We are going to learn how to find number of factors of any natural number and various applications of number of factors. Before starting with the main theory, we should be acquainted with some basic terms like factors, factorisation, prime number and composite number.
Factors: are those which completely divide a number. For e.g. if factors of 4 are asked, we can say that 1, 2 and 4 are factors of 4, since no other number will completely divide 4.
Factors of 2: 1, 2 Factors of 3: 1, 3
Factors of 4: 1, 2, 4 Factors of 5: 1, 5
Factors of 6: 1, 2, 3, 6. Similarly we can guess the factors of other number.
One point worth noting is that 1 is the smallest factor of any natural number and the number itself is the greatest factor of that number.
Also, all the factors of a natural number are either equal or less than that particular number.
Prime number: All the numbers having exactly two distinct factors i.e. 1 and itself are called Prime numbers. 2, 3, 5, 7, 11,…. and so on are examples of prime numbers.
Composite number: All the numbers having more than 2 distinct factors are called Composite numbers. 4, 6, 8, 9, 10,…. are examples of composite numbers.
1 is neither prime nor composite.
Factorisation: Breaking a number or resolving a number into its prime factors is called Factorisation.
For example, 20 = 4 × 5. This is not the factorised form since 4 is not a prime number. So, if we want to factorize 20, we can write it as 20 = 22 × 5.
So, to find the number of factors, we need to factorize that number first. Now, we will learn how to find number of factors of a particular number.
Method of finding number of factors:
Lets say if i ask you to find number of factors of 12, we will have to factorize it first.
So, 12 = 22 × 31. If we observe carefully, we just learnt that factors are those which completely divide a number and all the factors of a number are either equal or less than the number.
So, all the factors of 12 = 22 × 31 will be the powers of 2 or of 3. Because if we take 51 or 72 or 11, they can’t divide 12, since 12 does not have a prime factor of 5, 7 or any other prime number except 2 and 3.
So, all the factors of 12 are either 20 , 21, 22, 30, 31 or a combination of power of 2 and 3. Observe that no factor of 12 will be 23 or 24 or 32 or so on. Since, those numbers will not divide 12.
So, there are 3 factors of 22 i.e. 20, 21, 22 and there are 2 factors of 31 i.e. 30, 31. One problem still persists is that how many total factors of 12. We multiply 3 factors of 22 with 2 factors of 31 or we add them.
Answer should be that we multiply them because a combination of factors of 22 and 31 will also be factors of 12. And whenever combinations are being made, we always multiply.
So, total number of factors of 12 = 22 × 31 = 3 × 2 = 6 factors. If its still not clear, think logically.
For every 2’s, we have two 3’s. So, we have in total three 2’s (20, 21, 22), so total number of factors should be 6.
We can also relate it with day-to-day life example. If i have 3 different shirts and 3 jeans, in how many ways can i wear them?
Most of you will answer in 6 ways. But think smartly. I can wear each shirt with all three jeans, so in total i have 3 shirts, so i can wear them in 9 different ways. Here also combinations are being made, that’s why we multiplied i.e. 3 × 3 = 9 ways.
So, for any number we can find the number of factors by factorizing it, then adding one to the power of all the prime numbers and then multiplying them. (We add one to the power of prime factors to account for 20 or 30 or 50 if 2, 3 and 5 are prime factors because 1 is the smallest factor of any number).
E.g. 1: Find the number of factors of 64 × 152.
Don’t be in a hurry to tell the answer as 5 × 3 = 15 factors, as the answer will be wrong. We need to factorize it first.
So, 64 × 152 = 24 × 34 × 32 × 52 = 24 × 36 × 52.
So, the number of factors will be (4 + 1) × (6 + 1) × (2 + 1) = 5 × 7 × 3 = 105 factors.
So, we can standardize the formula to find the number of factors of any natural number.
If N is a natural number having P1, P2 and P3 as its prime factors.
So, N = P1a × P2b × P3c, then number of factors of N will be (a + 1) × (b + 1) × (c + 1).
Now, we will look at some examples before proceeding to exercise.
E.g. 2: If N = 25 × 34 × 52, then
a) Find the total even factors of N.
b) Find the total odd factors of N.
c) Find total prime factors of N.
d) Find total composite factors of n.
e) Find total factors of N which are multiples of 4.
f) Find total factors of N which are multiples of 6.
g) Find total factors of N which are multiples of 8.
h) Find total factors of N which are multiples of 12.
i) Find total factors of N which are multiples of 30.
j) Find total factors of N which will end with at least one trailing zero.
k) Find total factors of N which are perfect squares.
Answers:
a) We have to find all the even factors of N. Now, we need to think how to identify an even number. Its pretty simple, absence or presence of powers of 2 makes a number odd or even. Because 2 multiplied with any number (either odd or even) always results in an even number.
For e.g., 21 × 32 × 11 is an even number, we don’t need to find the value of the number since one 2 is present in the number.
Similarly, 20 × 52 × 71 is an odd number since no 2’s are present.
So, you should have got the logic that an even factor will be that which has at least one power of 2. If it has more than one 2, then also that number will be an even factor. So, an even factor is dependent on power of 2 and not on powers of any other prime factors.
So, coming back to the original question, N = 25 × 34 × 52, to find the even factors, we just need to see how many factors of 25 will work. All the factors of 25 will work except 20, since 20 is 1 and 1 is an odd number. In total, 25 has 6 factors of which only 5 factors will work if we want even factors. All the factors of 34 and 52 will work since 2 multiplied with anything results in even.
