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128 LearnersLast updated on December 11, 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 schedule events. In this topic, we will learn about the GCF of 30 and 80.
The greatest common factor of 30 and 80 is 10.
The largest divisor of two or more numbers is called the GCF of the number.
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 30 and 80, a few methods are described below -
Steps to find the GCF of 30 and 80 using the listing of factors
Step 1: Firstly, list the factors of each number Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30. Factors of 80 = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.
Step 2: Now, identify the common factors of them Common factors of 30 and 80: 1, 2, 5, 10.
Step 3: Choose the largest factor The largest factor that both numbers have is 10.
The GCF of 30 and 80 is 10.


To find the GCF of 30 and 80 using the Prime Factorization Method, follow these steps:
Step 1: Find the prime factors of each number Prime Factors of 30: 30 = 2 x 3 x 5 Prime Factors of 80: 80 = 2 x 2 x 2 x 2 x 5 = 2^4 x 5.
Step 2: Now, identify the common prime factors The common prime factors are: 2 x 5.
Step 3: Multiply the common prime factors 2 x 5 = 10.
The Greatest Common Factor of 30 and 80 is 10.
Find the GCF of 30 and 80 using the division method or Euclidean Algorithm Method.
Follow these steps:
Step 1: First, divide the larger number by the smaller number Here, divide 80 by 30 80 ÷ 30 = 2 (quotient), The remainder is calculated as 80 − (30×2) = 20 The remainder is 20, not zero, so continue the process.
Step 2: Now divide the previous divisor (30) by the previous remainder (20) Divide 30 by 20 30 ÷ 20 = 1 (quotient), remainder = 30 − (20×1) = 10.
Step 3: Now divide the previous divisor (20) by the previous remainder (10) 20 ÷ 10 = 2 (quotient), remainder = 20 − (10×2) = 0 The remainder is zero, the divisor will become the GCF.
The GCF of 30 and 80 is 10.
Finding the GCF of 30 and 80 looks simple, but students often make mistakes while calculating the GCF.
Here are some common mistakes to be avoided by the students.
A baker has 30 cupcakes and 80 muffins. She wants to organize them into equal sets with the largest number of items in each set. How many items will be in each set?
We should find the GCF of 30 and 80 GCF of 30 and 80 2 x 5 = 10.
There are 10 equal groups 30 ÷ 10 = 3 80 ÷ 10 = 8.
There will be 10 groups, and each group gets 3 cupcakes and 8 muffins.
As the GCF of 30 and 80 is 10, the baker can make 10 groups.
Now divide 30 and 80 by 10.
Each group gets 3 cupcakes and 8 muffins.
A school has 30 red balloons and 80 blue balloons. They want to decorate them in rows with the same number of balloons in each row, using the largest possible number of balloons per row. How many balloons will be in each row?
GCF of 30 and 80 2 x 5 = 10.
So each row will have 10 balloons.
There are 30 red and 80 blue balloons.
To find the total number of balloons in each row, we should find the GCF of 30 and 80.
There will be 10 balloons in each row.
A tailor has 30 meters of yellow fabric and 80 meters of green fabric. She wants to cut both fabrics 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 30 and 80 .
The GCF of 30 and 80 is 2 x 5 = 10.
The fabric is 10 meters long.
For calculating the longest length of the fabric, first, we need to calculate the GCF of 30 and 80, which is 10.
The length of each piece of fabric will be 10 meters.
A carpenter has two wooden planks, one 30 cm long and the other 80 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 30 and 80 is 2 x 5 = 10.
The longest length of each piece is 10 cm.
To find the longest length of each piece of the two wooden planks, 30 cm and 80 cm, respectively, we have to find the GCF of 30 and 80, which is 10 cm.
The longest length of each piece is 10 cm.
If the GCF of 30 and โbโ is 10, and the LCM is 240. Find โbโ.
The value of ‘b’ is 80.
GCF x LCM = product of the numbers
10 × 240 = 30 × b
2400 = 30b
b = 2400 ÷ 30 = 80

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.






