2020 AIME I Problems/Problem 4
Let be the set of positive integers with the property that the last four digits of are and when the last four digits are removed, the result is a divisor of For example, is in because is a divisor of Find the sum of all the digits of all the numbers in For example, the number contributes to this total.
We note that any number in can be expressed as for some integer . The problem requires that divides this number, and since we know divides , we need that divides 2020. Each number contributes the sum of the digits of , as well as . Since can be prime factorized as , it has factors. So if we sum all the digits of all possible values, and add , we obtain the answer.
Now we list out all factors of , or all possible values of . . If we add up these digits, we get , for a final answer of .
Solution 2 (Official MAA)
Suppose that has the required property. Then there are positive integers and such that . Thus , which holds exactly when is a positive divisor of The number has divisors: , and The requested sum is therefore the sum of the digits in these divisors plus times the sum of the digits in which is
Note that for all , can be written as for some positive integer . Because must be divisible by , is an integer. We now let , where is a multiple of . Then . We know and are integers, so for to be an integer, must be an integer. For this to happen, must be a multiple of . is prime, so . Because is a multiple of , . So . Be know that all end in , so the sum of the digits of each is the sum of the digits of each plus . Hence the sum of all of the digits of the numbers in is .
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