1987 AIME Problems/Problem 8


What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$?

Solution 1

Multiplying out all of the denominators, we get:

\begin{align*}104(n+k) &< 195n< 105(n+k)\\ 0 &< 91n - 104k < n + k\end{align*}

Since $91n - 104k < n + k$, $k > \frac{6}{7}n$. Also, $0 < 91n - 104k$, so $k < \frac{7n}{8}$. Thus, $48n < 56k < 49n$. $k$ is unique if it is within a maximum range of $112$, so $n = 112$.

Solution 2

Flip all of the fractions for

\[\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\  105n &>& 56 (k + n)& >& 104n\\  49n &>& 56k& >& 48n\end{array}\]

Continue as in Solution 1.

Solution 3

Flip the fractions and subtract one from all sides to yield \[\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.\] Multiply both sides by $56n$ to get \[49n>56k>48n.\] This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$. If $n=112,$ then $98>k>96$ and $k=97$ is the unique value. For $n\geq 113,$ there is at least $(49\cdot 113-48\cdot 113)-1=112$ possible numbers for $k$ and there is one $k$ every 56 numbers. Hence, there must be at least two values of $k$ that work. So, the largest value of $n$ is $\boxed{112}$.

Solution 4

Notice that in order for $k$ to be unique, $\frac{n}{n + k+ 1} \le \frac{8}{15}$ and $\frac{n}{n+ k-1} \ge \frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)$.

Thus, we want to find $k$ such that $\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)$. Solving this inequality yields $k \le 97$, and plugging this into $\frac{n}{n+k} < \frac{7}{13}$ in the original equation yields $n \le 112$ so the answer is $\boxed{112}$.

See also

1987 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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