Suppose a number \(N\) can be written as \({a^m}{b^n}{c^p} \ldots ,\) where \(a,b,c\) are prime numbers and \(m,n,p\) are natural numbers. Then 

If   where \(a,b,c\) are prime and \(m, n\) and \(p\) are natural numbers

Number of factors ( or divisors)

 \(= (m+1)(n+1)(p+1)……\)

Factors are formed by taking combinations of powers of \(a\), \(b\) and \(c\), we know that \({a^0},{a^1},{a^2}, \ldots {a^m}\) can be the factors, similarly \({b^0},{b^1},{b^2}, \ldots {b^n}\) and \({c^0},{c^1},{c^2}, \ldots {c^p}.\) Hence total number of combinations \(= (m + 1)((n + 1)(p + 1)\).

Sum of all these factors

 \[= \left[ {\frac{{{a^{m + 1}} - 1}}{{a - 1}}} \right]\left[ {\frac{{{b^{n + 1}} - 1}}{{b - 1}}} \right]\left[ {\frac{{{c^{p + 1}} - 1}}{{c - 1}}} \right]....\]

Example 12: Find the number of factors of the number 360.

Solution:  360 can be written as \({2^3}{3^2}{5^1}\)

Hence number of factors = 4.3.2 = 24

Example 13: Find sum of the factors  of the number 360.

Solution: 360 can be written as \({2^3}{3^2}{5^1}\). Hence sum  of the factors 

 = \(\left[ {\frac{{{2^{3 + 1}} - 1}}{{2 - 1}}} \right]\left[ {\frac{{{3^{2 + 1}} - 1}}{{3 - 1}}} \right]\left[ {\frac{{{5^{1 + 1}} - 1}}{{5 - 1}}} \right]\)=15.13.6 =1170

Example 14: Find the number of even factors of the number 360.

Solution:  The factors of 360 are given as:

Since we have to take only even factors, we can not select \({2^0}\). we can select only \({2^1},{\rm{ }}{2^2},{\rm{ }}{2^3}\). Similarly \({3^0},{\rm{ }}{3^1},{\rm{ }}{3^2}\) and \({5^0},{\rm{ }}{5^1}\) can be selected.

Total number of ways = 3.3.2 = 18 ways.

Total number of factors =18.

Example 15: Find sum of perimeters of all possible rectangles whose area is 360 square units and sides are integers.

Solution: To solve this problem let us assume area of the triangle is 10. Possible rectangles are (1, 10), (2, 5). Sum of perimeters = 2{1+10+2+5} 

= 2 (sum of factors).

Hence required sum of the perimeters 

= 2(sum of all factors)

= 2. \(\left[ {\frac{{{2^{3 + 1}} - 1}}{{2 - 1}}} \right]\left[ {\frac{{{3^{2 + 1}} - 1}}{{3 - 1}}} \right]\left[ {\frac{{{5^{1 + 1}} - 1}}{{5 - 1}}} \right]\) =2340.

Sum of reciprocals of the factors of a number N =\(\frac{{{\rm{sum}}\,\,{\rm{of}}\,{\rm{factors}}}}{N}\)

Example 16: A number \(N\) has \(x\) factors, \(2N\) has \(1.5x\) factors, \(6N\) has \(2.25x\) factors, \(30N\) has \(4.5x\) factors. How many factors does the number \(3600N\) have?

Solution:  Let the number \(N\) is of the form  \({2^m}{3^n}{5^p} \ldots ..\)

Hence \(2N = {\rm{ }}{2^{m + }}^1{3^n}{5^p} \ldots ..\)

\(6N = {\rm{ }}{2^m}^{ + 1}{3^n}^{ + 1}{5^p} \ldots ..\)

\(30N = {2^m}^{ + 1}{3^n}^{ + 1}{5^p}^{ + 1} \ldots ..\)

Ratio of number of factors of \(2N\) and \(N\) would be \(\frac{{m + 2}}{{m + 1}}\) = 1.5 (given) \( \Rightarrow \) \(m = 1\)

Similarly \(n = 1, p = 0\)

Now \(3600N\)  =  \({2^4}{.3^2}{5^2}.N\) =  \({2^m}^{ + {\rm{ }}4}{3^n}^{ + {\rm{ }}2}{5^p}^{ + {\rm{ }}2} \ldots ..\)

Hence number of factors

  \(=  (m + 5)(n + 3)(p + 3)…\)

Ratio of number of factors of \(3600N\) and \(N\) is \(\frac{{(m + 5)(n + 3)(p + 3)}}{{(m + 1)(n + 1)(p + 1)}}\, = \frac{{6.4.3}}{{2.2.1}} = 18\).

Product of all factors of a number \(N\)  

= \({\left[ N \right]^{\frac{{Number\,\,of\,\,factors}}{2}}}\)