2020 AIME I Problems/Problem 4
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Problem
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.
Solution 1
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 .
-molocyxu
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
Solution 3
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 divisor of . Then . We know and are integers, so for to be an integer, must be an integer. For this to happen, must be a divisor of . is prime, so . Because is a divisor 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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See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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