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Question 1

The three-digit number $$3a4$$ is added to 278 to obtain another three-digit number $$6b2$$, where $$a$$ and $$b$$ are digits. If $$6b2$$ is divisible by 9, then $$a + b = $$

Solution

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Since $$6b2$$ is divisible by 9, the digit sum $$6 + b + 2 = 8 + b$$ must be a multiple of 9, which forces $$b = 1$$. Then $$6b2 = 612$$ and $$3a4 = 612 - 278 = 334$$, so $$a = 3$$. Hence $$a + b = 3 + 1 = 4$$.

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