Official Solution: First of all any positive integer can yield only three remainders upon division by 3: 0, 1, or 2.

Since, the sum of the digits of \(10^m\) and \(10^n\) is always 1 then the remainders of \(\frac{10^m + n}{3}\) and \(\frac{10^n + m}{3}\) are only dependent on the value of the number added to \(10^m\) and \(10^n\). There are 3 cases:

If the number added to them is: 0, 3, 6, 9, ... then the remainder will be 1 (as the sum of the digits of \(10^m\) and \(10^n\) will be 1 more than a multiple of 3);

If the number added to them is: 1, 4, 7, 10, ... then the remainder will be 2 (as the sum of the digits of \(10^m\) and \(10^n\) will be 2 more than a multiple of 3);

If the number added to them is: 2, 5, 8, 11, ... then the remainder will be 0 (as the sum of the digits of \(10^m\) and \(10^n\) will be a multiple of 3).

(1) \(m \gt n\). Not sufficient.

(2) The remainder of \(\frac{n}{3}\) is \(2\). So, \(n\) could be: 2, 5, 8, 11, ... so we have the third case. Which means that the remainder of \(\frac{10^m + n}{3}\) is 0. Now, the question asks whether the remainder of \(\frac{10^m + n}{3}\),

which is 0, is greater than the reminder of \(\frac{10^n + m}{3}\),

which is 0, 1, or 2. Obviously it cannot be greater, it can be less than or equal to. So, the answer to the question is NO. Sufficient.

Answer: B

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