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

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222 ^222 divided by 7 wt is remainder

Asked In TCS manikanta (9 years ago)

Unsolved Read Solution (15)

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Anuj Yadav  (9 years ago)

222=31*7+5
31*1+5 when divided by 7 gives 5 as remainder

=>5^222
now 5^2^111
=>25^111
(7*3+4)^111 when devided by 7 gives 4 as remainder

=>4^2^55*4^1=16^55*4^1
=>(7*2+2)^55 when devided by 7 gives 2 as remainder
=>2^55*4=2^54*2*4=2^54*8=8^18*8

hence 8^18*8=(7+1)^18*8 when divided by 7 gives 1^18*8
=>8

and when 8 is divided by 7 gives 1 as remainder.

Hence answer will be 1.

Correct me if I am Wrong

Is this solution Helpfull? Yes (40) | No (1)  

Md.Sultan Khichi  (9 years ago)

222^222/7
=(5)^222/7
=(5^6)^37/7
(1)^37/7
=1
Ans remainder=1.

Is this solution Helpfull? Yes (21) | No (3)  

rajeev chaubey  (9 years ago)

222/7=5
now our question turns to-5^222/7
now we have to find cyclicity of 5
5^1/7=5(remainder)
5^2/7=4(remainder)
5^3/7=6
5^4/7=3
5^5/7=1(remainder)
cyclicity is 6
divide power 222 by 6
remainder 0
means unit digit is 1(last line of cycle of 5)

Is this solution Helpfull? Yes (10) | No (5)  

MILY SAXENA  (9 years ago)

222^1/7=5
222^2/7=4 => (take above remainder)- (5*222)/7=4
222^3/7=6 => (take above remainder)- (4*222)/7=6
222^4/7=2 => (take above remainder)- (6*222)/7=2
222^5/7=3 => (take above remainder)- (2*222)/7=3
222^6/7=1 => (take above remainder)- (3*222)/7=1
222^7/7=5 => (take above remainder)- (1*222)/7=5
5,4,6,2,3,1 is the remainder value
hence 222^222 divided by 7 has cyclicity=6 because repitition occurs after sixth power.
therefore,222/6 gives 0 remainder .
So,we check at 222^6/7 remainder which is equal to 1
ans. 1

Is this solution Helpfull? Yes (4) | No   

Saptarshi Das  (9 years ago)

For 222^222, the unit digit must be 4,if we go through the multiples of 7 then we will find 7*9=63 Whose unit digit is as close as the 4 like no other.so remainder is 1.

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

Rishi Pandey  (9 years ago)

222/7=5
(5*5)/7=4
(4*4)/7=2
(2*2*2)/7=1
no more remainder for proceed.
1 is ans.

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

abhisek  (9 years ago)

222^222/7
->5^222/7
->(5^3)^74/7
->6^74/7
->36^37/7
1^37/7
->ans is 1...

Is this solution Helpfull? Yes (1) | No   

upasana acharya  (9 years ago)

4
222 leaves 5
5^5=25
remainder 4

Is this solution Helpfull? Yes | No (2)  

Pratik sharma  (9 years ago)

Answer is 121112227as 1117 leave remainder -1 and as power is even thus it become 1so2-1222722227now as 238so237478747Now 87 leave remainder 1 soanswer is 1

Is this solution Helpfull? Yes | No (1)  

Rajat Jain  (9 years ago)

222^222/7
firstly divide 222 with 7 and we get 5 remainder now
(217+5)^222/7 now 217 will give 0 as remainder so left with 5 only so now we have to solve 5^222/7 now take square of 5 so ((5^2)^111)/7 it will make 25^111/7 so this equation will give 4 as remainder
now take cube of 4 which is 64 1 more than 63 so equation will become (4^3)37/7
64^37/7
(63+1)^37/7
so we get 1 as remainder

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Dhiren Paryani  (9 years ago)

(7*31+5)^222/7
(5^2)^111/7 => (7*4+1)^111/7
Hence Remainder is 1

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Ankit RAJPUT  (9 years ago)

(222)^222/7 = (224 - 2)^222/7
=> (-2)^222/7 = (2)^222/7
=> (2^3)^37 x 2^111/7
=> (8)^37 x 2^111/7 = (7+1)^37 x 2^111/7
=> 1^37 x 2^111/7 = 2^111/7
=> (2^3)^37/7 = (7+1)^37/7
=> 1^37/7 = 1 remainder Ans...

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Rishabh Nigam  (9 years ago)

((2^3)^74)/7
=((7+1)^74)/7
=(1^74)/7
=1/7
=1

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rahul sadhukhan  (8 years ago)

222^222=(222^6)^37%7=1^37%7=7
as 7 is a prime no, so 222^(7-1)/7=1

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Susil Pal  (8 years ago)

7 is a prime number so, divided by 6(7-1)
(222^(222%6))
((222^0)%7)=0 so, result=0

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