2016 AIME I Problems/Problem 5
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
Solution 4
Notice and . Also, note the sum of an arithmetic series is , where is our first term, is our final term, and is the number of terms. Since we know both sequences of and have the same length, and since is prime and shared by both and , we deduce that . Thus from here we know and by using our other factors and . Finally, we add the two systems up and we get . But, notice that , since the first term has , and our last term has . Plugging this back into our equation we get
Solution 5
We list two equations: \begin{align*} n+(n+1)+...+(n+k)&=374\\ t+(t+1)+...+(t+k)&=319. \end{align*} Subtracting the two, we get: Manipulating the first and second equation, we get: \begin{align*} n(k+1)+\frac{k(k+1)}{2}&=374 \\ t(k+1)+\frac{k(k+1)}{2}&=319. \end{align*} We factor out the common factor : \begin{align*} (k+1)\left(n+\frac{k}{2}\right)&=374 \\ (k+1)\left(t+\frac{k}{2}\right)&=319. \end{align*} Note that and have a GCD of now combining this with our equation that we see that has to equal Thus, we get: Note: We see that because n-t=5, it becomes impossible for n+k/2 and t+k/2 to both be multiples of 11. Thus, this satisfies our condition. Thus k+1 must be 11 to satisfy the common factor 11 constraint. mathboy282
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 |
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