Last updated on August 5th, 2025
The GCF is the largest number that can divide two or more numbers without leaving any remainder. GCF is used to share items equally, to group or arrange items, and to schedule events. In this topic, we will learn about the GCF of 330 and 180.
The greatest common factor of 330 and 180 is 30. The largest divisor of two or more numbers is called the GCF of the numbers. If two numbers are co-prime, they have no common factors other than 1, so their GCF is 1. The GCF of two numbers cannot be negative because divisors are always positive.
To find the GCF of 330 and 180, a few methods are described below
Steps to find the GCF of 330 and 180 using the listing of factors
Step 1: Firstly, list the factors of each number
Factors of 330 = 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165, 330.
Factors of 180 = 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.
Step 2: Now, identify the common factors of them
Common factors of 330 and 180: 1, 2, 3, 5, 6, 10, 15, 30.
Step 3: Choose the largest factor
The largest factor that both numbers have is 30.
The GCF of 330 and 180 is 30.
To find the GCF of 330 and 180 using the Prime Factorization Method, follow these steps:
Step 1: Find the prime factors of each number
Prime Factors of 330: 330 = 2 × 3 × 5 × 11
Prime Factors of 180: 180 = 2 × 2 × 3 × 3 × 5
Step 2: Now, identify the common prime factors
The common prime factors are: 2 × 3 × 5
Step 3: Multiply the common prime factors
2 × 3 × 5 = 30.
The Greatest Common Factor of 330 and 180 is 30.
Find the GCF of 330 and 180 using the division method or Euclidean Algorithm Method. Follow these steps:
Step 1: First, divide the larger number by the smaller number
Here, divide 330 by 180
330 ÷ 180 = 1 (quotient), The remainder is calculated as 330 − (180×1) = 150
The remainder is 150, not zero, so continue the process
Step 2: Now divide the previous divisor (180) by the previous remainder (150)
Divide 180 by 150
180 ÷ 150 = 1 (quotient), remainder = 180 − (150×1) = 30
Step 3: Divide the previous divisor (150) by the previous remainder (30)
150 ÷ 30 = 5 (quotient), remainder = 150 − (30×5) = 0
The remainder is zero, the divisor will become the GCF.
The GCF of 330 and 180 is 30.
Finding the GCF of 330 and 180 looks simple, but students often make mistakes while calculating the GCF. Here are some common mistakes to be avoided by the students.
A gardener has 330 tulip bulbs and 180 daffodil bulbs. She wants to plant them in rows with the largest number of bulbs in each row. How many bulbs will be in each row?
We should find the GCF of 330 and 180
GCF of 330 and 180
2 × 3 × 5 = 30.
There are 30 bulbs in each row.
330 ÷ 30 = 11
180 ÷ 30 = 6
There will be 30 bulbs in each row, with 11 rows of tulips and 6 rows of daffodils.
As the GCF of 330 and 180 is 30, the gardener can make rows with 30 bulbs each.
Now divide 330 and 180 by 30.
There are 11 rows of tulip bulbs and 6 rows of daffodil bulbs.
A concert hall has 330 seats in the orchestra section and 180 seats in the balcony. They want to arrange them in blocks with the same number of seats in each block, using the largest possible number of seats per block. How many seats will be in each block?
GCF of 330 and 180
2 × 3 × 5 = 30
So each block will have 30 seats.
There are 330 seats in the orchestra section and 180 in the balcony.
To find the total number of seats in each block, we should find the GCF of 330 and 180.
There will be 30 seats in each block.
A painter has 330 meters of blue paint tape and 180 meters of red paint tape. He wants to cut both tapes into pieces of equal length, using the longest possible length. What should be the length of each piece?
For calculating the longest equal length, we have to calculate the GCF of 330 and 180.
The GCF of 330 and 180
2 × 3 × 5 = 30
The length of each piece is 30 meters.
For calculating the longest length of the paint tape, first, we need to calculate the GCF of 330 and 180, which is 30.
The length of each piece of tape will be 30 meters.
A carpenter has two wooden planks, one 330 cm long and the other 180 cm long. He wants to cut them into the longest possible equal pieces, without any wood left over. What should be the length of each piece?
The carpenter needs the longest piece of wood GCF of 330 and 180 2 × 3 × 5 = 30.
The longest length of each piece is 30 cm.
To find the longest length of each piece of the two wooden planks, 330 cm and 180 cm, respectively, we have to find the GCF of 330 and 180, which is 30 cm.
The longest length of each piece is 30 cm.
If the GCF of 330 and ‘b’ is 30, and the LCM is 1980, find ‘b’.
The value of ‘b’ is 180.
GCF × LCM = product of the numbers
30 × 1980 = 330 × b
59400 = 330b
b = 59400 ÷ 330 = 180
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.