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Factors of a Number

The concept of factors of a number and its application in the questions is very important concept while preparing number system for the CAT examination. Factors of a number N are the numbers which divide N. So, if we talk about a number say, 12, then the factors of 12 are those numbers which will divide 12 completely. Now remember that 1 and the number itself are the trivial factors of every number. Therefore, when we talk about the numbers which divide 12, then 1 and 12 will definitely be those numbers. Now, what are the other numbers which can divide 12 completely?

By simply thinking about the numbers, we can find that the numbers which can divide 12 are 1, 2, 3, 4, 6 and 12. These numbers are called the factors of 12.

Now what if the number is a larger one? Do we need to calculate the number of factors manually like above?

Well, the answer is no. We have the formula to find the number of factors of a number. Let us now discuss it in detail.

Let us take a number say 108. Write down the prime factorisation of 108.

So, 108 = 33 × 22.

Now, if 33 is a factor of 108, then 30, 31, 32 will also be the factors of 108. Similarly, as 22, is a factor of 108, therefore, 20 and 21 will also be the factors of 108. So, the combinations of these factors will also be the factors of 108. The different combinations are 20 × 30, 20 × 3, 20 × 32, 20 × 33, 2 × 30, 2 × 3, 2 × 32, 2 × 33, 22 × 30, 22 × 31, 22 × 32, 22 × 33 or the factors are 1, 3, 9, 27, 2, 6, 18, 54, 4, 12, 36, 108. Therefore, 108 has 12 different factors.

Now, we can think about this in simple way also.

We have 22 ⇒ 20, 21, 22 = 3 factors

And 33 ⇒ 30, 31, 32, 33 = 4 factors.

Therefore, the total factors will be 4 × 3 = 12.

Let us generalize this result.

Let the number be N. First of all, write the prime factorisation of N.

N = pa × qb × rc × ……. where p, q, r….. are different prime numbers.

The number of factors of N will be (a + 1) × (b + 1) × (c + 1) × ……..

Let us take another number say 144. We have 144 = 24 × 32.

The number of factors of 144 = (4 + 1) × (2 + 1) = 5 × 3 = 15.

Ex. Find the number of factors of 300.

Sol: Here 300 = 22 × 3 × 52.

So, the number of factors of 300 = (2 + 1) × (1 + 1) × (2 + 1) = 18.

Sum and product of factors of a number N: Let the number be N. Prime factorize the number as done above.

Let N = pa × qb × rc × ……. where p, q, r, ….. are different prime numbers.

The sum of all factors of N = [(pa+1 – 1)/(p -1)] × [(qb+1– 1)/(q -1)] × [(rc+1 – 1)/(r -1)] × ….

The product of factors of number N = N(Total factors of N)/2

Ex. Find the sum and product of factors of 200.

Sol: We have 100 = 23 × 52

Here the number of factors = (3 + 1) × (2 + 1) = 12

Sum of factors = [(23+1 – 1)/(2 -1)] × [(52+1
– 1)/(5 -1)] = 15 × 124/4 = 465

Product of factors = 10012/2 = 1006.

Let us discuss the following practice questions:

1. Let us consider the number N = 23 × 34 × 53. Find the

(a) Total number of factors of N

(b) Number of factors of N which are perfect square

(c) Number of factors of N which are perfect cube

(d) Number of factors of N which are even

(e) Number of factors of N which are odd

(f) Number of factors of N which are multiples of 10

Sol:

(a) Here N = 23 × 34 × 53

The total number of factors = (3 + 1) × (4 + 1) × (3 + 1) = 4 × 5 × 4 = 80.

(b) Here N = 23 × 34 × 53.

The numbers which are perfect squares will have even powers of prime numbers. Here 23 is a factor of N. The perfect squares that we can have out of 23 are 20, 22. Similarly, the perfect squares that we can have out of 34 are 30, 32 and 34 and out of 53 are 50 and 52.

