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

Answer 1

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.

Complete step-by-step answer:
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.