Latest TCS Aptitude Question SOLUTION: 222^222 divided by 7.remainder is

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(#M40011651) TCS Numerical Ability question tcs Keep an EYE

222^222 divided by 7.remainder is

Asked In TCS sweta shaw (12 years ago)

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PARAG KAMRA  (12 years ago)

222^222 mod 7

(31*7+5)^222 mod 7

5^222 mod 7

(25)^111 mod 7


(7*3+4)^111 mod 7

(4)^111 mod 7

(16)^55.(4)^1 mod 7

(7*2+2)^55.4 mod 7

(2)^55.4 mod 7

(8)^18.8 mod 7

(7+1)^18.8 m0d 7

(1)^18.8 mod 7

1*8 mod 7=1

1 (ans)

Is this solution Helpfull? Yes (49) | No (5)  

spandana inti  (12 years ago)

222/7=(31+5/7)..(mixed fraction)
so its enuf if we check the remainder of(5/7)^222
remainder of
(5/7)^0=1
(5/7)^1=5
(5/7)^2=4
(5/7)^3=6
(5/7)^4=2
(5/7)^5=3
(5/7)^6=1... so on
the cycle repeats for every 6 times
now divide 222 by 6, the remainder will be 0
now check the cycle, the remainder for (5/7)^0 i.e 1
aswer is 1

Is this solution Helpfull? Yes (19) | No (1)  

Bankim  (12 years ago)

using euler theorem
7 and 222 are coprime to each other

phi(7) = 6
therefore 222^6 % 7 = 1
((222)^6)^37 % 7 = 1

so remainder is 1 when 222^222 is divided by 7.

Is this solution Helpfull? Yes (8) | No   

sameer khan  (12 years ago)

srry guys after (4)^222/7 we convert it (64)^74/7
then (63+1)^74/7
63/7 reminder is zero and (1)^74/7
so reminder is 1 its a right ans................

Is this solution Helpfull? Yes (7) | No (1)  

jayasri  (12 years ago)

(2)^2*111=4^111
((4^2)^55)*4
=((16)^55)*4
first digit of ((16)^55) is 6
so,6*4=24
4 is d ans.

Is this solution Helpfull? Yes (3) | No (15)  

nagashree  (11 years ago)

222^222 divided by 7
222 mod 7 gives remainder 5
problem reduces to 5 ^222 mod 7
now find out cycle
5 mod 7 gives rem 5
5^2 mod 7 gives rem 4
5^3 mod 7 gives rem 6 or -1
so it gives +1 rem at 5^6
so cycle here is 6
now 222 mod 6 gives 0 remainder
which means last term of seris
whihc is 1

Is this solution Helpfull? Yes (3) | No   

tejreddy  (12 years ago)

apply eulers remainder theroem..
since 222 and 7 are co primes,the totient function f(x)=7(1-1/7)=6
i.e,rem(222^6/7)=1=rem(222^12/7)=rem(222^18/7)=.......
since 222 is a multiple of 6
rem(222^222/7)=1...once refer eulers theroem to understand it better

Is this solution Helpfull? Yes (2) | No   

RAHUL KUMAR  (12 years ago)

(2^222)/7 can be written as ((2^3)^24)/(2^3-1)
then the remainder will be 1^24=1.

Is this solution Helpfull? Yes (2) | No (4)  

ankita khare  (12 years ago)

2^222 the cyclisity of 2 is 4 dan 4/7 rimender is 1 so answer is 1

Is this solution Helpfull? Yes (1) | No (2)  

Hari  (7 years ago)

(222^222)%7=1
how
step1 222%7=5
step2 (5^222) % 7
((5^3)^74) % 7
(125^74) % 7 125%7 = -1
step3 (-1)^74 % 7
1 (answer)

Is this solution Helpfull? Yes (1) | No   

siya  (12 years ago)

answer is 4

Is this solution Helpfull? Yes | No (9)  

sameer khan  (12 years ago)

222^222/7
(217+5)^222/7
in 217/7 reminder is zero so we take 5^222/7
now (25)^111/7 and then convert (21+4)^111/7
from 21/7 reminder is zero
now (4)^111/7
(2^2)111/7
(2)^222/7
so reminder is 2

Is this solution Helpfull? Yes | No (12)  

hargun  (12 years ago)

plz suggest from where to prepare this type of question..if any any e stuff plz send to hargun.15@gmail.com

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Smruti Patnaik  (12 years ago)

Ans. 4
A B A*B C (A*B)/C
222 222 49284 7 7040.571429

7*7040=49280
49284-49280=4

So the remainder is 4.

Is this solution Helpfull? Yes | No (2)  

siddharth  (11 years ago)

222(base) is divided by 7=31 r5.so we can write as 5^222=25^111. 25 divided by 7=3 r4. so 4^111=16^55 x4. dividing 16 by 7=2 r2. so 2^55 x4 = 32^5 x4. dividing 32 by 7=4 r4. so 4^5 x 4 = 4^6 = 64^2. dividing 64 by 7=9 r1. so 1^2=1

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ISHANTOSH DEY  (8 years ago)

correct answer is 5.....bcz if u divide 222 by 7 u will get 5 as remainder.....so u convert the given sum as (222)*(222)^221....and divide the single 222 with the 7...u will get remainder as 5

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