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Remainder problem.
by gayatri khasiram - Monday, 22 October 2012, 01:32 AM
  imagewhen divided by 9 gives the remainder

please help solve with steps..
Re: Remainder problem.
by naman malik - Monday, 22 October 2012, 10:59 AM
  (25)^625/9
this can be written as (-2)^625/9 by the concept of negative remainders
//keep -1 aside
(2)^625/9
which is nothing but..
((8)^208 *2^1)/9
on solving you will get 2 as remainder but don't forget you just left -1 aside in the very beginning so the ans should be -2
which is nothing but 7..

hope u understood......
smile smile smile
Re: Remainder problem.
by Tapesh Kumar - Saturday, 19 January 2013, 04:26 PM
 

Hi,

I was just trying to understand that the solution, but couldn't understand few things, pls can you help explain on the following.

1. Since we had 25^25^25 - How we took 25^625/9?

2. how at the end we got remainder of 2? I understand that 8^208 will lead to remainder of 1?

Pls help explain. Thanks!

Re: Remainder problem.
by minhaz pathan - Monday, 29 April 2013, 07:23 PM
  I have the same doubt. How can that be 25^625. Please explain.

Re: Remainder problem.
by aman bhatnagar - Tuesday, 28 May 2013, 10:29 AM
 

 

First of all its given 25^25^25 and not ((25)^25)^25.

Difference is that the topmost 25 in the exponent term is for the other 25 in the exponent only and not for the base 25.

 

Then we first divide the base 25 by 9 which gives remainder 7. Now we consider the exponent term, i.e., 25^25 and divide it with 6, i.e, the phi (N) (read as phi of N) value for 9.

 

phi (N)= Nos.less than or equal to N and prime to it.

To calculate phi (N) we us the following Euler's formula:-

 

If N= a^x*b^y*c^z.... then phi (N)= N(1-1/x)(1-1/y)(1-1/z).........

where a,b,c.... are the prime factors for N.

 

Note:- To use the above method we need to ensure that N (in this case 9) is co-prime with p (in this case 25).

 

Thus, on dividing the exponent 25^25 by 6, we get 1^25=1.

 

Thus now we are left with 7^1/9 which is 7 only.

 

So Answer is 7.

Re: Remainder problem.
by debaditya khan - Tuesday, 4 June 2013, 01:38 PM
  use eulers theorem
as 9 & 25 are co-prime
so=
9=3square=9(1-1/3)=6
so 25 to the power 6 will give remainder 1.

now divide 25 to the power 25 by 6

again 25 to the power 5 gives R=1
so we have=25/9 gives remainder= 7

therefore we get 7 as remainder.
Re: Remainder problem.
by Neeraj Nayan - Tuesday, 18 June 2013, 07:12 PM
  E(9)=6, Now, 25^25^25 = 25^(24+1)^25= 25^(6K1+1)^25 = 25^(6K+1)%9
25^6K*25%9= 1*7=7
Re: Remainder problem.
by Nitish Chopra - Thursday, 20 June 2013, 02:31 AM
  25^25^25 = 25^625 = (18+7)(18+7)........625Times = 18k + 7^625

therefore, remainder when 7^625 is divided by 9 will remain the same.

Now 7^625 = 7^3 x 7^3 x 7^3.....(208times)x 7

and remainder when 7^3 = 343 is divided by 9 is 1 and remainder when the left 7 is divided by 9 will be 7.