Difference between revisions of "2015 AIME II Problems"
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==Problem 3== | ==Problem 3== | ||
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| + | Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>. | ||
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| + | [[2015 AIME II Problems/Problem 3 | Solution]] | ||
==Problem 4== | ==Problem 4== | ||
Revision as of 16:58, 26 March 2015
| 2015 AIME II (Answer Key) | AoPS Contest Collections • PDF | ||
|
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
Let 
be the least positive integer that is both 
percent less than one integer and 
percent greater than another integer. Find the remainder when is divided by 
.
Problem 2
In a new school, 
percent of the students are freshmen, 
percent are sophomores, 
percent are juniors, and 
percent are seniors. All freshmen are required to take Latin, and 
percent of sophomores, 
percent of the juniors, and percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is 
, where 
and 
are relatively prime positive integers. Find 
.
Problem 3
Let be the least positive integer divisible by 
whose digits sum to . Find .
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
| 2015 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by 2014 AIME I, 2014 AIME II |
Followed by 2016 AIME | |
| 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. 
