In this article, we will explore the divisor concepts, including the number of factors, prime factors, and the fascinating properties of the sum and product of factors. By understanding these concepts, you will gain valuable insights into the relationships between numbers and unlock the secrets hidden within their divisors.
Divisors or Factors
The divisor or factor of any number divides it without leaving any remainder. For example, let’s consider the number 12. The divisors or factors of 12 are 1,2,3,4,6 and 12. (Divisor Concepts)
How to find the number of divisors
To find the total number of divisors of any number, we use the prime factorization method given below.
For example, the prime factorization of any number (N) is pa×qb×rc, Here p,q, and r are the prime numbers (2,3,5,7,11,…etc.) and a,b, and c are their respective exponents.
The total number of divisors of N= (a+1)×(b+1)×(c+1)
Example: Find the number of divisors of 360
360=23×32×51
The number of factors= (3+1)×(2+1)×(1+1)=24
24 factors= {1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360}
Que 1: Find the number of prime factors of 360.
Solution: Three Prime factors={2,3,5}
Que 2: Find the number of odd factors of 360.
Solution: Remove 23 from the prime factorization (Divisor Concepts)
Remaining Part= 32×51
Number of odd factors= (2+1)×(1+1)=6
Que 3: Find the number of even factors of 360.
Solution: 360=23×32×51=2×[22×32×51]
Even Factors= 2K (Divisor Concepts)
K=22×32×51
Number of even factors= (2+1)×(2+1)×(1+1)=18
Que 4: Find the number of factors of 360 that are divisible by 12.
Solution: 360=23×32×51=22×3×[21×31×51]
Required factor=12K
K=[21×31×51]
Number of required factors= (1+1)×(1+1)×(1+1)=8
Que 5: Find the number of factors of 360 which are perfect squares.
Solution: 360=23×32×51
Possible pairs={(20, 22)×(30, 32)×(50)}
Number of required factors= 2×2×1=4
The sum of all divisors
N=pa×qb×rc
The sum of all the divisors= (p0+p1+p2+…+pa)×(q0+q1+q2+…+qb)×(r0+r1+r2+…+rc)
If N=360 (Divisor Concepts)
360=23×32×51
The sum= (20+21+22+23)×(30+31+32)×(50+51)
=15×13×6
=1170
Que 1: Find the sum of all odd factors of 360.
Solution: Odd part= 32×51
The sum=(30+31+32)×(50+51)
=13×6
=78
Que 2: Find the sum of all even factors of 360.
Solution: 360=23×32×51=2×[22×32×51]
The sum=2×[(20+21+22)×(30+31+32)×(50+51)]
=2×[7×13×6]
=1092 (Divisor Concepts)
Que 3: Find the sum of all the factors of 360 which are divisible by 12.
Solution: 360=22×3×[21×31×51]
The sum=12×[(20+21)×(30+31)×(50+51)]
=12×[3×4×6]
=864
Que 4: Find the sum of all the factors of 360 which are perfect squares.
Solution: Required sum=(20+22)×(30+32)×(50)
=5×10×1
=50
The product of all divisors
N=pa×qb×rc
The number of all factors= t
t=(a+1)×(b+1)×(c+1)
Case 1: N is not a Perfect Square (Divisor Concepts)
The product of all factors= N^{\frac{t}{2}}.
Case 2: N is a Perfect Square
The product of all factors= N^{\frac{t-1}{2}}\times\sqrt{N}.
If N=360
360=23×32×51
The number of factors (t)= (3+1)×(2+1)×(1+1)=24
Since 360 is not a perfect square
Therefore, the product of all the factors= N^{\frac{t}{2}}=36012
Que 1: Find the product of all the factors of 144
Solution: 144=24×32
The number of factors (t)=(4+1)×(2+1)=15
Since 144 is a perfect square
Therefore, the product of factors= N^{\frac{t-1}{2}}\times\sqrt{N}.
=144^{7}\times\sqrt{144}.
=1214×12
=1215
Note: If the total number of divisors of a number is odd, then that number is a perfect square (Divisor Concepts)
Que 2: Find the product of all the even factors of 360.
Solution: 360=23×32×51=2×[22×32×51]
Even factor=2K
Number of divisors of K=t= (2+1)×(2+1)×(1+1)=18
Here K is not a perfect square
Product of even factors= 2^t\times K^{\frac{t}{2}}=218×[22×32×51]9
=236×318×59
Que 3: Find the product of all the odd factors of 360.
Solution: 360=23×32×51
Odd part= 32×51
Number of divisors of odd part= (2+1)×(1+1)=6
Product of odd factors= [32×51]3=36×53
Question Bank of Divisor Problems with Quick Solutions: Click Here
