Latest TCS Aptitude Question SOLUTION: 32^32^32/7 what is the remainder? 4^32^32/7 as 32%7=4 now, 4^3/7 gives us remainder 1; we have to write 32^32 in the form of 4^3n+x where x is the unit digit of 32^32 ie 6 so, we

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32^32^32/7 what is the remainder?

4^32^32/7 as 32%7=4
now, 4^3/7 gives us remainder 1;
we have to write 32^32 in the form of 4^3n+x where x is the unit digit of 32^32 ie 6
so,
we can re write the problem as
4^3n+6/7
=>4^3n/7 always gives remainder 1
and 4^6/7 gives remainder 1.
so final remainder should be 1...what is wrong in it ? as the ans given is 4.

Asked In TCS aniruddha Ghosh (9 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

Saraswathy  (9 years ago)

32^32^32 mod 7
4^32^32 mod 7
E(7) = 6
32^32 mod 6
2^32 mod 6
2^1 mod 6 = 2... 2^2 mod 6 = 4... 2^3 mod 6 = 2 again...
Cyclicity is 2... So rem[2^32/6] = 2k+'n'th value...
32 => 2*16+0..
2^32 mod 6 = 4
Now 4^32^32 mod 7 is reduced to 4^4 mod 7
16*16 mod 7 => 2*2 mod 7
4 mod 7

Ans : 4

Is this solution Helpfull? Yes (13) | No (1)  

Saraswathy  (9 years ago)

Sorry @aniruddha... I didnt understand what you have done...
I think it might be applicable for (32^32)^32...
Remember (32^32)^32 is not equal to 32^32^32...
(2^3)^2 = 2^6 = 64
2^3^2 = 2^9 = 512...
E(7) is nothing but Euler value of 7...
E(N) can be calculated by N*(1-1/a)*(1-1/b)*(1-1/c)... where a,b,c are prime factors of N...
So 7 => 7^1 [a^x]
So E(N) = 7*(1-1/7) = 6

Is this solution Helpfull? Yes (3) | No   

aniruddha Ghosh  (9 years ago)

@saraswathy what does E(7) stand for in your solution? Can you tell me what is wrong in the above mentioned process that I have done

Is this solution Helpfull? Yes (1) | No   

aniruddha Ghosh  (9 years ago)

@ saraswathy can you give me the link where a^b^c/7 or any no. type problems are explained thoroughly with the theory behind it?? It will be a great help if u can give me the link. I cant understand why we follow these steps.

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purushottam sah  (9 years ago)

32^32 its unit place is 6
now (2^5)32^32=2^2(7+1)^last digit will be 6(even number)---equationn no. (1)
we know that
(aX+1)^n/X= 1 remainder
if we devide equation no (1) by 7 then we get
4(1)/7=4

Is this solution Helpfull? Yes | No (1)  

Priya Pandey  (9 years ago)

(32^ 32^32) / 7 = (4^32^32) / 7 .......................(as 32/7 rem is 4)
now
search for cyclicity of 4/7... in this way.......
4/7 = (rem) 4
4^2 / 7 = (rem) 2
4^3 / 7 = (rem) 1............hence cycle of 3

now divivide 32^32 / 3
same way as above..
32^32 / 3 => 2^32 / 3
2 / 3 = 2
2^2 / 3 = 1 cycle of 2

32 / 2 = 0(rem)

0 means last term of cycle..that is 1 in 2^2 /3 and then 1st term of cycle in 4 / 7
that is 4 as remainder

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