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Re: What is the largest digit n for which the number 123,45n is divisible
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20 Feb 2019, 19:08
A number is divisible by 3 if the sum of its digits is divisible by 3.
In the above question. 1+2+3+4+5+n should be a number that is divisible by 3... And looking into the answer choices, the largest number that can take the value of n is 9, Because 1+2+3+4+5+9= 24 which is divisible by 3.
What is the largest digit n for which the number 123,45n is divisible
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21 Feb 2019, 09:28
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.
Re: What is the largest digit n for which the number 123,45n is divisible
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22 Feb 2019, 00:17
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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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86% (00:49) correct 
