y=x^4+4
x=any 5digit no..
find probability that y is divisible by 5??

• N can be any number so it will be 0.1,2,...9
  x=-----(5 digit no)
  y=(x^4+4)/5;
  so the value of x can be as a last digit 1,2,3,4,6,7,8,9
  total possible value is (0 to 9)=10
  so probability is =8/10=4/5
  i.e.
  (0)^4+4=4(not divisible)
  (1)^4+4=5
  (2)^4+4=20;
  (3)^4+4=85;
  (4)^4+4=260;
  (5)^4+4=___9(not divisible)
  (6)^4+4=__0;
  (7)^4+4=__5;
  (8)^4+4=__0;
  (9)^4+4=__5;
• 9 years agoHelpfull: Yes(51) No(1)
• Number divisible by 5 when last digit ends with either 0 or 5
  Last digit becomes 5 when x^4 ends with 1
  Last digit becomes 0 when x^4 ends with 6
  probability of 1 or 6 =>2/10=> 1/5
• 9 years agoHelpfull: Yes(7) No(10)
• 4/5 is correct ans
• 9 years agoHelpfull: Yes(5) No(0)
• Ans: 4/5
  
  Total possible 5 digit numbers=90000
  y=(x^4+4) & y%5=0
  So unit digit must be 1 or 2,3,4,6,7,8,9.
  Possible numbers=9*10*10*10*8=72000
  So probability is =72000/90000=4/5
• 8 years agoHelpfull: Yes(5) No(0)
• N is natural number so it will be 1,2,...9
  x=-----(5 digit no)
  y=(x^4+4)/5;
  so the value of x can be as a last digit 1,2,3,4,6,7,8,9
  total possible value is (1 to 9)=9
  so probability is =8/9;
  i.e (1)^4+4=5
  (2)^4+4=20;
  (3)^4+4=85;
  (4)^4+4=260;
  (6)^4+4=__0;
  (7)^4+4=__5;
  (8)^4+4=__0;
  (9)^4+4=__5;
  prob=8/9;
  
• 9 years agoHelpfull: Yes(4) No(4)
• answer : the five digit no. X will be divisible only when it has 5 "or" 0 in the one's position .. and the a five digit no. is possible when the one's position is 0, 1, 2, 3, 4 ,5 ,6 ,7 , 8, 9 ..i.e, in 10 ways ..but for divisibility from 5 there are 5 cases possible when in these digits are in one's position 1=> 1+4=5 , 3=>1+4=5,7=>1+4=5,9=>1+4=5
  So, 4/10 = 2/5 Answer
• 9 years agoHelpfull: Yes(0) No(6)