Least Common Denominator Calculator

The least common denominator (LCD) calculator determines the lowest number that can be used as the denominator for all input values. Input values can be represented by integers, fractions, and mixed numbers.

Directions for use

To use the LCD calculator, enter all given values separated by commas. The values can be both positive and negative. When entering a mixed number, separate the whole number part from the fractional part with a space, for example: \$5 \frac{1}{2}\$. Then press “Calculate.” The calculator will return the least common denominator of all input numbers, as well as the detailed solution algorithm.

Definitions

The least common denominator, or the lowest common denominator, is the lowest number that can be used as a denominator for a set of given values. Finding LCD is necessary if you want to perform addition or subtraction operations with fractions or mixed numbers.

How to find the least common denominator

To find LCD of a set of numbers, follow the steps below:

1. Convert all numbers into fractions.
2. Find the least common multiple (LCM) of the denominators of all fractions.
3. LCM of the denominators will be LCD for the original fractions. Re-write original fractions with LCD as the denominator.

Positive values

For example, let’s find LCD of the following numbers: 3, \$\frac{3}{8}\$, \$1 \frac{1}{2}\$, \$\frac{5}{4}\$. Following the steps of the above algorithm, we get:

1. Converting all numbers into fractions:

• 3 = \$\frac{3}{1}\$
• \$\frac{3}{8}\$ = \$\frac{3}{8}\$
• \$1 \frac{1}{2}\$ = 1 + \$\frac{1}{2}\$ = \$\frac{2}{2}\$ + \$\frac{1}{2}\$ = \$\frac{3}{2}\$
• \$\frac{5}{4}\$ = \$\frac{5}{4}\$

2. The fractions have the following denominators: 1, 8, 2, 4. Therefore, we need to find LCM of 1, 2, 4, 8. Let’s find LCM (1, 2, 4, 8) by listing multiples:

• Multiples of 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10…
• Multiples of 2: 2, 4, 6, 8, 10, 12…
• Multiples of 4: 4, 8, 12, 16…
• Multiples of 8: 8, 16, 24

LCM (1, 2, 4, 8) = 8

3. LCM (1, 2, 4, 8) = LCD (3, \$\frac{3}{8}\$, \$1 \frac{1}{2}\$, \$\frac{5}{4}\$) = 8.

Re-writing the original fractions, we get:

• 3 = \$\frac{3}{1}\$ = \$\frac{3 × 8}{1 × 8}\$ = \$\frac{24}{8}\$
• \$\frac{3}{8}\$ = \$\frac{3}{8}\$
• \$1 \frac{1}{2}\$ = \$\frac{3}{2}\$ = \$\frac{3 × 4}{2 × 4}\$ = \$\frac{12}{8}\$
• \$\frac{5}{4}\$ = \$\frac{5 × 2}{4 × 2}\$ = \$\frac{10}{8}\$

Negative values

The algorithm described above can also be used for finding LCD, if one or more of the given values are negative. For example, let’s find LCD (- 4, \$\frac{2}{3}\$):

• -4 = - \$\frac{4}{1}\$
• \$\frac{2}{3}\$ = \$\frac{2}{3}\$

2. The fractions have the following denominators: 1, 3. Therefore, we need to find LCM (1, 3). Let’s find LCM (1, 3) by listing multiples:

• Multiples of 1: 1, 2, 3, 4, 5…
• Multiples of 3 = 3, 6, 9…

LCM (1, 3) = 3

3. LCD (- \$\frac{4}{1}\$, \$\frac{2}{3}\$) = LCM (1, 3) = 3.

Re-writing the fractions with the new denominator, we get:

• -4 = - \$\frac{4}{1}\$ = - \$\frac{12}{3}\$
• \$\frac{2}{3}\$ = \$\frac{2}{3}\$

Calculation example

Cooking

You are baking a cake, for which you need:

• \$2 \frac{2}{3}\$ cups of flour,
• 2 cups of milk,
• 1 cup of sugar, and
• \$\frac{1}{2}\$ cup of melted butter.

The problem is, you only have 1 mixing bowl with a volume of \$6 \frac{1}{2}\$ cups. Will your bowl fit all required ingredients?

Solution

To solve the problem, we need to sum up the volumes of all given ingredients, and compare the final value with the volume of the mixing bowl.

