGIT IN UNITS PLACE.SO LET US TRY TO LOOK
FOR A PATTERN IN THIS.
POWER......UNITS DIGIT
7^1=.......................7
7^2=.......................9
7^3=.......................3
7^4-........................1
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7^5=......................7...SO IT REPEATS NOW..
HENCE WE HAVE TO FIND 1111 POWER IS UNDER WHICH GROUP..SO DIVIDE IT
WITH 4 AS THE ABOVE CYCLE REPEATS AFTER FREQUENCY OF 4.IT GIVES US A
REMAINDER OF 3
1111=4*277+3....SO IT FALLS UNDER GROUP OF 7^3...SO ITS UNITS DIGIT IS
3 AS SHOWN ABOVE
7^1111={7^(4*277)}{7^3)=IT HAS SAME ENDING AS 7^3 WHICH IS 3
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IF YOU KNOW RESIDUAL CLASSES,CONGRUENCE MODULO,WE CAN USE THAT METHOD
WE HAVE TO FIND X IN
7^1111=X(MOD10)
WE KNOW THAT
7^2=49=-1(MOD10)
RAISING TO 555 POWER BOTH SIDES,WE GET,
(7^2)^555=(-1)^555(MOD10)
7^1110=-1(MOD10)
MULTIPLYING WITH 7 NOW WE GET
7^1111=-7(MOD10)
7^1111=(10-7)(MOD10)=3(MOD10)
HENCE 3 IS THE REMAINDER WHEN DIVIDED WITH 10 OR UNITS PLACE HAS 3
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