text: Question
  
  # Mark the correct alternative of the following.If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of $a+b$is?(a) 12(b) 13(c) 14(d) 15
  
  Hint: We can use the divisibility rule for the numbers 3 and 5 to solve this question. In the case of number 5, the value of b can only be either 0 or 5. Similarly, we can apply conditions for number 3 also and get to a conclusion.
  
  Before proceeding we should know the divisibility rule of 3 as well as 5. These are as follows:
  Divisibility rule of 3: A number is divisible by 3 if and only if the sum of the digits of a number is divisible by 3.
  
  Divisibility rule of 5: A number is divisible by 5 if and only if the last digit of a number is either 0 or 5.
  
  Therefore, the number 2345 a 60b is divisible by 5 only if the value of $b=0\text{ or 5}$.
  
  We have to take the value of b as 5 as we have been asked the maximum value of $a+b$.
  
  Then the sum of the digits of the given number 2345 a 60b is $25+a$.
  
  Therefore, the numbers greater than 25 and divisible by 3 are 20, 30, 35.
  
  Hence, the value of a maybe 2, 5, 8 .
  
  By comparing, the possible values of a, we get that the maximum value of $a=8$
  
  Therefore, we have $a=8,b=5$.
  
  Hence, the maximum value of $a+b$ is $13$.
  
  Hence, the answer is option (b).
  
  Note: Always remember we have been asked for the maximum value of $a+b$, therefore for the number to be divisible by 5, the last digit must be 5 not 0 as we have been asked the maximum value of $a+b$. Make sure that after getting the maximum values of $a+b$, check at least once the number is divisible by both 3 and 5.
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