how many no. are between 1100 - 1300
which are divisible by 3
but condition is that all 4-digits of no. should be odd
for example-1131

• 1300-1100=200 now divide it by 2 we have 100 even number hence 100 odd no's now we see 50 are in each 1100-1200 and 1200-1300 now we see in 1200 to 1299 2 will always be there hence no number frm here in 1100-1199 we have 1113,1119,1131,1137,1155,1173,1179,1191,1197 hence
  9 no's
• 10 years agoHelpfull: Yes(45) No(8)
• ans:34
  1st no divisible by 3 in b/w 1100-1300 is 1101.
  so if we see the pattern which is divisible by 3 and shoulb be odd like this
  1101,1107,1113,1119,1125,1131..............................1299.
  d=6
  an=1299
  so now,1299=1101+(n-1)6
  =>n=34
• 10 years agoHelpfull: Yes(9) No(22)
• Amit bhai, total no are-9.
  1113,1119,1131,1137,1155,1173,1179,1191,1197.
• 10 years agoHelpfull: Yes(6) No(3)
• so according to given condn we hv to find only d no. bw 1100 to 1200 because after that all no. have 2 which is even 1201 to 1299. so nos which setesfies above condn r 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1195,.
  so there r total 12 no.s.
  so ans will be 12.
• 10 years agoHelpfull: Yes(3) No(11)
• Sorry the answer will be 13.
  I missed out d 2nd coditon in my 1st answer(All four digits shall be odd).
• 10 years agoHelpfull: Yes(2) No(6)
• Meghna baby 0 odd ni h...so answer is 9
• 10 years agoHelpfull: Yes(2) No(0)
• sorry ans will be 13 i missed the no. 1197.
• 10 years agoHelpfull: Yes(1) No(2)
• sorry ans will be 12 nd no.s will be 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1197
• 10 years agoHelpfull: Yes(1) No(5)
• answer is 13
  no's are 1101,1107,1110,1113,1119,1131,1137,1155,1170,1173,1179,1191,1197
  this satisfy's both the conditions
• 10 years agoHelpfull: Yes(1) No(2)
• 67 no.s are divisible
• 10 years agoHelpfull: Yes(0) No(12)
• Amit please explain your ans.I know my ans is wrong.
• 10 years agoHelpfull: Yes(0) No(8)
• 34 nos
  11_ _ last digit 1 (2nd pos 0 3 6 9)= 4 sum 3
  last digit 3 (2nd pos 1 4 7)= 3 sum 5
  last digit 5 (2nd pos 2 5 8)= 3 sum 7
  last digit 7 (2nd pos 0 3 6 9)= 4 sum 9
  last digit 9 (2nd pos 1 4 7)=3 sum 11 tot 17
  12_ _ last digit 1 (2nd pos 2 5 8)= 3 sum 4
  last digit 3 (2nd pos 0 3 6 9 )= 4 sum 6
  last digit 5 (2nd pos 1 4 7)= 3 sum 8
  last digit 7 (2nd pos 2 5 8)= 3 sum 10 tot 13
  last digit 9 (2nd pos 0 3 6 9)=4 sum 11 tot 17
  
• 10 years agoHelpfull: Yes(0) No(0)
• considering the second condition,numbers between 1200 to 1300 are of no use as they contain even digits.
  so between 1100 and 1199....no of numbers divisible by 3 are=30 (since 3 nos in between 1111 & 1120..similarly in each series of 10 nos,there will be 3 nos divisible by 3...so,3*9=27......also 1110,1140,1170 are divi by 3...hence,27+3=30).
  now,the 3 nos in each series with even nos is eliminated i.e 1121-1129,1141-1149,1161-1169,1181-1189. so,4*3=12 nos are eliminated.
  also 1110,1140,170 i.e 3 nos are eliminated.
  hence,30-12-3=15.
  now 1116,1134,1152,1158,1176,1194 contain even so,6 nos are eliminated.
  total=15-6=9.
• 10 years agoHelpfull: Yes(0) No(0)
• I think is must be 34
  Between 1100-1300
  no. divisible by 3 and also should be odd..
  i.e. starting from 1101 ....next will be 1107..
  So series will be 1101, 1107, 1113...upto 1299
  total terms will be 34..agree with Rupesh
• 10 years agoHelpfull: Yes(0) No(1)
• ans should be 19
• 10 years agoHelpfull: Yes(0) No(0)