7. Determine the largest integer $n$ with this property: there exist $n$ distinct positive integers $a_{1}, \ldots a_{n}$ such that, for any two of them chosen, neither their sum nor their difference is divisible by 100.
(A) 49
(B) 50
(C) 51
(D) 99
(E) 100.
7. The answer is (C). A number is divisible by 100 if and only if its decimal expression ends with 00.

We divide the integers $a_{1}, \ldots a_{n}$ into groups as follows: in the first group, we put those whose decimal expression ends with 00, in the second group those that end with 01 (or have only one digit and that digit is 1) or with 99, in the third group those that have the last two digits 02 or 98, and so on, until we reach the fifty-first group, in which we put those that end with the digits 50.

If there are two integers in the same group, then necessarily their difference, or their sum, will be divisible by 100: if indeed the two integers end with the same pair of digits, then their difference ends with 00 and is divisible by 100; if instead they end with different pairs of digits, the fact that they are in the same group means that their sum is divisible by 100. Since we have exactly 51 groups, $n$ is at most 51: if we had 52 or more integers, at least two would fall into the same group and, as stated, their difference, or their sum, would be divisible by 100.

On the other hand, it is easy to see that $100, 101, \ldots, 150$ is precisely a set of 51 positive integers with the property described in the text, and thus 51 is the required number.