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Difference between revisions of "1987 AIME Problems/Problem 8"

 
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== Problem ==
 
== Problem ==
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What is the largest positive integer <math>n</math> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?
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== Solution 1==
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Multiplying out all of the [[denominator]]s, we get:
  
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== Solution ==
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<cmath>\begin{align*}104(n+k) &< 195n< 105(n+k)\\
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0 &< 91n - 104k < n + k\end{align*}</cmath>
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Since <math>91n - 104k < n + k</math>, <math>k > \frac{6}{7}n</math>. Also, <math>0 < 91n - 104k</math>, so <math>k < \frac{7n}{8}</math>. Thus, <math>48n < 56k < 49n</math>. <math>k</math> is unique if it is within a maximum [[range]] of <math>112</math>, so <math>n = 112</math>.
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== Solution 2==
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Flip all of the fractions for
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<cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\
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105n &>& 56 (k + n)& >& 104n\\
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49n &>& 56k& >& 48n\end{array}</cmath>
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Continue as in Solution 1.
 +
 
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==Solution 3==
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Flip the fractions and subtract one from all sides to yield <cmath>\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.</cmath>  Multiply both sides by <math>56n</math> to get <cmath>49n>56k>48n.</cmath> This is equivalent to find the largest value of <math>n</math> such that there is only one multiple of 56 within the open interval between <math>48n</math> and <math>49n</math>.  If <math>n=112,</math> then <math>98>k>96</math> and <math>k=97</math> is the unique value. For <math>n\geq 113,</math> there is at least <math>(49\cdot 113-48\cdot 113)-1=112</math> possible numbers for <math>k</math> and there is one <math>k</math> every 56 numbers. Hence, there must be at least two values of <math>k</math> that work.  So, the largest value of <math>n</math> is <math>\boxed{112}</math>.
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==Solution 4==  
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Notice that in order for <math>k</math> to be unique, <math>\frac{n}{n + k+ 1} \le \frac{8}{15}</math> and <math>\frac{n}{n+ k-1} \ge \frac{7}{13}</math> must be true. Solving these inequalities for <math>k</math> yields <math>\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)</math>.
 +
 
 +
Thus, we want to find <math>k</math> such that <math>\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)</math>. Solving this inequality yields <math>k \le 97</math>, and plugging this into <math>\frac{n}{n+k} < \frac{7}{13}</math> in the original equation yields <math>n \le 112</math> so the answer is <math>\boxed{112}</math>.
  
 
== See also ==
 
== See also ==
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* [[1987 AIME Problems]]
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{{AIME box|year=1987|num-b=7|num-a=9}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 15:59, 16 December 2020

Problem

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 (Problems • Answer Key • Resources)
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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