Summarize this article:
Last updated on September 24, 2025
The GCF is the largest number that can divide two or more numbers without leaving any remainder. GCF is used to share the items equally, to group or arrange items, and schedule events. In this topic, we will learn about the GCF of 625 and 1000.
The greatest common factor of 625 and 1000 is 125. 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 625 and 1000, a few methods are described below
Steps to find the GCF of 625 and 1000 using the listing of factors
Step 1: Firstly, list the factors of each number
Factors of 625 = 1, 5, 25, 125, 625.
Factors of 1000 = 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000.
Step 2: Now, identify the common factors of them Common factors of 625 and 1000: 1, 5, 25, 125.
Step 3: Choose the largest factor The largest factor that both numbers have is 125.
The GCF of 625 and 1000 is 125.
To find the GCF of 625 and 1000 using the Prime Factorization Method, follow these steps:
Step 1: Find the prime factors of each number
Prime Factors of 625: 625 = 5 × 5 × 5 × 5 = 54
Prime Factors of 1000: 1000 = 2 × 2 × 2 × 5 × 5 × 5 = 23 × 53
Step 2: Now, identify the common prime factors The common prime factors are: 5 × 5 × 5 = 53
Step 3: Multiply the common prime factors 53 = 125.
The Greatest Common Factor of 625 and 1000 is 125.
Find the GCF of 625 and 1000 using the division method or Euclidean Algorithm Method. Follow these steps:
Step 1: First, divide the larger number by the smaller number
Here, divide 1000 by 625 1000 ÷ 625 = 1 (quotient),
The remainder is calculated as 1000 − (625×1) = 375
The remainder is 375, not zero, so continue the process
Step 2: Now divide the previous divisor (625) by the previous remainder (375)
Divide 625 by 375 625 ÷ 375 = 1 (quotient), remainder = 625 − (375×1) = 250
The remainder is 250, not zero, so continue the process
Step 3: Now divide the previous divisor (375) by the previous remainder (250)
Divide 375 by 250 375 ÷ 250 = 1 (quotient), remainder = 375 − (250×1) = 125
The remainder is 125, not zero, so continue the process
Step 4: Now divide the previous divisor (250) by the previous remainder (125)
Divide 250 by 125 250 ÷ 125 = 2 (quotient), remainder = 250 − (125×2) = 0
The remainder is zero, the divisor will become the GCF.
The GCF of 625 and 1000 is 125.
Finding GCF of 625 and 1000 looks simple, but students often make mistakes while calculating the GCF. Here are some common mistakes to be avoided by the students.
A teacher has 625 pencils and 1000 erasers. She wants to group them into equal sets, with the largest number of items in each group. How many items will be in each group?
We should find the GCF of 625 and 1000 GCF of 625 and 1000 53 = 125.
There are 125 equal groups 625 ÷ 125 = 5 1000 ÷ 125 = 8
There will be 125 groups, and each group gets 5 pencils and 8 erasers.
As the GCF of 625 and 1000 is 125, the teacher can make 125 groups.
Now divide 625 and 1000 by 125.
Each group gets 5 pencils and 8 erasers.
A school has 625 red chairs and 1000 blue chairs. They want to arrange them in rows with the same number of chairs in each row, using the largest possible number of chairs per row. How many chairs will be in each row?
GCF of 625 and 1000 53 = 125.
So each row will have 125 chairs.
There are 625 red and 1000 blue chairs.
To find the total number of chairs in each row, we should find the GCF of 625 and 1000.
There will be 125 chairs in each row.
A tailor has 625 meters of red ribbon and 1000 meters of blue ribbon. She wants to cut both ribbons 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 625 and 1000
The GCF of 625 and 1000 53 = 125.
The ribbon is 125 meters long.
For calculating the longest length of the ribbon first we need to calculate the GCF of 625 and 1000 which is 125.
The length of each piece of the ribbon will be 125 meters.
A carpenter has two wooden planks, one 625 cm long and the other 1000 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 625 and 1000 53 = 125.
The longest length of each piece is 125 cm.
To find the longest length of each piece of the two wooden planks, 625 cm and 1000 cm, respectively.
We have to find the GCF of 625 and 1000, which is 125 cm.
The longest length of each piece is 125 cm.
If the GCF of 625 and ‘a’ is 125, and the LCM is 5000. Find ‘a’.
The value of ‘a’ is 1000.
GCF × LCM = product of the numbers
125 × 5000 = 625 × a
625000 = 625a
a = 625000 ÷ 625 = 1000
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.