what is the probability of dividing a randomly selected number (x^4 + 1) by 5,where as x is a four digit number?

• (x^4 + 1) is divisible by 5 when last digit is 0 or 5.
  if we replace X by any of d no: from 0 to 9, non of d no: contain last digit 0 or 5.
  hence probability is 0.
• 11 years agoHelpfull: Yes(42) No(7)
• unit place of x could be 0-9;
  unit place of x^4 will be any of these
  0^4=0;1^4=1;2^4=6;3^4=1;4^4=6;5^4=5;6^4=6;7^4=1;8^4=6;9^4=1
  unit plase of x^4+4 would be respectively
  4, 5, 0, 5, 0, 9, 0, 5, 0, 5
  
  
  from above no those who is 0 or 5 can be divided by 5
  hense
  probability=8/10=0.8
• 11 years agoHelpfull: Yes(11) No(13)
• any integer^4 gives 1 or 6 as unit digit except an integer containing 5 as a unit digit. according to the given condition, in all the possible cases last digit is 2,6 or 7.
  Hence probability is 0.
• 11 years agoHelpfull: Yes(9) No(0)
• The probability is 0 since X^4 will either end 1,6, or when the power is of 5 then it will end in 5 and if then x^4+1 is never divisble by by5 hence probability is 0
• 11 years agoHelpfull: Yes(3) No(1)
• multiplication by 4 gives 4,8,2,6,0 as unit digits. for a no. being divisible by 5 by adding 1 the unit digit must be either 9 or 4 . chances of meeting such condition by multiplication by 4 are 1/5 as 9 is the only no of our interest.. so.. answer is 1/5
• 11 years agoHelpfull: Yes(2) No(4)
• probability will be zero as for any no x^4 last digit will not be either 4 0r 9
  so 0 is answer
  
• 11 years agoHelpfull: Yes(2) No(0)
• 4/5 is the only ans
• 9 years agoHelpfull: Yes(1) No(0)
• ans is 1/5.
  the last digit can be from 1 to 9...only if the last digit is 5 or 0...the last digit will be 0 or 5....divisibility by 5 is possible...hence probability is 2/10 or 1/5
• 11 years agoHelpfull: Yes(0) No(14)
• The unit digit of any number to the power 4 is either 1 or 6.. and even if we add 1, the resulting number will not be divisible by 5..
• 11 years agoHelpfull: Yes(0) No(2)
• for (x^4 + 1) to be divisible by 5, than it should have 0 or 5 as the last digits.
  unit place of x^4 will be any of these
  0^4=0;1^4=1;2^4=6;3^4=1;4^4=6;5^4=5;6^4=6;7^4=1;8^4=6;9^4=1
  hence if we replace X by any of d no: from 0 to 9, non of d no: contain last digit 0 or 5.
  hence probability is 0.
• 10 years agoHelpfull: Yes(0) No(0)