which is the product of all the factors
Answers
Answer:
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=pa×qb×rc×… N = p a × q b × r c × …
Step-by-step explanation:
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:
The number of factors
The sum of factors
Product of factors
Odd and Even Factors
Number of Factors which are Perfect Squares
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
×
3
2
×
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
×
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
×
3
2
or
2
2
×
3
1
or
2
1
×
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
×
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
×
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.