The given volumes are:

• Flour – \$2 \frac{2}{3}\$ cups
• Milk – 2 cups
• Sugar – 1 cup
• Butter – \$\frac{1}{2}\$ cup

To add these volumes, let’s first convert the given values into fractions with a common denominator, following the algorithm described above.

1. Converting all values into fractions, we get:

• \$2 \frac{2}{3}\$ = 2 + \$\frac{2}{3}\$ = \$\frac{6}{3}\$ + \$\frac{2}{3}\$ = \$\frac{8}{3}\$
• 2 = \$\frac{2}{1}\$
• 1 = \$\frac{1}{1}\$
• \$\frac{1}{2}\$ = \$\frac{1}{2}\$

2. The fractions have the following denominators: 1, 2, 3. Therefore, we need to find LCM of 1, 2, 3.

Let’s find LCM (1, 2, 3) by listing multiples:

• Multiples of 1: 1, 2, 3, 4, 5, 6, 7, 8…
• Multiples of 2: 2, 4, 6, 8, 10…
• Multiples of 3: 3, 6, 9, 12…

LCM (1, 2, 3) = 6

3. LCD (\$\frac{8}{3}\$, \$\frac{2}{1}\$, \$\frac{1}{1}\$, \$\frac{1}{2}\$) = LCM (1, 2, 3) = 6.

Re-writing the original fractions, we get:

• \$2 \frac{2}{3}\$ = \$\frac{8}{3}\$ = \$\frac{8 × 2}{3 × 2}\$ = \$\frac{16}{6}\$
• 2 = \$\frac{2}{1}\$ = \$\frac{2 × 6}{1 × 6}\$ = \$\frac{12}{6}\$
• 1 = \$\frac{1}{1}\$ = \$\frac{1 × 6}{1 × 6}\$ = \$\frac{6}{6}\$
• \$\frac{1}{2}\$ = \$\frac{1 × 3}{2 × 3}\$ = \$\frac{3}{6}\$

Now we can find the total volume of all ingredients:

Volume of ingredients = \$2 \frac{2}{3}\$ + 2 + 1 + \$\frac{1}{2}\$ = \$\frac{8}{3}\$ + \$\frac{2}{1}\$ + \$\frac{1}{1}\$ + \$\frac{1}{2}\$ = \$\frac{16}{6}\$ + \$\frac{12}{6}\$ + \$\frac{6}{6}\$ + \$\frac{3}{6}\$ = \$\frac{16 + 12 + 6 + 3}{6}\$ = \$\frac{37}{6}\$ = \$6 \frac{1}{6}\$

We know that the volume of the bowl is \$6 \frac{1}{2}\$ cups. Let’s compare these two values: \$6 \frac{1}{6}\$ and \$6 \frac{1}{2}\$. To compare the values, we need to re-write them as fractions with a common denominator:

1. Converting into fractions, we get:

• \$6 \frac{1}{6}\$ = \$\frac{37}{6}\$
• \$6 \frac{1}{2}\$ = \$\frac{13}{2}\$

2. The fractions have the following denominators: 2, 6. Therefore, we need to find LCM of 2 and 6. Let’s find LCM (2, 6) by listing multiples:

• Multiples of 2: 2, 4, 6, 8, 10…
• Multiples of 6: 6, 12, 18…

LCM (2, 6) = 6

3. LCD (\$\frac{37}{6}\$, \$\frac{13}{2}\$) = LCM (2, 6) = 6. Re-writing the original fractions, we get:

• \$6 \frac{1}{6}\$ = \$\frac{37}{6}\$
• \$6 \frac{1}{2}\$ = \$\frac{13}{2}\$ = \$\frac{13 × 3}{2 × 3}\$ = \$\frac{39}{6}\$

Finally, we see that the volume of all ingredients is \$\frac{37}{6}\$ cups, and the volume of the bowl is \$\frac{39}{6}\$ cups.

39 > 37, therefore, \$\frac{39}{6}\$ > \$\frac{37}{6}\$. This means that your bowl fits all necessary ingredients, and you can start baking the cake!

Answer

The volume of the ingredients can be expressed as \$\frac{37}{6}\$ cups, while the volume of the bowl can be expressed as \$\frac{39}{6}\$ cups. Therefore, the bowl will fit all necessary ingredients.