2020 AIME I Problems/Problem 4

Revision as of 17:00, 4 May 2020 by Avn (talk | contribs) (Solution)


Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42,020$ is in $S$ because $4$ is a divisor of $42,020.$ Find the sum of all the digits of all the numbers in $S.$ For example, the number $42,020$ contributes $4+2+0+2+0=8$ to this total.


We note that any number in $S$ can be expressed as $a(10,000) + 2,020$ for some integer $a$. The problem requires that $a$ divides this number, and since we know $a$ divides $a(10,000)$, we need that $a$ divides 2020. Each number contributes the sum of the digits of $a$, as well as $2 + 0 + 2 +0 = 4$. Since $2020$ can be prime factorized as $2^2 \cdot 5 \cdot 101$, it has $(2+1)(1+1)(1+1) = 12$ factors. So if we sum all the digits of all possible $a$ values, and add $4 \cdot 12 = 48$, we obtain the answer.

Now we list out all factors of $2,020$, or all possible values of $a$. $1,2,4,5,10,20,101,202,404,505,1010,2020$. If we add up these digits, we get $45$, for a final answer of $45+48=\boxed{093}$.


Video solution:


See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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