what is the remainder of (169376^30)/31

• simply apply the fermat theorems that is a^(P-1)mod(P)=1 where a and p are prime numbers..so remainder is 1 ans
  
• 9 years agoHelpfull: Yes(11) No(0)
• when we defide 169376 by 31 , it will give reminder 23
  means we can write 169376 as (31*n +23)
  now 169376^30/31= (31*n +23)^30/31
  after binomial expainsion of numerator we get all the term divisible by 31 except last term 23^30
  Now think about 23^30
  we can write it as (23^2)^15
  23^2=529
  now we have 529^15
  when we devide 529 by 31 ,it will give reminder as 2
  (529^15)/31=(31^m+2) ,after binomial expension of numerator we get all term divisible by 31 except last that is 2^15
  now think about 2^15
  that is 32^3=(31+1)^3
  apply binomial expansion on numerator all term will be divisible by 31 except last
  that is 1^3=1
  hence reminder will be 1
  
  Question is not so tough ,it follows traditional process but we have to solve this problem by dividing it into sub problem
• 9 years agoHelpfull: Yes(4) No(0)
• correct ans is 1
  previously i did small mistake in calculation
• 9 years agoHelpfull: Yes(2) No(0)
• 169376 and 31 are co-prime to each other.
  here m=16936, and n=31,,,
  [m^p/n]=1,,,always where p=n(1-1/a)(1-1/b)......where n=a^x*b^y*c^z....
  so here p=31(1-1/31)=30
  So we can write...[169376^30]/31=1..
  
  1 is the right answer.
  
• 9 years agoHelpfull: Yes(1) No(0)
• Ans is 8
  169376^30/31
  169376%31=23is remainder
  so there 23 comes 30 times
  23*30%31=690%31=8 remainder
• 9 years agoHelpfull: Yes(0) No(6)
• where a is a prime no???
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 1
  
  at the end (2^5)^3 remains which give 1 remainder.
• 9 years agoHelpfull: Yes(0) No(0)
• according to remainder theorem: [(x+a)^n]/x= a^n/x
  so, 169376 mod 31=23
  now, [(p*31+23)^30]/31= 23^30/31
  = [(23^2)^15]/31
  =529^15/31
  applying the theorem again,
  [(p*31+2)^15]/31
  =2^15/31
  =[(2^5)^3]/31
  =32^3/31
  =[(p*31+1)^3]/31(applying again)
  =1^3/31
  =1/31
  so, remainder= 1
• 9 years agoHelpfull: Yes(0) No(0)