**Various Types of Factors**

** Even factors**:

Even factors are the factors of number, which are divisible by 2.

**Example 1**: **Find the number of even factors of 58800?**

**Solution**: We first factorize 58800.

58800 = 2^{4 }3^{1}5^{2}7^{2}

In this case we have to find number of even factors, an even factor is divisible by 2 or we can say smallest power of 2 should be 1 not 0.

Hence a factor must have

2^{(1 or 2 or 3 or 4) }Â — 4 factors

3^{(0 or 1 )Â }Â Â Â Â Â — 1+1=2 factors

5^{(0 or 1 or 2)} — 1+2=3 factors

7^{(0 or 1 or 2) }— 1+2=3 factors

Hence total number of even factors = (4)(2)(3)(3) = 72

**Hence, number of even factors of a number **

$N={{\mathbf{2}}^{\mathbf{p}}}{{\mathbf{a}}^{\mathbf{q}}}{{\mathbf{b}}^{\mathbf{r}}}{{\mathbf{c}}^{\mathbf{s}}}\mathbf{are}\text{ }\mathbf{p}\left( \mathbf{q}\text{ }+\text{ }\mathbf{1} \right)\left( \mathbf{r}\text{ }+\text{ }\mathbf{1} \right)\left( \mathbf{s}\text{ }+\text{ }\mathbf{1} \right)$

__Odd factors__**: **

Odd factors are those factors, which are not divisible by 2.

**Example 2**: **Find the number of odd factors of 58800.**

**Solution: **We first factorize 58800.

58800 = 2^{4 }3^{1}5^{2}7^{2}

Total number of factors of 58800 is 5x2x3x3=90 and in previous example we calculated total number of even factors is 72.

Number of odd factors = Total number of factors – Number of even factors

=90- 72 = 18

** Alternate way**: Since odd factors should have power of 2 as 0. Hence odd factor must have

2

^{(0 ) }Â ———- 1 factor

3^{(0 or 1 )Â }Â Â Â — 1+1=2 factors

5^{(0 or 1 or 2)} — 1+2=3 factors

7^{(0 or 1 or 2) }— 1+2=3 factors

Total number of odd factors = (1 )(2)(3)(3) = 18

Hence, number of odd factors of a number

$N={{\mathbf{2}}^{\mathbf{p}}}{{\mathbf{a}}^{\mathbf{q}}}{{\mathbf{b}}^{\mathbf{r}}}{{\mathbf{c}}^{\mathbf{s}}}\mathbf{are}\text{ }\mathbf{p}\left( \mathbf{q}\text{ }+\text{ }\mathbf{1} \right)\left( \mathbf{r}\text{ }+\text{ }\mathbf{1} \right)\left( \mathbf{s}\text{ }+\text{ }\mathbf{1} \right)$

__Perfect square factors:__

If a number is perfect square then its prime factors must have even powers.

**Example 3: Find the number of factors of 58800 which Â are perfect square?**

**Solution:**

We know that for a number to be a perfect square, its factor must have the even number of powers.

We first factorize 58800.

58800 = 2^{4 }3^{1}5^{2}7^{2}

Hence perfect square factors must have

2^{(0 or 2 or 4)}—– 3 factors

3^{( 0 )Â }Â Â Â —–Â Â 1Â factor

5^{(0 or 2)} ——- 2 factors

7^{(0 or 2) }— 2 factors

Hence, the number of factors which are perfect square are 3x1x2x2=12

**Remember **

**-If number of factors is odd then the number is a perfect square and vice versaÂ is also true i.e. if a number is a perfect square then number of factors is Â odd.**

**This is because if number is a perfect square then p, q, and r are even and hence **

**(p + 1) (q + 1) and (r+ 1) are odd and so product of these numbers is also an odd number. **

**-If number of factors is even then number is not a perfect square. **