Difference between revisions of "2020 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
| + | Let <math>S</math> be the set of positive integers <math>N</math> with the property that the last four digits of <math>N</math> are <math>2020,</math> and when the last four digits are removed, the result is a divisor of <math>N.</math> For example, <math>42,020</math> is in <math>S</math> because <math>4</math> is a divisor of <math>42,020.</math> Find the sum of all the digits of all the numbers in <math>S.</math> For example, the number <math>42,020</math> contributes <math>4+2+0+2+0=8</math> to this total. | ||
== Solution == | == Solution == | ||
Revision as of 08:23, 13 March 2020
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
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. 
