Difference between revisions of "2020 AIME I Problems"
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==Problem 4== | ==Problem 4== | ||
| − | + | 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. | |
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[[2020 AIME I Problems/Problem 4 | Solution]] | [[2020 AIME I Problems/Problem 4 | Solution]] | ||
Revision as of 15:37, 12 March 2020
| 2020 AIME I (Answer Key) | AoPS Contest Collections • PDF | ||
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Instructions
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| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
to 
, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.Contents
Problem 1
In 
with 
point 
lies strictly between 
and 
on side 
and point 
lies strictly between and 
on side 
such that 
The degree measure of 
is 
where 
and 
are relatively prime positive integers. Find 
Problem 2
There is a unique positive real number 
such that the three numbers 
and 
in that order, form a geometric progression with positive common ratio. The number can be written as where and are relatively prime positive integers. Find
Problem 3
A positive integer 
has base-eleven representation 
and base-eight representation 
where 
and 
represent (not necessarily distinct) digits. Find the least such expressed in base ten.
Problem 4
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.
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
| 2020 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by 2019 AIME II |
Followed by 2020 AIME II | |
| 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. 
