Difference between revisions of "2016 AIME I Problems/Problem 5"

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Problem 5

Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ 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 $374$ page book. It took her a total of $319$ minutes to read the book. Find $n + t$ .

Solution 1

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 $374-319=55$ . Also note that the number of pages Anh reads is $dn+\frac{d(d-1)}{2}$ . Similarly, the number of minutes she reads for is $dt+\frac{d(d-1)}{2}$ . 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, $d=11$ .

Using this, we find that she reads $55$ "additional" pages and $55$ "additional" minutes. Therefore, $n=\frac{374-55}{11}=29$ , while $t=\frac{319-55}{11}=24$ . The answer is therefore $29+24=\fbox{053}$ .

Solution 2

We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$ , the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\left \lfloor{\frac{11}{2}}\right \rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=\boxed{053}$ .

Solution 3

If we let $k$ be equal to the number of days it took to read the book, the sum of $n$ through $n+k$ is equal to $(2n+k)(k+1)=748$ Similarly, $(2t+k)(k+1)=638$ We know that both factors must be integers and we see that the only common multiple of $748$ and $638$ not equal to $1$ that will get us positive integer solutions for $n$ and $t$ is $11$ . We set $k+1=11$ so $k=10$ . We then solve for $n$ and $t$ in their respective equations, getting $2n+10=68$ . $n=29$ We also get $2t+10=58$ . $t=24$ . Our final answer is $29+24=\boxed{053}$

@@ Line 10: / Line 10: @@
 ==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>.  <math>t=24</math>. Our final answer is <math>29+24=\boxed{053}</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>(2t+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 ==
 {{AIME box|year=2016|n=I|num-b=4|num-a=6}}
 {{MAA Notice}}

2016 AIME I (Problems • Answer Key • Resources)
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