. ab313ab when divided by 12 leaves no remainder. What is the value of a+b?

• as it is divided by 12 then the digit should be divided by by 3& 4 respectively.now for divisible by 4 last 2 digit should be divisible by4 & sum of the no.s is divisible by 3..so sum is(3+1+3+2a+2b)/3 remainder should be 0.so(7+2a+2b)=should be 9,12,15..for 9..2(a+b)=2so (a+b)=1 so last 2 digit is 10..but it is not divisible by 4
  so, we take 7+2a+2b=15,a+b=8/2=4..now ab may be 40 which is divisible by 4...so the solution is (a+b)=4
• 11 years agoHelpfull: Yes(27) No(1)
• 2(a+b)+7 mod 3 =0
  a+b=16
• 11 years agoHelpfull: Yes(2) No(2)
• ans is 4
  for divisible by 12,the no is divisible by 3,4
  now for 3
  a+b+3+1+3+a+b=2(a+b)+7
  and for 4
  last 2 digit should be divided by 4
  it means ab/4
  so,a+b=4
• 11 years agoHelpfull: Yes(2) No(0)
• a=6,b=4.....a+b=10
• 11 years agoHelpfull: Yes(1) No(2)
• a=1,b=6
  1631316 divided by 0 leaves no remainder
  So a+b=1+6
  =7
• 11 years agoHelpfull: Yes(1) No(0)
• a+b=4
  since a+b+3+1+3+a+b= 2(a+b)+7: now let a+b=x
  then on trial let x=4 then if a=4 b=0
  then the required no is divisible by 3,4,and 6 so it divisble by 12 and leaves no reamainder
• 11 years agoHelpfull: Yes(0) No(0)
• for completely divisible by 12 this number have to divisible by 4 and 3 individually.
  for divisible by 3
  sum of the number is divisible by 3
  (a+b+3+1+3+a+b)/3 gives rem=0;
  for divisible to be 4-
  last 2 digit is divisible by 4
  ab=4
  then a is either 1 or 3
  and b is either 1 or 3
  so a+b=4
  4 is the ans
  
• 11 years agoHelpfull: Yes(0) No(0)
• ans -(a+b)=7, a=1 n b=6
• 11 years agoHelpfull: Yes(0) No(1)
• a+b=10
  a=2
  b=8
• 11 years agoHelpfull: Yes(0) No(1)
• 12=3*4 so,ab313ab has to be divided 3 and 4.(a+b+3+1+3+a+b)mod 3=0 and (a+b)mod 4=0. (a+b)=4 satisfy these conditions
• 11 years agoHelpfull: Yes(0) No(0)
• Which is the CRT ans
• 8 years agoHelpfull: Yes(0) No(0)
• as it is divided by 12, it must be divided by 3 and 4.
  hence last two digits are divisible by 4.
  total sum of all digits are divisible by 3.
  (a+b+3+1+3+a+b)%3=0
  (7+2(a+b))%3=0
  and
  ab is divisible by 4.
  let us take ab=12 which is divisible by 4.
  ==> a+b=3
  ==> (7+2(3))=13 which is not divisible by 3. so ab is not equal to 12.
  let us take ab=16 which is divisible by 4.
  ==> a+b=7
  ==> 7+14 =21 which is divisible by 3.
  so ab="16" ==> a+b=7.
  let us take another example. ab=76 which is divisible by 4.
  ==> a+b=13
  ==> 7+2(13) = 33 which is divisible by 3.
  so ab= "76" ==> a+b= 13
• 7 years agoHelpfull: Yes(0) No(0)
• as it is divided by 12, it must be divided by 3 and 4.
  hence last two digits are divisible by 4.
  total sum of all digits are divisible by 3.
  (a+b+3+1+3+a+b)%3=0
  (7+2(a+b))%3=0
  and
  ab is divisible by 4.
  let us take ab=12 which is divisible by 4.
  ==> a+b=3
  ==> (7+2(3))=13 which is not divisible by 3. so ab is not equal to 12.
  let us take ab=16 which is divisible by 4.
  ==> a+b=7
  ==> 7+14 =21 which is divisible by 3.
  so ab="16" ==> a+b=7.
  let us take another example. ab=76 which is divisible by 4.
  ==> a+b=13
  ==> 7+2(13) = 33 which is divisible by 3.
  so ab= "76" ==> a+b= 13
• 7 years agoHelpfull: Yes(0) No(0)