Difference between revisions of "2016 AIME I Problems/Problem 5"
Aleph null (talk | contribs) m (→Solution: Fixed latex) |
(→Solution) |
||
| Line 4: | Line 4: | ||
Let <math>d</math> 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, <math>d</math> must be a factor of the total difference, which is <math>374-319=55</math>. Also note that the number of pages Anh reads is <math>dn+\frac{d(d-1)}{2}</math>. Similarly, the number of minutes she reads for is <math>dt+\frac{d(d-1)}{2}</math>. When <math>d</math> is odd (which it must be), both of these numbers are multiples of <math>d</math>. Therefore, <math>d</math> must be a factor of <math>55</math>, <math>319</math>, and <math>374</math>. The only such numbers are <math>1</math> and <math>11</math>. We know that Anh reads for at least <math>2</math> days. Therefore, <math>d=11</math>. | Let <math>d</math> 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, <math>d</math> must be a factor of the total difference, which is <math>374-319=55</math>. Also note that the number of pages Anh reads is <math>dn+\frac{d(d-1)}{2}</math>. Similarly, the number of minutes she reads for is <math>dt+\frac{d(d-1)}{2}</math>. When <math>d</math> is odd (which it must be), both of these numbers are multiples of <math>d</math>. Therefore, <math>d</math> must be a factor of <math>55</math>, <math>319</math>, and <math>374</math>. The only such numbers are <math>1</math> and <math>11</math>. We know that Anh reads for at least <math>2</math> days. Therefore, <math>d=11</math>. | ||
| − | Using this, we find that she reads <math>55</math> "additional" pages and <math>55</math> "additional" minutes. Therefore, <math>n=\frac{374-55}{11}=29</math>, while <math>t=\frac{319-55}{11}=24</math>. The answer is therefore <math>29+24=\fbox{ | + | Using this, we find that she reads <math>55</math> "additional" pages and <math>55</math> "additional" minutes. Therefore, <math>n=\frac{374-55}{11}=29</math>, while <math>t=\frac{319-55}{11}=24</math>. The answer is therefore <math>29+24=\fbox{053}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=4|num-a=6}} | {{AIME box|year=2016|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:42, 5 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 
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, must be a factor of the total difference, which is 
. Also note that the number of pages Anh reads is 
. Similarly, the number of minutes she reads for is 
. When is odd (which it must be), both of these numbers are multiples of . Therefore, must be a factor of 
, , and . The only such numbers are 
and 
. We know that Anh reads for at least 
days. Therefore, 
.
Using this, we find that she reads "additional" pages and "additional" minutes. Therefore, 
, while 
. The answer is therefore 
.
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. 
