A six-digit number is written as 3a5b2c where a, b, c are digits it is divisible by 4, 5 and 9 what is the maximum value of a+b?

Answer: D. 17

Since the number (3a5b2c) is divisible by 5, its last digit \\(c\\) must be either 0 or 5. As it is also divisible by 4, its last two digits must be divisible by 4. Therefore, \\(c=0\\), because 20 is divisible by 4 whereas 25 is not. Now, using the divisibility rule of 9, the sum of all digits must be divisible by 9. Hence, \\(3+a+5+b+2+0=10+a+b\\). Since \\(a+b\\) is at most 17, the maximum possible multiple of 9 is 27. Therefore, \\(10+a+b=27\\), giving \\(a+b=17\\). **Correct answer: Opt

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