2020 AIME I Problems/Problem 4
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
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
![\[(1+2+4+5+1+2+2+4+8+10+2+4)+12\cdot4 = 93.\]](http://latex.artofproblemsolving.com/f/f/c/ffc71b7c3c68ff3b82207ca3f41adf1ced0699be.png)
Video solution:
See Also
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