Number system is one of the most important topics in the quantitative aptitude section of CAT. If we look at Previous Question papers of CAT, we can see that around 4-6 questions are asked from this topic every year. Hence it becomes important to prepare this topic well. In this post, we will discuss factors of a number which is one of the important concepts under number systems.

There are only two concepts in factors. These are:

1. Finding the total factors of a number.

2. Finding the sum of the factors of a number.

All questions on factors make use of these two concepts only. Sometimes the question may involve the direct application of the concept while in other cases the questions may not be so direct. We will discuss both the cases with examples.

# FINDING THE TOTAL FACTORS OF A NUMBER

Finding the total number of factors for a given number is fairly straight forward.

If the number N is of the form $a^p*b^q*c^r$, then the total number of factors of N is given by $(p+1)*(q+1)*(r+1)$. Let us understand the logic behind this formula with the help of an example

### EXAMPLE

Find the total number of factors of 720.

The prime factorization of 720 would be $720 = 2^4*3^2*5$

Now we have to find the total number of factors of this number. There are 5 powers of 2 which are available. These are $2^0, 2^1, 2^2, 2^3, 2^4$. Hence we can choose the powers of 2 in 5 ways. i.e 1 or 2 or 3 or 4 or 5.

Similarly, we can choose the power of 3 in 3 ways. i.e. $3^0, 3^1, \textrm{ or } 3^2$.

In the same way, we can choose the power of 5 in 2 ways.

Hence the total number of factors of 720 would be 5*3*2 = 30

Thus, we can see the logic behind the above formula.

We have discussed the formula for the total number of factors. However, the questions are not always direct. Sometimes the question asks to find the total number of even or odd factors.

## TOTAL NUMBER OF EVEN/ODD FACTORS

If the prime factorization of a number does not have â€˜2â€™, then all of its factors would be odd.

If the prime factorization of a number has 2 then we can find the odd factors by excluding the term corresponding to the power of 2 in the formula.For example,

For example, lets say that the number N is of the form N = $2^p *a^q*b^r*c^s$

If the factors are to be odd then there shouldnâ€™t be any 2 in them. Hence the powers of 2 can only be selected in 1 way i.e. $2^0$. Hence we can exclude the 2 and simply find the total factors of the number Nâ€™ = $ a^q*b^r*c^s$.The total number of even factors of a given number can be obtained as follows

Total even factors = Total factors â€“ total no. of odd factors

Thus, we have discussed the basics related to finding the total number of factors for a given number. Letâ€™s now move on to the sum of the factors.

## FINDING THE SUM OF THE FACTORS

SinceÂ we have covered the logic behind the formula for the factors of a number, it will be much easier to understand the formula for the sum of the factors.

For N = $a^p*b^q*c^r$, the sum of the factors of a number is given by

S = $\frac{a^{p+1} – 1}{p-1}*\frac{b^{q+1}- 1}{q-1}*\frac{c^{r+1}-1}{r-1}$

Lets understand the logic behind the formula with the help of an example.

### EXAMPLE

Find the sum of all the factors of 360360 = $2^3*3^2*5$

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

We can see that the sum of the powers of 2 would be $2^0 + 2^1 + 2^2 = 7$

This is nothing but $\frac{2^4 â€“ 1}{2-1}$.

Similarly, sum of the powers of 3 would be $3^0 + 3^1 + 3^2 = 13$

We can write it as$\frac{3^3 â€“ 1}{3-1}$

Thus, we can see that for a general $a^n$, the sum of all powers(less than or equal to n) of a would be given by $\frac{a^{n+1} â€“ 1}{a-1}$.

Hence using this logic, we get the formula to find the sum of all factors of a particular number.

## SUM OF EVEN/ODD FACTORS

We may not get questions which directly ask for the sum of all the factors. The questions may ask to find the sum of all even or all odd factors. We have to use the same logic in that case which we used in the case of even and odd factors.

If we are asked to find the sum of all odd factors of a number, then we know that there should not be any even factor. Hence we can ignore the term containing 2 from the formula.

So the sum of odd factors of the number N = $2^4*a^p*b^q*c^r$ would be given by $S_{odd} = \frac{a^{p+1} â€“ 1}{ a â€“ 1}*\frac{b^{q+1} â€“ 1}{b-1}$Sum of even factors can be obtained by

$S_{even}$ = Sum of all factors â€“ Sum of the even factors

These are the basic concepts using which all the questions related to factors can be solved.

Letâ€™s have a look at a couple of examples to see the application of these concepts.

### EXAMPLE

How many factors of 540 are divisible by 6 but not divisible by 18?

We know that 540 = $2^2*3^3*5$

Since the number has to be divisible by 6, so there must be one 2 and 1 three in the factors. Moreover, since the number is not divisible by 18, the power of three cannot be more than 1. So, we can choose the power of 2 in only 2 ways $2^0 \textrm{or} 2^1$. Similarly, we can choose the power of 3 in only 1 way. i.e. $3^1$. We can choose the power of 5 in 2 ways ($5^0, 5^1$)

Hence the required number of factors = 2*1*2 = 4

Thus, there are 4 factors of 18 which are divisible by 6 but not divisible by 18.

Letâ€™s consider another example

### EXAMPLE

Find the sum of all the factors of 960, which are multiple of 4 but not multiples of 16.

We know that

960 = $2^6*3*5$

We have to find the sum of the factors which are multiples of 4 but not multiples of 16.

Hence the only powers of 2 which will be available will be $2^2$ and $2^3$

Hence the required sum would be

S = $(2^2 + 2^3)*(3^0 + 3^1)*(5^0 + 5^1) = 12*4*6 = 288$.

The example shows that questions on factors are very easy and you just need to be aware of the basics.

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