*Monday, September 21st, 2020*

let me first list down the topics that I am going to cover in this particular blogpost:

- Number of factors of a given number
- Number of
*even*factors or*odd*factors of a given number - Sum of all factors of a given number
- Sum of all
*even*factors or*odd*factors of a given number

We know that a number N can be written as a product of its factors as given below

- N = a
^{p}x b^{q }x c^{r}…

Here a,b,c… are prime factors of N

& p,q,r … are the powers of the prime factors of N.

**In such a case the number of factors of N are given by the formula**

The obvious question which arises is, why this formula in particular. It is actually a game of choices and options available. Let us see how.

Suppose you are going on a camping trip and you have packed all the necessary items. Now you are wondering how many movie DVDs, music CDs & MP3 players you should take. (*Please do not judge me by the example. I have no experience about camping trips but I guess that is obvious by the example mentioned*). In your collection, you have 5 movie DVDs, 4 music CDs and 2 MP3 players.

Try to answer these questions:

- How many choices / options do you have for the number of movie DVDs that you can take?
- How many choices / options do you have for the number of music CDs that you can take?
- How many choices / options do you have for the number of MP3 players that you can take?
- Are the answers to all the above questions independent of each other?

Your answers for the above questions should be:

- 6 (
*You can take either 1 or 2 or 3 or 4 or 5 Movie DVDs. You also have the option of not taking a movie DVD at all.)* - 5 (
*You can take either 1 or 2 or 3 or 4 Music CDs. You also have the option of not taking a music CD at all.)* - 3 (
*You can take either 1 or 2 or 3 or 4 or 5 MP3 players. You also have the option of not taking a MP3 player at all.)* - Yes. All answers are independent of each other.

So, what is the total number of options / choices do you have while packing for the camping trip?

Answer = 6 x 5 x 3 = 90

If you understood the above example, the number of factors formula should be a piece of cake.

Consider a number **N = 2 ^{5 }x 3^{4} x 5^{2}**

It means that you have: 2,2,2,2,2 && 3,3,3,3 && 5,5

Any combination of the above will make a factor. You have 6 choices for picking up the number of 2s in a factor, 5 choices for picking up the number of 3s in a factor and 3 choices for picking up the number of 5s in a factor.

So, the total number of factors = 6 x 5 x 3 = 90.

Now, for finding out the number of even factors consider the camping trip case.

If one of your friends insists that he wants to watch Twilight while on the camping trip and you should bring the movie DVD because you own a copy. (*Once again I will restrain myself and not pass a judgement on you or your friend or the choice of the movie)*

So, in this case the number of choices for picking up the movie DVD has reduced from 6 to 5 because you HAVE TO bring the Twilight DVD.

So, the total number of choices that you have now is 5 x 5 x 3 = 75

Same logic can be extended to finding out the number of even factors.

Number of even factors = 5 x 5 x 3 = 75 {Because your factor has to contain at least one 2, analogous to the twilight DVD}

Number of odd factors = 5 x 3 = 15 {In this case, your factor cannot contain any 2s, analogous to not being allowed to take a movie DVD}

As a matter of fact, if you have the total number of factors and the total number of even factors; their difference would directly give you the total number of odd factors.

Such logic can be extended to find out the number of factors divisible by a particular number.

Number of factors divisible by 12 = 4 x 4 x 3 = 48 {In this case, your factor will have to contain at least two 2s and one 3}

We know that a number N can be written as a product of its factors as given below

- N = a
^{p}x b^{q }x c^{r}…

Here a,b,c… are prime factors of N

& p,q,r … are the powers of the prime factors of N.

**In such a case the sum of factors of N are given by the formula**

The logic remains the same in this case also. If you expand the above expression, you will end up with all the factors.

For **N = 2 ^{5 }x 3^{4} x 5^{2}**

Sum of all factors

Sum of all the even factors

Sum of all the odd factors

Sum of all the factors divisible by 12

Divisibility Rules for CAT Quantitative Aptitude Preparation

Baisc Idea of Remainders

Application of LCM (Lowest Common Multiple) in solving Quantitative Aptitude Problems

How to Find Number of Trailing Zeros in a Factorial or Product

Dealing With Factorials

All questions from CAT Exam Quantitative Aptitude – Number Systems

Quantitative Aptitude – Number Systems – Q1: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

Quantitative Aptitude – Number Systems – Q2: If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

Quantitative Aptitude – Number Systems – Q3: The numbers 1, 2,…,9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

I hope that this post was helpful and you will not face any problems in finding out the number of factors and related stuff.

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