If the number 2345a60b is exactly divisible by 3 and 5, then the maximum value of a+b is ____.

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Detailed Solution

If the number 2345a60b is exactly divisible by 3 and 5, then the maximum value of a+b is 13.
Given data,
2345a60b is exactly divisible by 3 and 5.
In the divisibility rule of 3 and 5,
The number is completely divisible by 3 if the sum of the digits is divisible by 3.
The number is completely divisible by 5 if the last digit of the given number is 0 or 5.
Therefore, the value of b must be 0 or 5, as b is the last digit of the given number.
a+b has to be maximum thus b will be 5.
Substituting the value of b,
We get 2345a605.
Now, adding the digits 2345a605,

\Rightarrow

2+3+4+5+a+6+0+5

\Rightarrow

25+a
If we add 2, 5 or 8 in 25 we get a number greater than 25 and divisible by 3.
As, we need the maximum value of a.
Thus, the value of ‘a’ will be 8.
Now, calculating the value of a+b.
Substituting the values,
a+b=8+5
a+b=13
Hence, the value of a+b is 13.

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