Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 CEMC Gauss (Grade 8) Problems/Problem 6

Revision as of 20:24, 30 September 2025 by Anabel.disher (talk | contribs) (welcome to the worst way to do problem 6!!1!)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

The area of a rectangle is $12$. Each of its side lengths is a whole number. What is the smallest possible perimeter of this rectangle?

$\text{ (A) }\ \frac{1}{4} + \frac{1}{5} \qquad\text{ (B) }\ \frac{1}{4} + \frac{1}{6} \qquad\text{ (C) }\ \frac{1}{4} + \frac{1}{3} \qquad\text{ (D) }\ \frac{1}{4} + \frac{1}{8} \qquad\text{ (E) }\ \frac{1}{4} + \frac{1}{7}$

Solution 1 (very slow)

We can convert each fraction to a common denominator in multiple steps, and compare the numerators to see which one is the largest.

$\frac{1}{4} + \frac{1}{5} = \frac{1 \times 5}{4 \times 5} + \frac{1 \times 4}{5 \times 4}$

$=\frac{5}{20} + \frac{4}{20} = \frac{5 + 4}{20} = \frac{9}{20}$

$\frac{1}{4} + \frac{1}{6} = \frac{1 \times 3}{4 \times 3} + \frac{1 \times 2}{6 \times 2}$

$=\frac{3}{12} + \frac{2}{12} = \frac{3 + 2}{12} = \frac{5}{12}$

$\frac{1}{4} + \frac{1}{3} = \frac{1 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4}$

$=\frac{3}{12} + \frac{4}{12} = \frac{3 + 4}{12} = \frac{7}{12}$. This has a larger numerator than $\frac{1}{4} + \frac{1}{6}$, so it must be the larger value. This means that we don't have to include $\frac{1}{4} + \frac{1}{6}$ in our comparison from now on.

$\frac{1}{4} + \frac{1}{8} = \frac{1 \times 2}{4 \times 2} + \frac{1 \times 1}{8 \times 1}$

$=\frac{2}{8} + \frac{1}{8} = \frac{2 + 1}{8} = \frac{3}{8}$

$\frac{1}{4} + \frac{1}{7} = \frac{1 \times 7}{4 \times 7} + \frac{1 \times 4}{7 \times 4}$

$=\frac{7}{28} + \frac{4}{28} = \frac{7 + 4}{28} = \frac{11}{28}$

We can now compare $\frac{9}{20}$ and $\frac{7}{12}$ using a common denominator:

$\frac{9}{20} = \frac{9 \times 3}{20 \times 3} = \frac{27}{60}$

$\frac{7}{12} = \frac{7 \times 5}{12 \times 5} = \frac{35}{60}$. This has the larger numerator here with the same denominator, so $\frac{1}{4} + \frac{1}{3}$ must be larger than $\frac{1}{4} + \frac{1}{5}$.

We can now compare $\frac{7}{12}$ and $\frac{3}{8}$ using a common denominator:

$\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}$

$\frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}$. This has the smaller numerator with the same denominator, so $\frac{1}{4} + \frac{1}{3}$ must be larger than $\frac{1}{4} + \frac{1}{8}$.

We can now compare $\frac{7}{12}$ and $\frac{11}{28}$.

$\frac{7}{12} = \frac{7 \times 7}{12 \times 7} = \frac{49}{84}$

$\frac{11}{28} = \frac{11 \times 3}{28 \times 3} = \frac{33}{84}$. This has the smaller numerator with the same denominator, so $\frac{1}{4} + \frac{1}{3}$ must be larger than $\frac{1}{4} + \frac{1}{7}$

Because all of the other answer choices have been eliminated, the answer must be $\boxed {\textbf {(C) } \frac{1}{4} + \frac{1}{3}}$.

~anabel.disher

Solution 2 (faster)

Without converting to like-terms and adding the fractions, we can simply compare the denominators and see which answer choice has the smallest denominator to find the largest sum. All of the answer choices contain $\frac{1}{4}$, so that part doesn't matter. However, each one of the answer choices has a different fraction that has the same numerator but different denominators.

The answer choice with the smallest denominator is $\boxed {\textbf {(C) } \frac{1}{4} + \frac{1}{3}}$.

~anabel.disher

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_CEMC_Gauss_(Grade_8)_Problems/Problem_6&oldid=261942"