We can write it as 23 ⇒ 20, 22 = 2 perfect squares

33 ⇒ 30, 32 and 34 = 3 perfect squares

53 ⇒ 50, 52 = 2 perfect squares

Therefore, total number of perfect square factors = 2 × 3 × 2 = 12.

(c) Here N = 23 × 34 × 53

The numbers which are perfect cubes will have prime numbers raised to powers which are multiples of 3.

From 23, we can have 20 and 23 as perfect cubes i.e., 2 in number.

From 33, we can have 30 and 33 as perfect cubes i.e., 2 in number.

From 53, we can have 50 and 53 as perfect cubes i.e., 2 in number.

Therefore, total number of factors which are perfect cubes = 2 × 2 × 2 = 8.

(d) In order to get the factors which are even, we can write N = 2 (22 × 34 × 53)

Now, we have to find the total number of factors of the number which are inside the bracket as these factors when are multiplied by 2 outside will become even.

The number of factors of the number inside the bracket i.e., 22 × 34 × 53 are 3 × 5 × 4 = 60.

Therefore, the number N has 60 factors which are even.

(e) We know that the total number of factors of N are 80 out of which 60 are even.

Therefore, the number of odd factors = 80 – 60 = 20.

Or, we can also say that in order to get the number of odd factors, we should drop all the 2s. Now, if we drop all the 2s from the number N, it will become 34 × 53. The number of factors of this number are (4 + 1) × (3 + 1) = 5 × 4 = 20.

(f) We have, N = 23 × 34 × 53

In order to get the number of factors which are multiples of 10, let us rewrite the N as

N = 2 × 5 × (22 × 34 × 52)

Now as 10 is multiplied with the number inside the bracket, so the factors of the number inside the bracket will be multiple of 10. So, we have to find the number of factors of 22 × 34 × 52 which are 3 × 5 × 3 = 45. Therefore, there are 45 factors of N which are multiples of 10.

2. Find the sum of the factors of 160.

Sol: We have 160 = 25 × 5

Sum of factors = [(26 – 1)/(2 -1)] × [(52 – 1)/(5 -1)] = 63 × 6 = 378

3. Find the number of factors of 777777.

Sol: Here 777777 = 777 × 1001 = 7 × 111 × 1001 = 7 × 37 × 3 × 7 × 11 × 13

⇒ 777777 = 3 × 72 × 11 × 13 × 37

The number of factors of 777777 are 2 × 3 × 2 × 2 × 2 = 48.

4. How many numbers from 1 to 50 have exactly 4 factors?

Sol: If a number has exactly four factors, then it will be either of the form a3, where ‘a’ is prime number or of the form a × b, where ‘a’ and ‘b’ are prime numbers.

 Case I: When the number is of the form a3, where ‘a’ is a prime number.

In this case the values that ‘a’ can take are 2 and 3 only as a = 5 will give a3 = 53 = 125 which is more than 50. Hence two numbers 23 = 8 and 33 = 27 are possible.

Case II: When the number is of the form a × b, where ‘a’ and ‘b’ are prime numbers.

Now if we take a = 2, then the possible values of b are 3, 5, 7, 11, 13, 17, 19 and 23. Therefore, 8 numbers are possible in this case.

If we take a = 3, then the possible values of b are 5, 7, 11 and 13. Therefore, 4 numbers are possible in this case.

If we take a = 5, then the possible value of b is 7. Therefore, only one number is possible in this case.

If we take a = 7, then no value of b is possible as even with b = 11, the product will become 7 × 11 = 77 which is more than 50.

Therefore, the total numbers from 1 to 50 which have exactly 4 factors = 2 + 8 + 4 + 1 = 15.

5. Find the sum of all the odd factors of 4500.

Sol: We have 4500 = 22 × 32 × 53

Now in order to get the sum of odd factors, we have to drop all the 2s. So, the answer will be obtained by just finding the sum of factors of 32 × 53 = [(33 – 1)/(3 -1)] × [(54 – 1)/(5 -1)] = 13 × 156 = 2028.