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Last updated on August 12, 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, group or arrange items, and schedule events. In this topic, we will learn about the GCF of 8 and 30.
The greatest common factor of 8 and 30 is 2. 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 8 and 30, a few methods are described below:
Steps to find the GCF of 8 and 30 using the listing of factors:
Step 1: Firstly, list the factors of each number Factors of 8 = 1, 2, 4, 8. Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30.
Step 2: Now, identify the common factors of them Common factors of 8 and 30: 1, 2.
Step 3: Choose the largest factor The largest factor that both numbers have is 2. The GCF of 8 and 30 is 2.
To find the GCF of 8 and 30 using the Prime Factorization Method, follow these steps:
Step 1: Find the prime factors of each number Prime factors of 8: 8 = 2 x 2 x 2 = 2³ Prime factors of 30: 30 = 2 x 3 x 5
Step 2: Now, identify the common prime factors The common prime factor is: 2
Step 3: Multiply the common prime factors 2 = 2 The Greatest Common Factor of 8 and 30 is 2.
Find the GCF of 8 and 30 using the division method or Euclidean Algorithm Method. Follow these steps:
Step 1: First, divide the larger number by the smaller number Here, divide 30 by 8 30 ÷ 8 = 3 (quotient), The remainder is calculated as 30 − (8×3) = 6 The remainder is 6, not zero, so continue the process
Step 2: Now divide the previous divisor (8) by the previous remainder (6) Divide 8 by 6 8 ÷ 6 = 1 (quotient), remainder = 8 − (6×1) = 2 The remainder is 2, not zero, so continue the process
Step 3: Now divide the previous divisor (6) by the previous remainder (2) Divide 6 by 2 6 ÷ 2 = 3 (quotient), remainder = 6 − (2×3) = 0 The remainder is zero, the divisor will become the GCF. The GCF of 8 and 30 is 2.
Finding the GCF of 8 and 30 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 8 red flowers and 30 yellow flowers. She wants to arrange them in bouquets with the largest number of flowers in each bouquet, using the same number of flowers per bouquet. How many flowers will be in each bouquet?
We should find the GCF of 8 and 30 GCF of 8 and 30 is 2. There are 2 flowers per bouquet 8 ÷ 2 = 4 30 ÷ 2 = 15
There will be 2 flowers per bouquet, with 4 bouquets of red flowers and 15 bouquets of yellow flowers.
As the GCF of 8 and 30 is 2, the gardener can arrange the flowers in bouquets with 2 flowers each. Now divide 8 and 30 by 2. There will be 4 bouquets of red flowers and 15 bouquets of yellow flowers.
A chef has 8 kg of flour and 30 kg of sugar. He wants to package them into bags with the same weight, using the largest possible weight per bag. How much weight will each bag have?
GCF of 8 and 30 is 2. So each bag will have 2 kg of either flour or sugar.
There are 8 kg of flour and 30 kg of sugar. To find the total weight in each bag, we should find the GCF of 8 and 30. Each bag will have 2 kg.
A baker has 8 loaves of bread and 30 croissants. She wants to pack them into boxes with an equal number of items, using the maximum number of items per box. How many items should be in each box?
For calculating the maximum number of items, we have to calculate the GCF of 8 and 30 The GCF of 8 and 30 is 2. Each box will have 2 items.
To calculate the maximum number of items per box, first, we need to calculate the GCF of 8 and 30, which is 2. Each box will have 2 items.
A landscaper has two lengths of garden hose, one 8 meters and the other 30 meters. He wants to cut them into the longest possible equal pieces, without any hose left over. What should be the length of each piece?
The landscaper needs the longest piece of hose. GCF of 8 and 30 is 2. The longest length of each piece is 2 meters.
To find the longest length of each piece of hose, 8 meters and 30 meters respectively, we find the GCF of 8 and 30, which is 2 meters. The longest length of each piece is 2 meters.
If the GCF of 8 and ‘b’ is 2, and the LCM is 120, find ‘b’.
The value of ‘b’ is 30.
GCF x LCM = product of the numbers
2 × 120 = 8 × b
240 = 8b
b = 240 ÷ 8 = 30
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.