Latest TCS Aptitude Question SOLUTION: what is the remainder when 5243 is devided by 13? a) 12 b) 5 c) 8 d) 1

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(#M40012236) TCS question Keep an EYE

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

Asked In TCS ATUL KUMAR SINGH (11 years ago)

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RAHUL VERMA  (11 years ago)

answer will be 8 when 5^243::
as 5^243%13:5^1%13=5
5^2=25%13=12
5^3=125%13=8
5^4=625%13=1
5^5=3125%13=5
so repetition is after 4 cycle...so
we will divide 243%4=3
so 5^243=((5^240)*(5^3))%13
5^240 will give remainder 1,, 5^3%3=8
so answer is 1*8=8

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chetan gupta  (11 years ago)

when 5243/13=403 with remainder 4

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Tabrez Siddiqui  (11 years ago)

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

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Mukta Yadav  (11 years ago)

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 .

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ATUL KUMAR SINGH  (11 years ago)

sorry chetan, but the answer would be 8.

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venkatesh  (11 years ago)

4 only when 5243/13

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Aarti Tewari  (11 years ago)

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.

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Akshay Prakash  (11 years ago)

Answer will be 4

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snehil  (11 years ago)

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

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