What should be subtracted from 4^15 so that the remaining is divisible by 31
1. 0
2. 1
3. 2
4. 3

• answer=1
  4^15= 2^30=2^5^6=32^6=(31+1)^6
  now (31+1)^6 / 31= 1^6 /31= 1(remainder)
  so we have to sbtract 1
• 11 years agoHelpfull: Yes(52) No(0)
• 4^15=(4^3)^5=(64)^5=(62+2)^5=(2)^5=32
  32-1=31
  so one(1) should be subtracted
• 11 years agoHelpfull: Yes(12) No(0)
• Using Remainder Theorem,
  4^15/31
  ==> (4*4*4)^5/31
  ==> (2)^5/31
  ==> 32/31
  Remainder is 1
  Ans is :- 1
• 11 years agoHelpfull: Yes(5) No(2)
• 1
  remainder theorm
  ((31 +1)^6)/31 = 1
• 11 years agoHelpfull: Yes(1) No(0)
• 4^15=4^3=64
  if 64-2=62
  62/2=31.
  hence 2 is substracted.
• 11 years agoHelpfull: Yes(1) No(1)
• we have to sbtract 1
• 11 years agoHelpfull: Yes(1) No(1)
• 4^15=2^30=(2^5)6=32^6/31=(31+1)^6/31
  by applying binomial expansion=1^6=1
• 11 years agoHelpfull: Yes(1) No(0)
• answer = 1 only
• 11 years agoHelpfull: Yes(0) No(0)
• the remainder is 1
• 11 years agoHelpfull: Yes(0) No(0)
• what is the tricks for solving this kind of solution...kindly describe.....
• 11 years agoHelpfull: Yes(0) No(0)