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 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
Video Solutions
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.