Factors of a number is an important sub-topic from number systems. In this article, we will discuss the fundamentals of factors of a number. Almost every competitive examination has 2-3 medium to difficult level questions based on factors of a number. Taking this into consideration, we will discuss the advanced application of this topic, to give you an edge over other candidates.

Factors of a number N refers to all the numbers which divide N completely. These are also called **divisors of a number.**

These are certain basic formulas pertaining to factors of a number N, such that,

N= p^{a}q^{b}r^{c}

Where, p, q and r are prime factors of the number n.

a, b and c are non-negative powers/ exponents

- Number of factors of N = (a+1)(b+1)(c+1)
- Product of factors of N = N
^{No. of factors/2} - Sum of factors: ( p
^{0}+p^{1}+...+p^{a}) ( q^{0}+ q^{1}+....+q^{b}) (r^{0}+r^{1}+...+r^{c})/ (p^{a}-1)(q^{b}-1)(r^{c}-1)

- Sum of factors
- Number of factors
- Product of factors

- Sum of factors = [(2
^{0}+2^{1}+2^{2}+2^{3})(3^{0}+3^{1})(5^{0}+5^{1})]/ [(2-1) (3-1)(5-1)] = 45 - Number of factors = (3+1)(1+1)(1+1) = 16
- Product of factors = 120
^{(16/2)}= 120^{8}

- Number of odd factors
- Number of even factors

Total number of factors = (2+1)(1+1)(1+1) = 12

**Number of odd factors**will be all possible combinations of powers of 3 and 5 (excluding any power of 2) . Hence number of odd factors = (1+1)(1+1) = 4

By manually checking, these factors are 1, 3, 7 and 21.**Number of even factors = total no. of factors - no. of even factors**= 12 - 4 = 8

By prime factorization, N = 2 × 3, and N^{2} = 2^{2}×3^{2}. No. of Factors of N^{2} = 9 which are 1, 2, 3, 4, 6, 9, 12, 18, 36. Now, out of these, the first four are less than N, that is, 1,2,3 and 4. But out of these, 1, 2 and 3 completely divide 6. That leaves us with 4. Hence, no. of factors that are less than N but do not divide N completely is **1**.

To arrive at this by a formula,

To arrive at this by a formula,

- Add 1 to the number of factors of N
^{2}No. of factors = (2+1)(2+1)= 9; by adding 1, 9+1 =10 - Divide this by 2, to get the number of pairs Number of pairs = 10/2 = 5
- From this number obtained, subtract the number of factors of N. No. of factors of N = (1+1)(1+1) = 4;
**5-4 = 1,**which is the answer.

Let us shift to the problem at hand by following the steps,

- N
^{2}= 3^{30}x 7^{86}; No. of factors = (30+1)(86+1) = 31x 87= 2697; adding 1 gives 2698 - Number of pairs= 2698/2 = 1349
- No. of factors of N= (15+1)(43+1) = 704; Answer is 1349-704 =
**645**

- This question appears across competitive exams every year. Hence, we have evolved a shortcut for the given problem. According to it, the answer to such a problem is given by the product of powers of primes of N.
- For the above problem, the answer can be directly obtained by 15 × 43 =
**645**

- Hence, to obtain the solution of x and y, we need to find the pairs whose product is 840.
- Prime factorization of 840 = 2
^{3}×3 × 5 ×7. So, the number of factors of 840 = (3+1)(1+1)(1+1)(1+1) = 32. - The number of pairs that will yield unique positive integral solutions for this equation = no. of factors /2 = 32/2 = 16. Because, for every pair, say, 4 x 210, we obtain unique solutions for x and y.

- x and y will only yield distinct integral values, provided the right hand side of the equation is either an odd number or a multiple of 4.

Suggested Action:

- Certain medium to difficult level questions of the same type are asked year after year in several competitive examinations. A thorough practice of sample questions (as given in the video) can give you a great competitive advantage for time management.