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Number of Factors Concepts and Formulas for CAT exam

number of factors for CAT exam

Number of factors based questions have regularly appeared in various competitive exams including CAT.  Primarily, these questions are based on the prime factorization of a number.  Generally,  factors based questions are of following types:

For certain kind of questions, we already have direct formula and shortcut techniques which we apply to get the answer. However,  if an aspirant wants to master this topic,  must also understand the conceptual approach to solve these types of questions.

In this article,  we will focus more on the concepts and approach to tackle question based on factors.

Prime factorisation

We all know that every composite number can be written as a product of some prime numbers. For example, we can write 90 as \(2 \times {3^2} \times 5\). This process is called prime factorisation, and it is the very first  step to solve any questions related to factors.

Number of factors of a number

Factors are those numbers that divide the given number completely. For example, below is the list of all factors of number 72.

Factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 48 and 72.

Observe that the number 1 is always a factor of every number, and the number itself will be the factor of the number. Here we can see that there are 12 factors of 72.

Now let us understand the formula for finding the number of factors of any composite number. We will take the above example of 72.

If the prime factorization 72 we get, \(72 = {2^3} \times {3^2}\)

Clearly, the number 72 is divisible by each of \({2^0},{2^1},{2^2},{2^3}\) but not by any higher power
of 2 like \({2^4},{2^5},…\)

Similarly, the number 72 is divisible by each of \({3^0},{3^1},{3^2}\) but not by \({3^3},{3^4},….\)

We should also observe that the number can be divisible by any ‘combination’ of one
of \({2^0},{2^1},{2^2},{2^3}\) and one of \({3^0},{3^1},{3^2}\) i.e. by numbers of the type \({2^1} \times {3^2}\) or \({2^2} \times {3^1}\) or \({2^1} \times {3^1}\) and so on.
Thus, with \({2^0}\), we could have a total of 3 ‘combinations’ i.e.

number of factors of a number 01

Each of these 3 numbers would divide \({2^3} \times {3^2}\) and thus would be a factor.

Similarly, with \({2^1}\), we could have 3 more combinations

number of factors of a number 02

And each of these 3 factors would be distinct from the earlier 3 factors.

Similarly with EACH of \({2^2}\) and \({2^3}\), we would get 3 more distinct factors and thus the

the total number of factors would be 4 × 3 = 12.

If observed, each of the factors is are distinct numbers because the power of 2 or 3 differs in each of the combinations.

Since the exponent of 2 could assume 4 different values (from 0 to 3), the exponent of 3 could assume 3 distinct values (from 0 to 2), the total number of combinations is 4  × 3 = 12.

From the above example, it is clear that all the factors of 72 would be in the form \({2^a} \times {3^b}\) where a could assume any value from 0 to 3 i.e., 4 different values, and b could consider any value from 0 to 2 i.e., 3 different values. Therefore, the number of different combinations and the number of factors would be  4 × 3 i.e. 12.

The above concepts is lucidly explained in the below video tutorials on number of factors.

Important Formula

Number of factors

If we have to find the number of factors of any number say N, then we should follow below steps:

Step 1: Prime factorize \(N = {p^a} \times {q^b} \times {r^c} \times …\)

Step 2: The number of factors of N= (a+1)(b+1)(c+1)…

Sum of factors

To find the sum of all the factors of a number (say N), we follow below two steps:

Step 1: Prime factorize \(N = {p^a} \times {q^b} \times {r^c} \times …\)

Step 2: Sum of factors =\(\left( {\frac{{{p^{a + 1}} – 1}}{{p – 1}}} \right)\left( {\frac{{{q^{b + 1}} – 1}}{{q – 1}}} \right)\left( {\frac{{{r^{c + 1}} – 1}}{{r – 1}}} \right) \ldots \)

Product of Factors

To find the product of all the factors of a number (say N), we follow below three steps:

Step 1: Prime factorize \(N = {p^a} \times {q^b} \times {r^c} \times …\)

Step 2: Let the number of factors of N be x. therefore, x= (a+1)(b+1)(c+1)…

Step 3: Product of factors =\({N^{\frac{x}{2}}}\)

Let us take an example to understand the working of the above formulas.

Question: Find the number of factors, the sum of factors and product of factors of 1800.


Prime factorization of \(1800 = {2^3} \times {3^2} \times {5^2}\)

Number of factors 1800= (3+1)(2+1)(2+1) = 36

Sum of factors of 1800

\(\begin{array}{l} = \left( {\frac{{{2^{3 + 1}} – 1}}{{2 – 1}}} \right)\left( {\frac{{{3^{2 + 1}} – 1}}{{3 – 1}}} \right)\left( {\frac{{{5^{2 + 1}} – 1}}{{5 – 1}}} \right)\\ = 15 \times 4 \times 31 = 1860\end{array}\)

Product of Factors of 1800=\({1800^{\frac{{36}}{2}}} = {1800^{18}}\)

Number of  factors of specific types

The questions which appear in competitive exams from factors are not restricted to the above three types only.  Variety of questions can be created based on finding the factors. The best way to approach such questions is to go by the conceptual way rather than remembering the formula for each kind.

Let us take some varieties of questions which appear frequently.

Number of odd and even factors

Question:  Find the number of Even and Odd factors of 1200.


Prime factorization of  \(1200 = {2^4} \times {3^1} \times {5^2}\)

We know the fact that all the factors of 1200 will be in the form of \({2^a} \times {3^b} \times {5^c}\).

Where a will range from 0 to 4, b from 0 to 1 and c from 0 to 2. However, it is quite evident that any factor which has 2  as one of the prime factors is always even.  similarly,  and your factors will not have 2 as its prime factor.

Therefore, for ODD factors, the exponent of 2 i.e., a has to be 0 always. Or, the number of ways using 2  for making the combination is 1, i.e., \({2^0}\). Also, the number of values which the exponent of 3 and 5, i.e., b, and c can take are 2 and 3 respectively.

Hence the number of odd factors of 1200 = 1x2x3=6.

Extending the logic, we can say that for a better to be Even,  it must contain 2 at least once.

So, the number of values a, b, and c can take are 4, 2, and 3 respectively.

Therefore, the total number of even factors of 1200 = 4x2x3=24.

Number of factors which are perfect squares

Question: Find the number of factors which are perfect squares of 10800.


Prime factorization of \(10800 = {2^4} \times {3^3} \times {5^2}\)

If we prime factorise any number which is a perfect square, we would observe that in all cases the exponent of all the prime factors of the number to be even only.

For example, 36 is perfect square \(36 = {2^2} \times {3^2}\). here we can see that the exponent of both 2 and 3 are even.

Again, any factor 10800 will be in the form of \({2^a}{ \times ^b} \times {5^c}\). For the factors to be perfect squares, all the values a, b, and c has to be even only.

Or, the possible values which a can take =  0, 2, 4, i.e. 3 values only. Similarly, b can take 0, 2 i.e. 2 values and c can take 0, 2 i.e. 2 values.

Therefore, the different combinations we can have = 3x2x2 = 12.

Hence, 10800 has 12 factors which are perfect squares.

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