let S be a set of postive integers such that every element n of S satisfies the condition
1-1000<=n<=1300
2-every digit in n is odd
then how many element of S are divisible by 3 ???
a-9
b-15

• a-9
  elements of S exists between 1110 and 1199.
  Elements are 1113, 1119, 1131, 1137, 1155, 1173, 1179 , 1191, 1197
• 10 years agoHelpfull: Yes(27) No(9)
• possible combinations are
  (1,1,1,3)= 2 ways
  (1,1,1,9)= 2 ways
  (1,1,3,7)= 2 ways
  (1,1,5,5)=1 way
  (1,1,7,9)= 2 way
  
  so total 9 numbers are possibles between 1000 and 1300 which are divisible by 3.
• 10 years agoHelpfull: Yes(19) No(3)
• n is between 1000-1300 and should be odd and divisible by 3..
  so first no. is 1005
  last no. is 1299
  
  after 1002 is even 1005 is odd and again 1008 is even
  so next number is 1011.. so after difference of 6 again we get odd and divisible by 3.
  
  so now using A.P
  
  1299=1005 + (n-1)6
  we get n=50
• 9 years agoHelpfull: Yes(14) No(10)
• as per the given question.
  n is between 1000 and 1300 and also each digit is odd no. so that no. is as follows
  1111,1113,1115,1117,1119
  1131,1133...................,1139
  .
  .
  .
  .
  .
  1191....................................1199
  
  we found the value of n now you have to see which no. is divisible by 3
  
  
  so that ans=9
• 10 years agoHelpfull: Yes(10) No(2)
• i think given options for answer is incorrect.
  Answer would be 50.
  possible elements of n={1005,1011,1017,...1299}
  so here we can see that the diff. between two number is 6.so total number divisible by 3
  =((1299-1005)/2) + 1
  =50
• 9 years agoHelpfull: Yes(6) No(7)
• Manju why u have not started with 1005 ,1017 this number are also divisible by 3 and all digits are odd.
• 9 years agoHelpfull: Yes(2) No(2)
• l=1299
  a=1002
  l=a+(n-1)d
  so that n=100
• 10 years agoHelpfull: Yes(1) No(8)
• n>=1000 so first such no. can be 1005 since it is odd and divisible by 3
  n
• 9 years agoHelpfull: Yes(1) No(4)
• given options are incorrect
  1st number that satisfies the condition is 1005 then 1011,1017....(common difference is 6)
  last number will be 1299 (1005,1011,1017............,1293,1299) this is the series
  using AP(a=1005, d=6) 1299=1005+(n-1)6
  answer is n=50
• 9 years agoHelpfull: Yes(1) No(2)
• 44 becoz 1000 to 1100 ->14
  1100 to 1200->16
  1200 to 1300->14
  14+14+16
• 8 years agoHelpfull: Yes(0) No(0)
• Those who are saying 9 are wrong.
  Here is the solution
  series will be (1005,1011,1017,1023........ 1239,1299) They all are the odd no. and divisible by 3
  So
  a=1005 d=6 last digit(l)=1299
  l=a + (n-1)d
  1299 = 1005 +(n-1) 6..solve this
  50 = n Answer
• 2 years agoHelpfull: Yes(0) No(0)