AIME 2016 I · 第 5 题

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 $29+24=\boxed{053}$

Solution 4

Notice $374=34\cdot 11$ and $319=29\cdot 11$. Also, note the sum of an arithmetic series is $\frac{2n+k}{2} \cdot b$, where $n$ is our first term, $n+k$ is our final term, and $b$ is the number of terms. Since we know both sequences of $n$ and $t$ have the same length, and since $11$ is prime and shared by both $319$ and $374$, we deduce that $b=11$. Thus from here we know $2n+k=68$ and $2t+k=58$ by using our other factors $34$ and $29$. Finally, we add the two systems up and we get $2t+2k+2n=126$. But, notice that $k=b-1$, since the first term has $k=0$, and our last term has $k=b-1$. Plugging this back into our equation we get $2n+2t=106 \implies n+t=\boxed{053}$

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:

$$(n-t)(k+1)=374-319=55.$$ 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 $k+1$: \begin{align*} (k+1)\left(n+\frac{k}{2}\right)&=374 \\ (k+1)\left(t+\frac{k}{2}\right)&=319. \end{align*} Note that $374$ and $319$ have a GCD of $11,$ now combining this with our equation that $(n-t)(k+1)=55,$ we see that $k+1$ has to equal $11.$ Thus, we get:

$$(n,t)=(29,24) \Rightarrow n+t=29+24=\boxed{053}.$$ 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