what is the remainder when 5243 is devided by 13 a 12 b 5 c 8 d 1

when 5243/13=403 with remainder 4
12 years agoHelpfull: Yes(12) No(6)
i think ques. should be -what is the remainder when 5^243 is devided by 13?
then remainder will be a)12.
last 2 digit of 5^243=25 so 25/13 leaves remainder 12
12 years agoHelpfull: Yes(9) No(14)
ans: C) 8
THE Question should be 5^243/13.
we know that 5^2=-1%13
and therefore 5^(2*121)= {(-1)^121}%13..........I
and 5 = 5%13.........II
ans according to the property of mod i.e. %
I X II : 5^243 = (-1)X 5 % 13
= -5 % 13
= 8 % 13
Hence, the remainder is 8 .
12 years agoHelpfull: Yes(8) No(1)
sorry chetan, but the answer would be 8.
12 years agoHelpfull: Yes(5) No(6)
4 only when 5243/13
12 years agoHelpfull: Yes(4) No(3)
See..5 raised to any power will give 5 at the last digit.
now..divide the power with 5..
243%5=3
now the eqn becomes=5^3 % 13
125 % 13 =8
ans is. 8

for such ques find the cyclicity..n divide the power with the cyclicity.
12 years agoHelpfull: Yes(4) No(0)
Answer will be 4
12 years agoHelpfull: Yes(1) No(3)
remainder will be 8
let use flater theorem
5^243/13
5^9 * 5^234 /13
for remaindar
(5^9)*(5^234)%13
(5^9)(5^18)^13 %13
(5^9)(5^18)(5^18)^(13-1) %13
according theorem
(5^18)^(13-1) mod 13=1
(5^9)(5^18) mod 13
(5^27) mod 13
(5)(5^2)^13 mod 13
(5)(5^2) (5^2)^(13-1) mod 13
(5^2)^(13-1) mod 13=1 again
so 5*25 mod 13
8 will be ans
12 years agoHelpfull: Yes(1) No(1)

what is the remainder when 5243 is devided by 13?
a) 12 b) 5 c) 8 d) 1