2020 AMC 10 A

Solution 1: We know that , so we can write . Simplifying, we get . Similarly, we can write or . Solving these two modular congruences, which we know is the only solution by CRT (Chinese Remainder Theorem). Now, since the problem is asking for the least positive integer greater than 1000 , we find the least solution is . However, we are have not considered cases where or . so we try so again we add 420 to . It turns out that does indeed satisfy the original conditions, so our answer is (C) 18

Solution 2 (bashing): We are given that and . This tells us that is divisible by but not . It also tells us that is divisible by but not . Starting, we find the least value of which is divisible by which satisfies the conditions for , which is , making . We then now keep on adding until we get a number which satisfies the second equation. This number turns out to be , whose digits add up to (C) 18