Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.
What is the largest positive integer n for which there is a unique integer k such that \(\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}\)?
Digits
Algebra
Numbers
Answer: is 112.
AIME I, 1987, Question 8
Elementary Number Theory by David Burton
Simplifying the inequality gives, 104(n+k)<195n<105(n+k)
or, 0<91n-104k<n+k
for 91n-104k<n+k, K>\(\frac{6n}{7}\)
and 0<91n-104k gives k<\(\frac{7n}{8}\)
so, 48n<56k<49n for 96<k<98 and k=97
thus largest value of n=112.
