Find the smallest no divisible by 5 ,3,7 and leave a remainder of 2 in each case and is divisible by 11
Option
a) 737
b) 735
c) 105
d) 107

• the smallest no divisible by 5,3 and 7 and leaving reminder 2
  and which is divisible by 11 is 737..
  737/11=67
  and when divided by 5,3 & 7 leave the reminder 2.....
  so ans is option a.737
• 10 years agoHelpfull: Yes(15) No(0)
• LCM of 5,3,7=105
  737%105=2
  737%11=0
  737
• 10 years agoHelpfull: Yes(9) No(0)
• to be divisible by 11, it is necessary that difference between the digit's sum at odd places and digit's sum at even place is 0 or divisible by 11.
  so here only 737 is the only digit which is divisible by 11.
  like:7+7-3=11, which is divisible by 11.
• 10 years agoHelpfull: Yes(7) No(0)
• 737 it is divisible by 3 , 5 and 7 leaves remainder 2 , also divisible by 11
• 10 years agoHelpfull: Yes(2) No(1)
• 7+7-3=11, divisible by 11. 737 is only number which is divisible by 11.
  LCM of 3,5,7 is 105 and 105*5+2=737. So, 2 is the remainder.
• 10 years agoHelpfull: Yes(1) No(1)
• Options b and c are not correct as they are divisible by 5. The remaining options are a and d. To be a multiple of 11 first take sum of the digits at odd place and even place and then take difference. In this case
  7+7-3=11, which is divisible by 11. So ans is a.
• 10 years agoHelpfull: Yes(0) No(0)
• Options b and c cannot be the answer as they are divisible by 5. Remaining options left are a and c. To be divisible by 11 the difference between the sum of digits on odd places and even places should be divisible by 11. In this case
  7=7-3=11
  Hence divisible by 11.
• 10 years agoHelpfull: Yes(0) No(0)
• a)737,reason divisible by 11 and divide by 5,3,7gives remainder 2
• 10 years agoHelpfull: Yes(0) No(0)
• 737
  737%5=2
  737%3=2
  737%7=2
  737%11=0
  no other option is possible
• 10 years agoHelpfull: Yes(0) No(0)
• a)737
  737/11=67 that is it is correctly divisble by 11
  in all other case it leaves remainder 2
• 10 years agoHelpfull: Yes(0) No(0)
• answer is a)737
  we know that divided by 3,7,5
  and gives reminder 2 nd which no is also divisible by 11
• 10 years agoHelpfull: Yes(0) No(0)
• 737=737/5+737/3+737/7 the all number leaves a remainder 2 and it is divisible by 11
  
• 10 years agoHelpfull: Yes(0) No(0)
• a , since 735 is only divisible by 11
  
• 10 years agoHelpfull: Yes(0) No(0)
• ans is option (a)
• 10 years agoHelpfull: Yes(0) No(0)
• IN CASE OF DIVISABILITY BY 11 , it is necessary that difference between the digit's sum at odd places and digit's sum at even place is 0 or divisible by 11.
  so here only 737 is the only digit which is divisible by 11.
  like:7+7-3=11, which is divisible by 11.
• 9 years agoHelpfull: Yes(0) No(0)