# Problems on Divisors of Number

In this post we will have a look at the problems on the number of divisors of a number. We will see how to find the number of divisors of a number, number of even and odd divisors of a number, sum of divisors of a number and product of divisors of a number.

The key to solving problems on divisors of a number is the prime factors of that number. Consider a number 360. 360 in terms of its prime factors is expressed as ${2^3}*{3^2}*{5}$. Using 360 and its prime factors lets look into the various kinds of problems on Divisors of a number

### Number of Divisors of a number

$360 = {2^3}*{3^2}*{5}$

In order to find the number of divisors of 360,

1. Consider the exponents of the prime factors of 360.
-> The exponents of prime factors 2, 3, and 5 are 3, 2, and 1 respectively.
2. Increment each of the exponent obtained in previous step by 1.
3. Product of the incremented exponents from step 2 gives the number of divisors of 360. i.e $(3+1)(2+1)(1+1) = 24$

Hence the number of divisors of 360 is 24.

In general, if N is a composite number such that
$N = a^p * b^q * c^r …$
where, a, b, and c are prime numbers, then the number of divisors of N is given by $(p + 1)(q + 1)(r + 1)…$

### Number of Even and Odd Divisors of a number

$360 = {2^3}*{3^2}*{5}$

In order to calculate the number of even divisors of 360, except for the exponent of 2, we increment the exponent of all other factors by 1 and take their product. Thus we have,

Number of even divisors of 360 => 3 x (2 + 1) x (1 + 1) = 18

To calculate the number of odd divisors of 360, ignore the exponent of 2 and take the product of the incremented(increment by 1) exponents of all other prime factors. Therefore,

Number of odd divisors of 360 => (2+1)(1+1) = 6

In general, if N is a composite number such that
$N = 2^p * b^q * c^r …$
where, a, b, and c are prime numbers, then
-> The number of Even divisors of N is given by $(p)(q + 1)(r + 1)…$
-> The number of Odd divisors of N is given by $(q + 1)(r + 1)…$

### Sum of Divisors of a number

For $N = a^p * b^q * c^r …$, Sum of the divisors of N is given by

$(\frac{a^{p+1}-1}{a-1})(\frac{b^{q+1}-1}{b-1})(\frac{c^{r+1}-1}{c-1})…$

Now, the Sum of divisors of 360 => $(\frac{2^{3+1}-1}{2-1})(\frac{3^{2+1}-1}{3-1})(\frac{5^{1+1}-1}{5-1})$ = 1170

### Product of Divisors of a number

$360 = {2^3}*{3^2}*{5}$

Product of divisors of a number can be obtained based on the number of divisors of that number. i.e, For $N = a^p * b^q * c^r …$, the product of divisors of N is given by

${N^\frac{x}{2}},\ where\ x\ =\ \frac{Number\ of\ divisors\ of\ N}{2}$

Hence, for 360, we have previously found the number of divisors of 360 to be 24. Now, the product of divisors of 360 is

${360^\frac{24}{2}} => {360^{12}}$

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