*Friday, July 10th, 2020*

If N is a composite number such that N =a^{p },b^{q },c^{r }… where a, b, c are prime factors of N and p,q,r ….. are positive integers, the number of factors of N is given by the expression (p+1) (q+1)(r+1).

For example: 140 = 2^{2 }x 5^{1 }x 7^{1 }

Hence, 140 has (2+1)(1+1)(1+1), i.e., 12 factors.

The basic principle is simple, every number has certain number of factors. You can find manually for the small numbers but for the large ones there is a need of a mathematical formula.

The formula is easy to learn, but you should know the basic concept behind this i.e. why we are adding 1 to each power. When we say that a^{m} is a factor of a number N, we understand that each of a^{m, }a^{m-1, }a^{m-2},…a^{1} , a^{0 } will also be a factor of N. Simply speaking, we have to count all the factors from 0 till m and that’s why we add 1 to each factors. Now, say for example we have to find all the factors that do not have a^{3} in them, then we will ignore all the factors starting from a^{3} to a^{m} and consider all a^{0}, a^{1} and a^{2}. A lot of questions are formed around this concept and if one is not careful, it can lead to loss of easy marks.

Please note that the figure arrived at by using the above formula also includes 1 and the given number N itself as factors. So if you want to find out the number of factors the given number has excluding 1 and the number itself, we find out (p+1) (q+1) (r+1) and then subtract 2 from that figure.

In the above example, the number 140 has 10 factors excluding 1 and itself.

Some more examples:

- Find the number of factors that the number 2025 has.

Sol: First, express 2025 as a product of its prime factors. (Note that to express a given number as a product of its prime factors, we first need to identify the prime factors the given number has by applying the rules of divisibility)

2025= 3 x3x3x3x5x5= 3^{4 }x 5^{2 }

Hence, the number of factors 2025 has is (4+1) (2+1) = 15

- How many divisors, (including 1 and itself) does the number 8625 have?

Sol: Note that the two terms factors and divisors are used interchangeably. First express 8625 in terms of its prime numbers.

8625= 3 x 5 x5 x 5 x 23 = 3^{1 }x 5^{3 }x 23^{1 }

Hence, 8625 has (1+1) (3+1) (1+1) = 16 factors

Excluding 1 and itself, the number has (16-2) = 14 factors

The given number N (which can be written as equal to a^{p }.b^{q }. c^{r }… where a, b, c are prime factors of N and p, q, r….. are positive integers) can be expressed as the product of two factors in different ways.

The number of ways in which this can be done is given by the expression ½ {(p+1)(q+1)(r+1)….}

So, 140 can be expressed as a product of two factors in 12/2 or 6 ways {because (p+1)(q+1)(r+1) in the case of 140 is equal to 12}

If p, q, r, etc. are all even, then the product (p+1)(q+1)(r+1)…becomes odd and the above rule will not be valid since we cannot take ½ of an odd number to get the number of ways. If p,q,r … are all even, it means that the number N is a perfect square. This situation arise in the specific cases of perfect squares because a perfect square can also be written as (square root x square root). So, two different cases arise in case of perfect squares depending on whether we would like to consider the writing number as (square root x square root) also as one way.

Thus, to find out the number of ways in which a perfect square can be expressed as a product of 2 factors, we have the following 2 rules.

- as a product of two DIFFERENT factors in ½ {(p+1)(q+1)(r+1)…. -1} ways excluding (√N x √N)
- as a product of two factors (including √N x √N) in ½ {(p+1)(q+1)(r+1)….+1} ways.

Examples:

- In how many ways can 14630 be written as the product of two factors?

**Sol:** 14630 = 2x5x7x11x19

Hence, 14630 can be written as the product of two factors in

½ {(1+1) (1+1) (1+1) (1+1) (1+1)} = 16 ways

- In how many ways can the number 44100 can be written as a product of two different factors?

**Sol:** First expressing 44100 a product of its prime factors. We get 44100 = 2^{2 }x 3^{2 }x 5^{2 }x 7^{2 }

Since all the powers are even, the given number is a perfect square. Since the question has asked us to find the number of ways of writing the number as a product of two “different” factors, we cannot consider (sq. root x sq. root).

So, the required number of ways is: ½ {(2+1) (2+1) (2+1) (2+1) -1} = ½ {81-1} =40.

A slightly more complicated concept involves finding the sum of factors. Again, it is the extension of the number of factors and the properties of geometric progression and extremely derivable.

If N = a^{m} x b^{n} x c^{p} x d^{q}…. where, a,b, c and d are prime factors of N.

Sum of all factors will be nothing but S= (a^{0}+ a^{1}+ a^{2}+ a^{3}+….. a^{m}) (b^{0}+ b^{1}+ b^{2}+ b^{3}+….. b^{m})………

Using sum of geometric progression, S= (a^{m+1} – 1) *(b^{n+1} – 1)* (c^{p+1} – 1)……./ (a-1)(b-1)(c-1)……

which is essentially the sum of the factors. Again, using some logic, we can figure it out that if a particular factor is not required, we can simply eliminate it from the bracket and leave the rest as it is and calculate the answer.

For e.g. Consider 120 and find the sum of its factors.

The prime factorization of 120 is 23x31x51. By applying the formulae,

Sum of factors = [(20+21+22+23)(30+31)(50+51)]/ [(2-1) (3-1)(5-1)] = 45

We shall begin by working out the product of all the factors of a given number.

Example: What is the product of all the factors of 180?

Solution: 180= 2^{2}3^{2}5^{1}. There are (2+1)(2+1)(1+1) or 18 factors.

If the given number is not a perfect square, at least one of the indices is odd and the number of factors is even. We can form pairs such that the product of the two numbers in each pair is the given number (180 in this example).

The required product is 180^{9}.

In general, if N=p^{a}q^{b}r^{c} (where at least one of a,b,c is odd) then the product of all the factors of N is N^{d/2} where d is the total number of factors.

Example: Consider the number 120, find the product of the factors.

The prime factorization of 120 is 2^{3}x3^{1}x5^{1}

Number of factors = (3+1)(1+1)(1+1) = 16

Product of factors = 120(16/2) = 120^{8}

The questions by asking the number of odd factors or even factors in which you can eliminate all powers of 2 except 2^{0 }(which is odd){for the case of odd factors}.

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.

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