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# Write all the factors of 125.

Last updated date: 18th Jun 2024
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Hint:
Hint: We solve this problem by using the divisibility of numbers with all numbers from 1 to 9 where we can divide the given number 125 exactly. For checking the divisibility we use the divisibility rules.
(1) The divisibility rule of says that every number is divisible by 1
(2) The divisibility rule of 2 says that if the unit digit of number is even then that number is divisible by 2
(3) If the sum of digits is equal to multiple of 3 then the number is divisible by 3
(4) If the number formed from last two digits of a number is divisible by 4 then the whole number is divisible by 4
(5) If the unit digit is 0, 5 then the number is divisible by 5
(6) If the number is divisible by 2 and 3 then the number is divisible by 6

We are given that the number as 125
We know that we can check the factors by using the divisibility rules from 1 to 9
(i) Let us check the divisibility of 125 with 1
We know that every number is divisible by 1
So, we can write the given number as
$\Rightarrow 125=1\times 125$
(ii) Let us check the divisibility of 125 with 2
We know that if the unit digit of number is even then that number is divisible by 2
But, here the unit digit in 125 is not a even number
So, we can say that 125 is not divisible by 2
(iii) Let us check the divisibility of 125 with 3
We know that if the sum of digits is equal to multiple of 3 then the number is divisible by 3
Now, by adding the digits of 125 we get
$\Rightarrow 1+2+5=8$
Here, we can see that the number 8 is not a multiple of 3
So, we can say that 125 is not divisible by 3
(iv) Let us check the divisibility of 125 with 4
We know that if the number formed from last two digits of a number is divisible by 4 then the whole number is divisible by 4
Here, we can see that the number formed from last two digits of 125 is 25 which is not divisible by 4
So, we can say that 125 is not divisible by 4
(v) Let us check the divisibility of 125 with 5
We know that if the unit digit is 0, 5 then the number is divisible by 5
Here, we can see that the unit digit of 125 is 5
So, we can say that the number 125 is divisible by 5
Now, we can write the given number after dividing with 5 as
$\Rightarrow 125=1\times 5\times 25$
We know that the 25 can be written as $25={{5}^{2}}$
By substituting this value in above equation we get
$\Rightarrow 125=1\times {{5}^{3}}$
So, we can take the possible factors by selecting at least 1 number from the above product.
Therefore, the possible factors of 125 are 1, 5, 25 and 125.

Note: We can use the prime factorisation method for solving this problem.
The prime factorisation method is nothing but rewriting the given number as the product of prime numbers.
Now, by writing the number 125 as product of prime numbers we get
$\Rightarrow 125={{5}^{3}}$
So, we can take the possible factors by selecting at least 1 number from the above product.
Therefore, the possible factors of 125 are 1, 5, 25 and 125.