Factor a Sum and Choose the GCF or the LCM | Grade 6 Math | LevelKite

You already know two tools. A factor of a number is a whole number that divides it with no remainder. A factor that two numbers share is a common factor, and the largest one is their greatest common factor (GCF). A multiple of a number is that number times a counting number such as $1$ , $2$ or $3$ . A multiple that two numbers share is a common multiple, and the smallest one is their least common multiple (LCM).

In this lesson you put both tools to work. You will:

rewrite a sum (an addition) as a product (a multiplication) by pulling out a common factor,
decide whether a word problem needs the GCF or the LCM,
check GCF and LCM work and find the mistake.

Picture $4$ equal rows of tiles. Each row has $9$ red tiles and $5$ blue tiles. You can count all the tiles in two ways.

Count by rows: $4$ rows of $9 + 5$ tiles is $4 \times (9 + 5)$ tiles.
Count by color: $4 \times 9$ red tiles plus $4 \times 5$ blue tiles is $4 \times 9 + 4 \times 5$ tiles.

Both ways count the same tiles, so the two expressions (the two ways of writing the total) are equal. This is the distributive property: multiplying a sum by a number gives the same result as multiplying each addend (each number being added) by that number and then adding.

Now read it from the sum back to the product. The sum $36 + 20$ is $4 \times 9 + 4 \times 5$ . Both addends have the factor $4$ , so you can write the $4$ once, in front, and put what is left in parentheses. This is factoring a sum: you rewrite the sum as a common factor times a smaller sum. Factoring changes how the sum looks, not its value. Both forms equal $56$ .

36 + 20 = 4 \times 9 + 4 \times 5 = 4 \times (9 + 5)

Worked example

Factor a sum with the GCF

Use the GCF to factor the sum $24 + 36$ .

Find the GCF of the two addends, 24 and 36. Their common factors are 1, 2, 3, 4, 6 and 12. The greatest is 12, so 12 is the factor to pull out.
$\text{GCF of } 24 \text{ and } 36 = 12$
Write each addend as 12 times another number. Divide by 12 to find that number: 24 divided by 12 is 2, and 36 divided by 12 is 3.
$24 = 12 \times 2 \qquad 36 = 12 \times 3$
Replace each addend with its product. The value stays the same, because 12 times 2 is still 24 and 12 times 3 is still 36.
$24 + 36 = 12 \times 2 + 12 \times 3$
Both products have the factor 12. The distributive property lets you write the 12 once, in front of the sum of the numbers that are left.
$12 \times 2 + 12 \times 3 = 12 \times (2 + 3)$
Check by working out both sides. They must match, because factoring changes how the sum looks and not its value.
$24 + 36 = 60 \qquad 12 \times (2 + 3) = 12 \times 5 = 60$

Answer

Factored with the GCF, $24 + 36 = 12 \times (2 + 3)$ .

Now for word problems. A word problem usually does not say whether to find the GCF or the LCM. You decide from what the numbers are doing. The same two numbers can lead to two very different questions. Look at $12$ and $18$ .

Two questions about the numbers 12 and 18

What happens	What is asked	Numbers that fit	Answer
You split $12$ and $18$ into the same number of equal groups, with nothing left over.	The greatest number of groups	Common factors: $1, 2, 3, 6$	The GCF, which is $6$
You count by twelves: $12, 24, 36, \dots$ You also count by eighteens: $18, 36, 54, \dots$	The first number in both counts	Common multiples: $36, 72, 108, \dots$	The LCM, which is $36$

Key concept

Factors split, multiples match

A common factor of two numbers splits both of them into equal groups with nothing left over. The GCF is the largest common factor. It is also the factor to pull out of a sum.

A common multiple is a number that two repeating counts both reach. The LCM is the first one they reach.

So decide what the problem does before you calculate. If it splits two amounts with nothing left over and asks for the greatest number of equal groups, or the largest group size that works for both, find the GCF. If it repeats two counts and asks for the first place they match, find the LCM.

Worked example

A splitting problem

A coach has $48$ oranges and $60$ granola bars. She packs snack bags so that every bag has the same number of oranges and the same number of bars. She uses every item. What is the greatest number of bags she can make, and what goes in each bag?

Decide which tool fits. The coach splits the oranges equally among the bags with none left over, so the number of bags is a factor of 48. She splits the bars the same way, so it is also a factor of 60. The greatest number of bags is the GCF of 48 and 60.
Find the GCF. The common factors of 48 and 60 are 1, 2, 3, 4, 6 and 12. The greatest is 12, so the coach can make 12 bags.
$\text{GCF of } 48 \text{ and } 60 = 12$
Divide each amount by the 12 bags to find how many of each item go in one bag.
$48 \div 12 = 4 \text{ oranges} \qquad 60 \div 12 = 5 \text{ bars}$
Check with a factored sum. The total number of items equals 12 bags times the 4 oranges and 5 bars in each bag. Both sides equal 108. Because 4 and 5 share no factor greater than 1, no larger number of bags works.
$48 + 60 = 12 \times (4 + 5)$

Answer

The coach can make at most $12$ bags. Each bag holds $4$ oranges and $5$ granola bars.

