Difference between revisions of "2016 AIME I Problems/Problem 5"
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Using this, we find that she reads 55 "additional" pages and 55 "additional" minutes. Therefore, n=(374-55)/11=29, while t=(319-55)/11=24. The answer is therefore 29+24=53. | Using this, we find that she reads 55 "additional" pages and 55 "additional" minutes. Therefore, n=(374-55)/11=29, while t=(319-55)/11=24. The answer is therefore 29+24=53. | ||
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+ | == See also == | ||
+ | {{AIME box|year=2016|n=I|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Revision as of 16:55, 4 March 2016
Problem 5
Anh read a book. On the first day she read pages in minutes, where and are positive integers. On the second day Anh read pages in minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the page book. It took her a total of minutes to read the book. Find .
Solution
Let d be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer, d must be a factor of the total difference which is 55. Also note that the number of pages Anh reads is dn+d(d-1)/2. Similarly, the number of minutes she reads for is dt+d(d-1)/2. When d is odd (which it must be), both of these numbers are multiples of d. Therefore, d must be a factor of 55, 319, and 374. The only such numbers are 1 and 11. We know that Anh reads for at least 2 days, therefore d is 11.
Using this, we find that she reads 55 "additional" pages and 55 "additional" minutes. Therefore, n=(374-55)/11=29, while t=(319-55)/11=24. The answer is therefore 29+24=53.
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.