If 12 divides ab313ab (in decimal notation, where a, b are digits > 0, the smallest value of a+b is

• let us consider a=1,b=2
  then 1231312 will not be divisible by 3 which is one of the factor of 12.
  let us conider a=1,b=6
  1631312 will be divisible by all factors of 12.hence a+b=1+6=7
• 5 years agoHelpfull: Yes(6) No(1)
• 4031340/12== 4+0=4
• 5 years agoHelpfull: Yes(4) No(7)
• ab313ab: To be divisible by 12 it must be divisible by 3 and 4.
  Now the divisibilty rule of 4 says that the last two-digit number must be divisible by four.
  ab=04,08 this cannot be true as (a and b>0)
  ab=12(divisibility rule for four is satisfied but 3+1+3+2(a+b)=7+6=13 cannot be divisible by 3
  ab=16(this can be divisible by both 3 and 4 so a=1,b=6 ,answer is a+b=7)
• 5 years agoHelpfull: Yes(3) No(0)
• 3231332/12==0;3+2=5;
• 5 years agoHelpfull: Yes(2) No(8)
• ab313ab is divisible by 12 ,means ab313ab is divisible by 3 and 4 .
  
  
  condition 1 :- we know, any number is divisible by 3 when sum of all digits of number is divisible by 3.
  so, a + b + 3 + 1 + 3 + a + b = 7 + 2(a + b)
  therefore, 7 + 2(a +b) is integral multiple of 3.
  e.g., 7 + 2(a+ b) = 3n
  if n = 3 , (a + b) = 1
  if n = 5 , (a + b) = 4
  if n = 7, (a + b) = 7
  hence, (a + b) = 1 + 3P , where P = {0, 1, 2, 3...}
  
  condition 2:- now, one more condition is ab313ab is divisible by 4 , it is possible only when last two digit ab is divisible by 4
  
  
  a , b > 0 , so we have to choose a and b in such a way that it a , b > 0 , (a + b) = 1+ 3P, where P is whole number and ab is divisible by 4.
  we can choose a = 1 & b = 6 , a = 5 & b = 2, etc..
  so, smallest value of (a + b) = 1 + 6 or (5+ 2) =7
  
  hence, answer should be 7.
• 5 years agoHelpfull: Yes(1) No(2)
• a, b >0
  ab313ab/12
  ab313ab/(3and4)
  ab/4 possible 12,16,24,28,32,36,44,48,52,56,64,68,72,76,84,88,92,96
  (a+b+3+1+3+a+b)=(2a+2b+7)/3
  Possible a and b=16,28,52,64,76
  a+b smallest value 7
• 5 years agoHelpfull: Yes(1) No(1)