Match numerators to bound the denominator · 12 practice problems

Test the edges: $n = 5$ gives $\frac{6}{5}$, just under $\frac{6}{4} = \frac{3}{2}$, which works; $n = 14$ gives $\frac{6}{14}$, just over $\frac{6}{15} = \frac{2}{5}$, which works. $n = 4$ and $n = 15$ give exactly $\frac{3}{2}$ and $\frac{2}{5}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{2}{5} < \frac{6}{n} < \frac{3}{2}$ flips to $\frac{2}{3} < \frac{n}{6} < \frac{5}{2}$, so $4 < n < 15$ after multiplying by $6$.

Test the edges: $n = 11$ gives $\frac{8}{11}$, just under $\frac{8}{10} = \frac{4}{5}$, which works; $n = 35$ gives $\frac{8}{35}$, just over $\frac{8}{36} = \frac{2}{9}$, which works. $n = 10$ and $n = 36$ give exactly $\frac{4}{5}$ and $\frac{2}{9}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{2}{9} < \frac{8}{n} < \frac{4}{5}$ flips to $\frac{5}{4} < \frac{n}{8} < \frac{9}{2}$, so $10 < n < 36$ after multiplying by $8$.

Test the edges: $n = 7$ gives $\frac{10}{7}$, just under $\frac{10}{6} = \frac{5}{3}$, which works; $n = 34$ gives $\frac{10}{34}$, just over $\frac{10}{35} = \frac{2}{7}$, which works. $n = 6$ and $n = 35$ give exactly $\frac{5}{3}$ and $\frac{2}{7}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{2}{7} < \frac{10}{n} < \frac{5}{3}$ flips to $\frac{3}{5} < \frac{n}{10} < \frac{7}{2}$, so $6 < n < 35$ after multiplying by $10$.

Test the edges: $n = 10$ gives $\frac{6}{10}$, just under $\frac{6}{9} = \frac{2}{3}$, which works; $n = 19$ gives $\frac{6}{19}$, just over $\frac{6}{20} = \frac{3}{10}$, which works. $n = 9$ and $n = 20$ give exactly $\frac{2}{3}$ and $\frac{3}{10}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{3}{10} < \frac{6}{n} < \frac{2}{3}$ flips to $\frac{3}{2} < \frac{n}{6} < \frac{10}{3}$, so $9 < n < 20$ after multiplying by $6$.

Test the edges: $n = 10$ gives $\frac{12}{10}$, just under $\frac{12}{9} = \frac{4}{3}$, which works; $n = 27$ gives $\frac{12}{27}$, just over $\frac{12}{28} = \frac{3}{7}$, which works. $n = 9$ and $n = 28$ give exactly $\frac{4}{3}$ and $\frac{3}{7}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{3}{7} < \frac{12}{n} < \frac{4}{3}$ flips to $\frac{3}{4} < \frac{n}{12} < \frac{7}{3}$, so $9 < n < 28$ after multiplying by $12$.

Test the edges: $n = 11$ gives $\frac{12}{11}$, just under $\frac{12}{10} = \frac{6}{5}$, which works; $n = 26$ gives $\frac{12}{26}$, just over $\frac{12}{27} = \frac{4}{9}$, which works. $n = 10$ and $n = 27$ give exactly $\frac{6}{5}$ and $\frac{4}{9}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{4}{9} < \frac{12}{n} < \frac{6}{5}$ flips to $\frac{5}{6} < \frac{n}{12} < \frac{9}{4}$, so $10 < n < 27$ after multiplying by $12$.

Test the edges: $n = 9$ gives $\frac{14}{9}$, just under $\frac{14}{8} = \frac{7}{4}$, which works; $n = 20$ gives $\frac{14}{20}$, just over $\frac{14}{21} = \frac{2}{3}$, which works. $n = 8$ and $n = 21$ give exactly $\frac{7}{4}$ and $\frac{2}{3}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{2}{3} < \frac{14}{n} < \frac{7}{4}$ flips to $\frac{4}{7} < \frac{n}{14} < \frac{3}{2}$, so $8 < n < 21$ after multiplying by $14$.

Test the edges: $n = 13$ gives $\frac{15}{13}$, just under $\frac{15}{12} = \frac{5}{4}$, which works; $n = 39$ gives $\frac{15}{39}$, just over $\frac{15}{40} = \frac{3}{8}$, which works. $n = 12$ and $n = 40$ give exactly $\frac{5}{4}$ and $\frac{3}{8}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{3}{8} < \frac{15}{n} < \frac{5}{4}$ flips to $\frac{4}{5} < \frac{n}{15} < \frac{8}{3}$, so $12 < n < 40$ after multiplying by $15$.

Test the edges: $n = 16$ gives $\frac{18}{16}$, just under $\frac{18}{15} = \frac{6}{5}$, which works; $n = 23$ gives $\frac{18}{23}$, just over $\frac{18}{24} = \frac{3}{4}$, which works. $n = 15$ and $n = 24$ give exactly $\frac{6}{5}$ and $\frac{3}{4}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{3}{4} < \frac{18}{n} < \frac{6}{5}$ flips to $\frac{5}{6} < \frac{n}{18} < \frac{4}{3}$, so $15 < n < 24$ after multiplying by $18$.

Test the edges: $n = 16$ gives $\frac{20}{16}$, just under $\frac{20}{15} = \frac{4}{3}$, which works; $n = 35$ gives $\frac{20}{35}$, just over $\frac{20}{36} = \frac{5}{9}$, which works. $n = 15$ and $n = 36$ give exactly $\frac{4}{3}$ and $\frac{5}{9}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{5}{9} < \frac{20}{n} < \frac{4}{3}$ flips to $\frac{3}{4} < \frac{n}{20} < \frac{9}{5}$, so $15 < n < 36$ after multiplying by $20$.

Test the edges: $n = 21$ gives $\frac{24}{21}$, just under $\frac{24}{20} = \frac{6}{5}$, which works; $n = 41$ gives $\frac{24}{41}$, just over $\frac{24}{42} = \frac{4}{7}$, which works. $n = 20$ and $n = 42$ give exactly $\frac{6}{5}$ and $\frac{4}{7}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{4}{7} < \frac{24}{n} < \frac{6}{5}$ flips to $\frac{5}{6} < \frac{n}{24} < \frac{7}{4}$, so $20 < n < 42$ after multiplying by $24$.

Test the edges: $n = 21$ gives $\frac{30}{21}$, just under $\frac{30}{20} = \frac{3}{2}$, which works; $n = 35$ gives $\frac{30}{35}$, just over $\frac{30}{36} = \frac{5}{6}$, which works. $n = 20$ and $n = 36$ give exactly $\frac{3}{2}$ and $\frac{5}{6}$, which the strict inequalities exclude.

Another way: Use reciprocals: since all three fractions are positive, $\frac{5}{6} < \frac{30}{n} < \frac{3}{2}$ flips to $\frac{2}{3} < \frac{n}{30} < \frac{6}{5}$, so $20 < n < 36$ after multiplying by $30$.