So, answer for even factors should be = 5 × 5 × 3 = 75 factors. (Remember we are multiplying since combinations will be made here also)
If the logic is clear, we can solve these problems orally.
b) To find odd factors, just the reverse logic of even factors should be considered. No factor of 2 will work except 20, since any other factor of 2 multiplied with any factor of 34 or 52 will result in even number. All the factors of 34 or 52 will be considered since all of them are odd only.
So, answer for odd factors of N will be = 1 × 5 × 3 = 15 factors.
Alternative method: We could have found the odd factors by subtracting the even factors of N calculated in the previous question from total number of factors of N.
c) Next is to find total prime factors of N. This should be a cakewalk as it is visible that N = 25 × 34 × 52, has 3 prime factors which are 2, 3 and 5.
d) To find the composite factors of N, we just learnt what are the prime numbers and composite numbers. So using that logic, if we subtract total prime factors of N and 1 (since 1 is neither prime nor composite) from total number of factors of N, we will get composite factors of N.
Total factors of N = 25 × 34 × 52 = 6 × 5 × 3 = 90 factors.
So, composite factors of N are 90 – 3 – 1 = 86 factors.
So, whenever composite factors are to need to be calculated, we just need to subtract the total prime factors present in the number and 1 from total number of factors.
e) To find out how many factors of N are multiples of 4, we need to see what the requirement is. The factors which are multiples of 4 must have a 22 or more in that number.
Any number of the form (4 × a natural number) is a multiple of 4. For e.g., 4 × 1, 4 × 2, 4 × 3, 4 × 4, and so on. So, our restriction or limitation is on power of 2 and not on 3 and 5.
So, if N = 25 × 34 × 52, All the factors of 34 and 52 will work but only 4 factors of 25 will work. (22, 23, 24, 25 will only work as we cannot form a 4 with 20 and 21).
So, total number of factors of N which are multiples of 4 are 4 × 5 × 3 = 60
f) Factors of N which are multiple of 6 can be found using the logic of previous question. To form a 6, we need at least one 2 and at least one 3.(As 6 = 21 × 31). So, there is limitation on the powers of 2 and 3 but not on the powers of 5.
So only 5 factors of 25 and 4 factors of 34 will work and all the factors of 52 will be valid. (20 and 30 will be discarded as 6 cannot be formed with 20 × 30).
So, answer is 5 × 4 × 3 = 60 factors are multiples of 6.
g) Multiples of 8: The factors which are multiples of 8 must have minimum of 23 or more than 23. It means that only 3 factors of 25 will work and those are 23, 24 and 25. 20, 21 and 22 will be discarded since we cannot make a 8 with those. Our limitation is just on the powers of 2 and not on 3 and 5, so all the factors of 34 and 52 will work.
So, answer is 3 × 5 × 3 = 45 factors of N are multiples of 8.
h) Multiples of 12: For a number to be multiple of 12, 22 × 31 should be present in it. It means that any higher power of 2 greater than or equal to 22 will be valid and any higher power of 3 greater than or equal to 31 will work. So, 22, 23, 24 and 25, i.e. 4 factors of 25 will work and 31, 32, 33 and 34 i.e. 4 factors of 34 will work. No restriction on power of 5, so all the three factors of 52 will work.
So, answer is 4 × 4 × 3 = 48 factors of N are multiples of 12.
i) Factors which are multiples of 30: Requirement for multiples of 30 means 21 × 31 × 51 is the basic need. We should have at least one 2, at least one 3 and at least one 5. That means the factors 20, 30 and 50 will be discarded. So, 5 factors of 25, 4 factors of 34 and 2 factors of 52 will work.
So, answer is 5 × 4 × 2 = 40 factors of N are multiples of 30.
j) Requirement for the factors of N ending with at least one trailing zero is 21 × 51. It means that minimum of one 2 and one 5 is required. There is no limitation on power of 3, so all the factors of 34 will work. But only 5 factors of 25 and 2 factors of 52 will work.
So, answer is 5 × 5 × 2 = 50 factors of N will end up with at least one trailing zero.
k) To find all the factors of N which are perfect squares, we should consider each prime factor separately. N is given as 25 × 34 × 52.
We will start with 25. How many factors of 25 are perfect squares?
All the numbers whose powers/ exponents are even or multiples of 2 are termed as PERFECT SQUARES. So, if apply the logic, we can say that only three factors i.e. 20, 22, 24 are perfect squares. Same thing will be applied for the factors of 34. Only three factors of 34 i.e. 30, 32 and 34 are perfect squares. And only 2 factors of 52 i.e. 50 and 52 are perfect squares.
So, answer is 3 × 3 × 2 = 18 factors of N are perfect squares.
Exercise:
1. If N = 152 × 142 × 62, then find
a) How many factors of N are odd?
b) How many factors of N are multiple of 14?
c) How many factors of N will end up with at least one trailing zero?
d) How many factors of N are perfect cubes?
e) How many factors of N will end up with 5 at the unit’s place?
f) How many factors of N are multiples of 22 × 71 × 51?
g) How many factors of N are composite?
2. How many factors of 162 × 202 are common with 64 × 153 × 212?
3. How many factors of 35 × 152 × 42 are not common to 51 × 122?
Answer key:
1. a) 45 b) 120 c) 120 d) 4 e) 30 f) 60 g) 220 2. 15 factors 3. 90 factors