When 335 is added to 5A7, the result is 8B2. | vedclass

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(D) Given the addition: $5A7 + 335 = 8B2$Looking at the units place: $7 + 5 = 12$, so the last digit is $2$ and there is a carry of $1$ to the tens pl

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$4$

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(D) Given the addition: $5A7 + 335 = 8B2$Looking at the units place: $7 + 5 = 12$, so the last digit is $2$ and there is a carry of $1$ to the tens place.Looking at the tens place: $A + 3 + 1 = B$ (where $1$ is the carry from the units place).So, $A + 4 = B$.Looking at the hundreds place: $5 + 3 = 8$, which matches the result.Since $8B2$ is divisible by $3$, the sum of its digits must be a multiple of $3$:$8 + B + 2 = 10 + B = 	ext{multiple of } 3$.Possible values for $B$ are $2, 5, 8$.Case $1$: If $B = 2$, then $A + 4 = 2 \Rightarrow A = -2$ (Not possible as $A$ is a digit).Case $2$: If $B = 5$, then $A + 4 = 5 \Rightarrow A = 1$.Case $3$: If $B = 8$, then $A + 4 = 8 \Rightarrow A = 4$.The largest possible value of $A$ is $4$.

When $335$ is added to $5A7$, the result is $8B2$. $8B2$ is divisible by $3$. What is the largest possible value of $A$?

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