Worked example

A matching problem

A green light blinks every $6$ seconds. A yellow light blinks every $8$ seconds. Both lights blink right now. After how many seconds will they next blink at the same moment?

Decide which tool fits. Each light repeats in equal steps of time. Counting seconds from now, the green light blinks at multiples of 6 and the yellow light at multiples of 8. They blink together at a common multiple, and the next time is the least one. Find the LCM of 6 and 8.
List the blink times for each light, in seconds from now. Stop at the first number that is in both lists. That number is 24, so the LCM is 24.
$\text{Green: } 6, 12, 18, 24 \qquad \text{Yellow: } 8, 16, 24$
Check that 24 seconds fits both lights. It is the 4th green blink after now and the 3rd yellow blink after now.
$24 \div 6 = 4 \qquad 24 \div 8 = 3$

Answer

The lights next blink at the same moment after $24$ seconds.

To find a mistake in GCF or LCM work, you do not have to start over. Put the answer through four checks. Every number here is a whole number greater than $0$ , and the size check depends on that.

Right tool? Splitting two amounts into equal groups with nothing left over needs a common factor. Finding where two repeating counts match needs a common multiple.
Right size? A factor of a number is never larger than that number, so a GCF is never larger than the smaller number. A multiple of a number is never smaller than that number, so an LCM is never smaller than the larger number.
Is it common? A GCF must divide both numbers with no remainder. An LCM must be a multiple of both numbers.
Greatest or least? For a GCF, look for a larger common factor. For an LCM, look for a smaller common multiple. If you find one, the answer fails.

When a check fails, look back through the work to see where the mistake began.

Three answers that fail a check

Answer given	Check that fails	Why it fails	Correct answer
The GCF of $12$ and $18$ is $36$ .	Right size?	$36$ is larger than $12$ . It is the LCM, not the GCF.	The GCF is $6$ .
The LCM of $6$ and $9$ is $27$ .	Is it common?	$27 \div 6$ leaves a remainder of $3$ , so $27$ is not a multiple of $6$ .	The LCM is $18$ .
The GCF of $16$ and $24$ is $4$ .	Greatest or least?	$8$ also divides $16$ and $24$ , and $8$ is larger than $4$ .	The GCF is $8$ .

Worked example

Find the mistake

Leo tries to factor $18 + 30$ using the GCF. He writes: the factors of $18$ are $1, 2, 3, 6, 9, 18$ . The factors of $30$ are $1, 2, 3, 5, 10, 15, 30$ . The GCF is $3$ , so $18 + 30 = 3 \times (6 + 10)$ . Find Leo's mistake and correct it.

Check the tool and the size. Factoring a sum needs a common factor, so the GCF is the right tool. Leo's answer, 3, is not larger than 18, the smaller addend. Both checks pass.
Check that 3 is common. It divides both addends with no remainder, so it is a common factor. Leo's equation is even true, because both sides equal 48.
$18 \div 3 = 6 \qquad 30 \div 3 = 10$
Check that 3 is the greatest. Inside Leo's parentheses, 6 and 10 are both even, so they still share the factor 2. That means 3 times 2, which is 6, divides both addends. A larger common factor exists, so 3 is not the GCF.
$18 \div 6 = 3 \qquad 30 \div 6 = 5$
Find where the mistake began. Leo's list for 30 leaves out 6, even though 6 times 5 is 30. Writing the factors in pairs makes a missing factor easier to catch.
$30 = 1 \times 30 = 2 \times 15 = 3 \times 10 = 5 \times 6$
Correct the work with the GCF, 6. The numbers left inside, 3 and 5, share no factor greater than 1, and both sides equal 48.
$18 + 30 = 6 \times 3 + 6 \times 5 = 6 \times (3 + 5)$

Answer

Leo left $6$ out of the factors of $30$ , so he chose a common factor that is not the greatest. The GCF of $18$ and $30$ is $6$ , and the correct factoring is $18 + 30 = 6 \times (3 + 5)$ .

The takeaway. Common factors split, and common multiples match. To factor a sum, pull out the GCF and write what is left in parentheses, as in $24 + 36 = 12 \times (2 + 3)$ . In a word problem, find the GCF when you split two amounts into the greatest number of equal groups with nothing left over. Find the LCM when two repeating counts first match. To check any answer, ask four things. Is it the right tool? Is it the right size? Is it truly common? Is it truly the greatest or the least?