Difference between revisions of "1987 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
+ | 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>? | ||
+ | == Solution 1== | ||
+ | Multiplying out all of the [[denominator]]s, we get: | ||
− | == Solution == | + | <cmath>\begin{align*}104(n+k) &< 195n< 105(n+k)\\ |
+ | 0 &< 91n - 104k < n + k\end{align*}</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 2== | ||
+ | Flip all of the fractions for | ||
+ | |||
+ | <cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\ | ||
+ | |||
+ | 105n &>& 56 (k + n)& >& 104n\\ | ||
+ | |||
+ | 49n &>& 56k& >& 48n\end{array}</cmath> | ||
+ | |||
+ | Continue as in Solution 1. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | 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 == | ||
− | + | {{AIME box|year=1987|num-b=7|num-a=9}} | |
− | {{ | + | [[Category:Intermediate Algebra Problems]] |
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:59, 16 December 2020
Problem
What is the largest positive integer for which there is a unique integer
such that
?
Solution 1
Multiplying out all of the denominators, we get:
Since ,
. Also,
, so
. Thus,
.
is unique if it is within a maximum range of
, so
.
Solution 2
Flip all of the fractions for
Continue as in Solution 1.
Solution 3
Flip the fractions and subtract one from all sides to yield Multiply both sides by
to get
This is equivalent to find the largest value of
such that there is only one multiple of 56 within the open interval between
and
. If
then
and
is the unique value. For
there is at least
possible numbers for
and there is one
every 56 numbers. Hence, there must be at least two values of
that work. So, the largest value of
is
.
Solution 4
Notice that in order for to be unique,
and
must be true. Solving these inequalities for
yields
.
Thus, we want to find such that
. Solving this inequality yields
, and plugging this into
in the original equation yields
so the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.