2019 AMC 10 B

We can figure out by noticing that will end with zeroes, as there are three in its prime factorization. Next, we use the fact that is a mutiple of both and . Their divisibilty rules (see Solution ) tell us that (mod ) and that (mod ) By inspection, we see that , is a valid solution. Therefore the answers .

We know that and are both factors of , Furthermore, we know that , because ends in three zeroes (see Solution ). We can simply use the divisibility rules for and for this problem to find and . For to be divisible by , the sum of digits must simply be divisible by . Summing the digits, we get that must be divisible by . This leaves either or as our answer choice. Now we test for divisibility by . For a number to be divisible by , the alternating sum must be divisible by (for example, with the number , , so is divisible by ). Applying the alternating sum test to this problem, we see that must be divisible by . By inspection, we can see that this holds if and The sum .