Y=(X^4)+4
Where X is a 4 digit number. What is the probability that Y is divided by 5
a)1/5
b)2/5
c)4/5
d)1

This is asked in today's(9/2/14) exam

• here is the silution .
  
  when the last digit of x is 1 , last digit of y=(x^4)+4 is 5( as last digit of x^4 is 1 and 1+4=5 last digit of y)
  
  similarly for 2 ..last digit of y is 0 (div by 5)
  for 3.................y is 5 (div by 5)
  for 4.................y is 0 (div by 5)
  for 5.................y is 9 (not div by 5 )
  for 6.................y is 0 (div by 5)
  for 7.................y is 5 (div by 5)
  for 8.................y is 0 (div by 5)
  for 9.................y is 5 (div by 5)
  for 0.................y is 4 (not div by 5)
  
  hence probality = 8/10
  = 4/5......is the answer.. :)
  
  
• 10 years agoHelpfull: Yes(35) No(5)
• Y=(X^4)+4
  y is divisible by 5 only if unit digit of y is 0 or 5
  
  unit digit of x=0 => last digit of y = 4
  unit digit of x=1 => last digit of y = 5
  unit digit of x=2 => last digit of y = 0
  unit digit of x=3 => last digit of y = 5
  unit digit of x=4 => last digit of y = 0
  unit digit of x=5 => last digit of y = 9
  unit digit of x=6 => last digit of y = 0
  unit digit of x=7 => last digit of y = 5
  unit digit of x=8 => last digit of y = 0
  unit digit of x=9 => last digit of y = 5
  
  i.e when unit digit of x is 0 & 5 then y is not divisible by 5
  
  x= total 4 digit no. = 9*10*10*10 = 9000
  y=(x^4+4),divided by 5= 9000-9*10*10-9*10*10 = 7200
  
  so p= y/x = 7200/9000 = 72/90 = 4/5
  
• 10 years agoHelpfull: Yes(31) No(1)
• y is divisible by 5 only if y has its last digit as "0" or "5"
  ;;;
  x may has its last digit 0-9.
  so to make y divisible by 5 we need last digit of x^4 as 1 or 6 so that
  y=x^ + 4= (...1)+4=...5 or
  y=x^ + 4= (...6)+4=...0
  
  so we have only 8 cases where we get y divisible by 5
  if last digit of x is 1,2,3,4,6,7,8,9
  u can get this from cyclicity
  1-> 1,1,1,1
  2-> 2,4,8,6
  3-> 3,9,7,1
  4-> 4,6,4,6
  6-> 6,6,6,6
  7-> 7,9,3,1
  8-> 8,4,2,6
  9-> 9,1,9,1
  so out of 10 we have 8 chances where y is divisible by 5 so probability is
  = 8/10 =4/5 ( option c)
  
• 10 years agoHelpfull: Yes(3) No(0)
• x is 4 digit no. if unit digit of(x^4) is 1 or 6 then it is divisible by 5.
  if unit place of 4 digit is 1,2,3,4,6,7,8,9 then we get output of unit place is 1 or 6.
  total 4 digit no is 9*10*10*10
  and divisible no is 9*10*10*8,
  hence probablity is 9*10*10*8/9*10*10*10= 72/90=4/5
• 10 years agoHelpfull: Yes(2) No(0)
• since X is a 4 digit no.
  i.e. let X = xxxx where x= 0 to 9,
  
  now ,to make y divisible by 5,
  which is possible if only last term of (x^4) is '1' or '6', in that case.
  
  => y = (xxx1)+ 4 = xxx5; or
  => y = (xxx6)+ 4 = xxx0
  
  therefore, to get last term as 5 or 0, last digit can only be substituted by '1'
  or '6'.
  it can be done in two (1,6) out of ten ways(0,1,2,3,4,5,6,7,8,9).
  so , prob. =2/10 or 1/5...
  
  1/5 is the answer.
  
• 10 years agoHelpfull: Yes(2) No(1)
• Last digit of X can be 1,2,3,4,6,7,8,9 (total 8 digits) so that expression(X^4 + 4) would return 0 or 5 as last digit (to be divisible by 5)
  we have total no of 4 digit numbers = 9000
  & no of 4 digit numbers having last digit among (1,2,3,4,6,7,8,9) is = 8*900
  probablity = 8*900/9000 = 4/5
• 10 years agoHelpfull: Yes(1) No(1)
• c)4/5
  as no should be divisible by 5 addition of (X^4)+4 should have units place as 0 or 5
  so nos will be 1,3,7,9,2,4,6,8
  so probability is 8/10=4/5
• 10 years agoHelpfull: Yes(1) No(0)
• possibility of 4 digit no's
  For unit place possible digits are 1,9,4,6
  For 10th place possible digits are 0 to 9
  For 100th place possible digits are 0 to 9
  For 1000th place possible digits are 1 to 9
  Hence, possible 4 digit nos = 9*10*10*4=3600
  total possible 4 digit nos = 9*10*10*10=9000
  Hence, Probability = 3600/9000=2/5.
  
• 10 years agoHelpfull: Yes(0) No(6)
• b) 2/5 because in the table of 4 only 20 40 is divided by 5 so it will come 2 times
• 10 years agoHelpfull: Yes(0) No(7)
• ans is c) 4/5
• 10 years agoHelpfull: Yes(0) No(0)
• @rakesh u r taking repition while takin total cases n non repeat while taking favorabl cases??
• 10 years agoHelpfull: Yes(0) No(0)
• for (x4 + 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)