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Number System x^y^z/D
by sunny sachdeva - Tuesday, 25 May 2010, 09:04 AM
  Hi ,
Does anyone knows how to go about problems where it is asked to find out the remainder where X^Y^Z is divided by D.

For example Remainder when 32^32^32 is divided by 7?

Thanks
Re: Number System x^y^z/D
by Total Gadha - Tuesday, 25 May 2010, 12:49 PM
  See the last two videos HERE. Find the remainder when 3232 is divided by Ø(7) = 6. Let this be R. Then find the remainder when 32R is divided by 7.

The thing to understand is that 3232 would be some number only, although big. So you are trying to find remainder when 32Big Number is divided by 7.
Re: Number System x^y^z/D
by Lakhan Mohta - Wednesday, 7 July 2010, 05:44 PM
  ans is 2.
Re: Number System x^y^z/D
by Abhishek Jain - Wednesday, 7 July 2010, 10:21 PM
  isn't it 4..?? surprise
Re: Number System x^y^z/D
by Himanshu Shekhar - Thursday, 8 July 2010, 02:02 PM
 

ANS: 4.

APPROACH 1:

LET REMAINDER, R = (32^32^32) MOD 7 = (4^32^32) MOD 7.

PATTERN METHOD , (43N) MOD 7 = 1, (43N+1) MOD 7 = 4, (43N-1) MOD 7 = 2. ONLY 3 REMAINDERS ARE POSSIBLE. SINCE (32^32) MOD 3 = (2^32) MOD 3 = 1, (32^32) CAN BE WRITTEN IN THE FORM 3N+1. ANSWER IS 4.

APPROACH 2:

(322) MOD 7 = (42) MOD 7 = 2; (324) MOD 7 = (22) MOD 7 = 4;

(328) MOD 7 = (42) MOD 7 = 2; (3216) MOD 7 = (22) MOD 7 = 4;

(3232) MOD 7 = (42) MOD 7 = 2;

(32^32^2) MOD 7 = (22) MOD 7 = 4; (32^32^4) MOD 7 = (42) MOD 7 = 2;

(32^32^8) MOD 7 = (22) MOD 7 = 4; (32^32^16) MOD 7 = (42) MOD 7 = 2;

(32^32^32) MOD 7 = (22) MOD 7 = 4.

 

Re: Number System x^y^z/D
by Rohit Saluja - Monday, 11 May 2015, 05:28 PM
  Well i went through only some of the answer and I didn't find any answer as convincing so here is my answer

we are given the no
32^32^32
okay so taking the above to power of the number which is 32^32 this means that the no is

32^(some number obtained by finding 32^32 )
now 32^32 is 32*32*32......32 times.

Also 32*32 is 1024 so 32^32 is 1024^16
which is raised to the power of 32

so the nu is 32^1024^16

When we divide 32 by t we get the remainder as 4 also

so


4^1024^16=== 4^(some huge even number)

now when this 1024^16 is a huge even no which can be completely represented in powers of two.

Now note this when we divide the powers of 2 with 3 we get the following pattern

2^1/3 = 2;
2^2/3= 1;
2^3/3= 2;
2^4/3=1;

So we can infer that when we divide a number which is 2 raiswed to the power of even no and then divide it by 3 we get
1 as remainder
Now coming back to our question

4^1024^16 so 4^3^(some even integer which we get after dividing 1024^16 and that will leave the remiainder as 1)so

4^1024^16==(4^3^(2k))*4/7

(4^3^(2k))=63^(2k)(4)=(-1)^2k)*(4)/7==4 is the answer.
Re: Number System x^y^z/D
by Sidhant Acharya - Wednesday, 20 May 2015, 11:20 AM
  Hi so according to the approach mentioned .. Had a query why would be divide the power that is 32 ^32 from 6 I.e Euler of 7 first...?...we do I it fr normal a/ b ryt....
BT in this we doing it for y^z first why is that....