Latest CTS Aptitude Question SOLUTION: 2^81 when divided by 6,what is the remainder

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(#M40034233) CTS Numerical Ability question Keep an EYE

2^81 when divided by 6,what is the remainder

Asked In CTS YAMSHI SAGIR (10 years ago)

Solved NAOREM DINESH SINGH Read Solution (13)

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Read Solution (13) :

Saraswathy  (10 years ago)

2^81/6 ====> (2^1*2^2*2^3*....81 times]/6

2^1/6 ----------> remainder=2
2^2/6 -->4/6 ---> remainder=4
2^3/6 -->8/6 ---> remainder=2
2^4/6 -->16/6 --->remainder=4
2^5/6 -->32/6 --->remainder=2
.
.
.
it goes on in a cyclic manner

so remainder is "2" for all odd powers of 2 nd "4" for all even powers of 2

here 2^81 which has the odd power

so remainder =2

Is this solution Helpfull? Yes (60) | No (1)  

Rohit Agarwal  (10 years ago)

((2^3)^27)/6
==>((8)^27)/6
==>(2^27)/6
==>((8)^9)/6
==>(2^9)/6
==>((8)^3)/6
==>(2^3)/6
==>8/6

so remainder is 2

Is this solution Helpfull? Yes (13) | No (2)  

Vivek  (10 years ago)

2^odd/6------->remainder= 2.
2^even/6------>remainder= 4.

So, the answer is 2.

Is this solution Helpfull? Yes (6) | No   

NAOREM DINESH SINGH  (10 years ago)

remainder= 1

proof:
2^81/6
= 4^80/6
=2^79/3
=2^(4+75)/3
=16^75/3...
& 3|16|5
15
----
1

since 16^n/3, remainder=1
16^75/3, remainder=1

hence proved...

Is this solution Helpfull? Yes (2) | No (8)  

Harwinder Singh  (10 years ago)

2^81=2^80*2^1
so 2^80*2/6=2^8*10/3
now just take 2^8/3=256/3
which leaves a reminder of 1..so 1 is the ans

Is this solution Helpfull? Yes (2) | No (3)  

rajesh  (10 years ago)

Ans: 1 ;(2^81)/(3x2)==>(2^80)/3==>even powers will give remainer 1 and odd powers will give remainder 2

Is this solution Helpfull? Yes (2) | No   

Sohini Nandi  (10 years ago)

2^81/(2*3)
=2^80/3
=(3-1)^80/3
=(-1)^80
=1

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

Rahul Das  (10 years ago)

let odd power of 2 is 3=8
8%6=2
next power 5=2^5=32
32%6=2
odd power of 2 when divided by 6 remainder is 2

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SATYA SHIV  (10 years ago)

2^81===2*4^40 =16^20*2/6 =rem2*4^20/6 2*16^10/6 =2*4^5/6 =8*16^2/6=2*4/6=2

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vinay kumar  (10 years ago)

for every 4 divisivbles we get 2 so
for 81 th number we get 2 so it is divide at 0 we get remainder as 2

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Pathmavathi  (10 years ago)

2^81
2 s unit sequence 2,4,6,8,0 so totally 5
81/5= 16 quot, 1 remaindr
in sequence list 1 ll be 2
so ans=2

Is this solution Helpfull? Yes (1) | No   

jagannadh  (10 years ago)

4 is the answer...

Is this solution Helpfull? Yes | No (5)  

a  (10 years ago)

2^81/6
->2^9/6=1
->2^9/6,2^9/6. . . . .2^9/6 total 9 times each one gives remainder one
so final remainder is 1

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