Web Reference: Since lcm (1, 2, ..., 100) is much larger than 100!, the number of such integers is 100!. Therefore, k = 100!. The question asks for the sum of the squares of the digits of k. Since k = 100!, this is not directly calculable. However, the question seems to have a flaw. Given: We need positive integers n ≤ L, where L = lcm (1,2,...,100), such that remainders of n mod m are all different for m = 2,3,...,100. Idea: For distinct remainders r_m = n mod m (for each m), the remainders must be a permutation of 0,1,...,m-1 for each m, but only one remainder per modulus. 1 pt for the fully correct answer, 2 pts for obtaining d + a = (c + d)(d − c + 1) and all its analogs (1 pt if only some are present), 2 pts for reduction to the final equation (1 pt if division is made by an expression that might be 0), 3 pts for applying AM-GM inequality (lose 1 pt if non-negativity is not checked), 2 pts for finishing.
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