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for a number to be divisible by 3 .

the sum of digita should be divisible by 3

here sum of digits = 15

so find the next biggest digit which can be added to 15 and the sum.is still divisible by 3

so the options are 3,6,9

9 is the biggest.

Option E

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We know that if sum of the digits are divisible by 3, the number is also divisible by 3.here, 4+5+9(largest one) = 18 which is divisible by 3. So answer is E.

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=>

Recall that a number is divisible by \(3\) if and only if the sum of its digits is divisible by \(3\). The sum of the digits of \(123,45n\) is \(1 + 2 + 3 + 4 + 5 + n = n + 15.\) This is divisible by \(3\) exactly when \(n\) is divisible by \(3\).
So, \(n\) must be a multiple of \(3\).
Thus, the largest digit, \(n\) is \(9\).

Therefore, the answer is E.
Answer: E
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Hello from the GMAT Club BumpBot!

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