sum of the digit of a five number is 41.find the probability that such a number is divisible by 11.

• in order to get the sum as 41, the following 5 digit combination exist:
  99995-> no.of 5 digits=5
  99986-> no.of 5 digit=20
  99977-> no.of 5 digit=10
  99887-> no.of 5 digit=30
  98888-> no.of 5 digit=5
  
  now, 70 such number exists.,
  now for a 5 digit number of form(pqrst) to be divisible by
  (p,r,t)+(q+s)=41
  (p,r,t)=(9,9,8) and (q,s)=(8,7)-----(1)
  or (p,r,t)=(9,9,8) and (q,s)=(9,6)----(2)
  
  using 1st eqn we can construct (3!/2!)*2! = 6 numbers
  using 2nd eqn we can construct (3!/2!)*2! = 6 numbers
  
  number of favble cases = 12.,
  
  hence required probability = 12/70 = 6/35
• 12 years agoHelpfull: Yes(33) No(12)
• since number divisible by 11 =sum of digit of number must be divisible by 11;
  
  here 41 is not divisible by 11.
  so probability of number divisible by 11 is 0;
• 14 years agoHelpfull: Yes(15) No(76)
• let the sum of digits at odd places be a and those at even places be b; such that a+b=41;a-b: divisible by 11; among all the possible combinations of (a,b) starting from (41,0) to (1,40) there lies only (26,15) and (37,4) only whose difference is divisible by 11 thus probability is (2/41)
• 9 years agoHelpfull: Yes(3) No(5)
• In order to get the sum as 41, the following 5 digit combination exist :
  
  99995--->(5!/4!)=5 numbers.
  
  99986--->(5!/3!)=20 numbers.
  
  99977--->5!/(3!2!)=10 numbers.
  
  99887--->5!/(2!2!)=30numbers.
  
  98888--->5!/4! =5 numbers.
  
  Thus, 5 + 20 + 10 +30 +5 i.e 70 such number exists.
  
  For a 5 digit number in the form (pqrst) to be divisible by 11, the condition is :-
  
  =>(p+r+t)−(q+s)=11. ---------------(a)
  
  Also, it is given :-
  
  =>(p+r+t)+(q+s)=41. ---------------(b)
  
  From (a) and (b), we get :-
  
  p+r+t=26, q+s=15.
  
  (p,r,t)=(9,9,8) and (q,s)=(8,7) ----------------(i)
  
  or (p,r,t)=(9,9,8) and (q,s)=(9,6) ----------------(ii)
  
  Using 1st equation, we have :-
  
  =>(3!/2!)x2!= 6.
  
  Using 2nd equation we have :-
  
  =>(3!/2!)x2! =6.
  
  Thus, the number of favourable cases = 12.
  
  Hence, the required probability = 12/70 = 6/35.
• 7 years agoHelpfull: Yes(3) No(0)
• sorry to say this krishna kumar guptha 0 is wrong answer because the question is the probability for that 5 digit number not for the 41.
• 9 years agoHelpfull: Yes(2) No(4)