Difference between revisions of "2016 AIME I Problems/Problem 5"
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==Solution 3== | ==Solution 3== | ||
| − | If we let <math>k</math> be equal to the number of days it took to read the book, the sum of <math>n</math> through <math>n+k</math> is equal to <math>(2n+k)(k+1)=748</math> Similarly, <math>(2n+k)(k+1)=638</math> We know that both factors must be integers and we see that the only common multiple of <math>748</math> and <math>638</math> not equal to <math>1</math> that will get us positive integer solutions for <math>n</math> and <math>t</math> is <math>11</math>. We set <math>k+1=11</math> so <math>k=10</math>. We then solve for <math>n</math> and <math>t</math> in their respective equations, getting <math>2n+10=68</math>. <math>n=29</math> We also get <math>2t+10=58</math> | + | If we let <math>k</math> be equal to the number of days it took to read the book, the sum of <math>n</math> through <math>n+k</math> is equal to <math>(2n+k)(k+1)=748</math> Similarly, <math>(2n+k)(k+1)=638</math> We know that both factors must be integers and we see that the only common multiple of <math>748</math> and <math>638</math> not equal to <math>1</math> that will get us positive integer solutions for <math>n</math> and <math>t</math> is <math>11</math>. We set <math>k+1=11</math> so <math>k=10</math>. We then solve for <math>n</math> and <math>t</math> in their respective equations, getting <math>2n+10=68</math>. <math>n=29</math> We also get <math>2t+10=58</math>. <math>t=24</math>. Our final answer is <math>29+24=\boxed{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 00:38, 20 May 2017
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 1
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 
.
Solution 2
We could see that both and are divisible by in the outset, and that 
and 
, the quotients, are relatively prime. Both are the 
number of minutes across the days, so we need to subtract 
from each to get 
and 
.
Solution 3
If we let 
be equal to the number of days it took to read the book, the sum of through 
is equal to 
Similarly, 
We know that both factors must be integers and we see that the only common multiple of 
and 
not equal to that will get us positive integer solutions for and is . We set 
so 
. We then solve for and in their respective equations, getting 
. 
We also get 
. 
. Our final answer is
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. 
