After learning how to find number of factors, now we will see some of its tricky applications. Let us start with the basic funda:
If we know a natural number in the factorised form, we can find out the number of factors by adding one to the power and multiplying them. But what if we know the number of factors, can we find the factorised form of the number?
If a number N = P1a × P2b × P3c (where P1, P2 and P3 are prime factors) has 4 factors, can we find out the factorised form of a number?
Yes, we can. We know (a + 1) × (b + 1) × (c + 1) = 4.
Now, 4 has to be written as product of natural numbers and there are fixed number of ways to be done. 4 can be written as 4 × 1 × 1 or 2 × 2 × 1. But you will argue that 1 × 4 × 1 or 1 × 2 × 2 are different ways. But all of them will lead to same factorised form.
If (a + 1) × (b + 1) × (c + 1) = 4 × 1 × 1, one of a , b, c is 3 and rest of them are zeroes and the number will be of the form P13 or P23 or P33. It doesn’t matter whether you take P1, P2 or P3. The factorised form remains the same.
If (a + 1) × (b + 1) × (c + 1) = 2 × 2 × 1, then any two of a, b, c has to be 1 and the third one will be zero. Thus the factorised form will be P11 × P21.
E.g. 1: If a natural number has 8 factors, how many possibilities of that number exist?
First of all, we have to break whatever number of factors is given as product of two natural numbers or three natural numbers or so on.
8 can be broken down as 1 × 8, 2 × 4, 2 × 2 × 2.
Now we can write the factorised form as: 1 × 8 = P17 or 2 × 4 = P11 × P23 or 2 × 2 × 2 = P11 × P21 × P31. So, the answer should be 3 possibilities of the number exist.
Note: When we are finding out number of factors, we add one to the power of the prime number and then multiply them. But when we are starting backwards, we need to subtract one to get the factorised form. Also note that we don’t normally write P10 as any prime number raised to zero will always be 1 and 1 is not a prime number.
E.g. 2: Which greatest two-digit number has 5 factors?
This is a pretty simple question. Do not worry about the two-digit number, if we just frame the factorised form, we will get the answer immediately.
If a number has 5 factors, we will split 5 as the product of two natural numbers or three natural numbers which will give us unique factorised form.
5 = 1 × 5 = P4. So, a number having 5 factors will be of the form P4.
Now, we know P is a prime number and it can take values starting from 2, 3, 5, 7, 11 onwards.
So, start plug-in values and see which one of them satisfies. 24 = 16, 34 = 81, 54 = 625. We don’t need to go any further since 74 or 114 will not be a two digit number.
So, answer is 81.
Note: Also, we should have observed that there is only one way of writing 5 as the product of two natural numbers since 5 is a prime number and all the prime numbers have only 2 distinct factors.
E.g. 3: If the number of factors of N is 13, then find the number of factors of N2.
Again if we can get the factorised form, we can get the answer orally. And once again, the number of factors is a prime number, so 13 can be written down as product of two natural numbers in just one way.
13 = 1 × 13 = P12 = N.
So, if N = P12, then N2 = P24, so number of factors of N2 will be 25.
E.g. 4: If the number of factors of N2 is 25, find the number of factors of N and N3.
Again the same type of question, but in this the factorised form will give us N2 and from there we can find the number N. After finding N, we can find N3 and its factors.
Lets find the factorised form of N2; 25 can be broken down into 1 × 25 or 5 × 5.
So, N2 = P24 or P14 × P24. We also need to check whether the factorised form satisfies the requirement for N2 as N2 is a perfect square. (All the powers of prime numbers of a perfect square must be multiple of 2 or even numbers). Both the factorised forms obtained for N2 satisfies. So, both the forms are valid.
If N2 = P24, then N = P12, so number of factors of N are 13.
If N2 = P14 × P24, then N = P12 × P22, so number of factors of N are 3 × 3 = 9 factors.
If N = P12, so N3 = (P12)3 = P36; N3 has 37 factors.
If N = P12 × P22, then N3 = (P12 × P22)3 = P16 × P26; N3 has 7 × 7 = 49 factors.
E.g. 5: If the number of factors of N are 15, find the number of factors of 16N,
a) If N is an odd number b) If no condition is given. (That means N can be odd or even)
First one can be solved very easily. Lets see how to do it. If N is an odd number, then it should immediately strike our mind that if N is an odd number, then when we write N in the form of factorised form, no prime factor will be 2. If any of the prime factors is 2, then that number becomes even.
So, if we want to find out number of factors of 16N, practically we are multiplying 16 with N. and 16 is an even number, so we can say that 16 and N are co-prime numbers. Those of you who do not know what a co-prime number is, here is is the definition.
Co-prime Numbers: A pair of number is said to be co-prime if they do not have any common factors except 1. All the odd and even numbers are co-prime because they don’t have any common factors except 1. For e.g. (4, 5), (9, 7), (16, 13) are pairs of co-prime numbers.
So, 16 and N are co-prime. Just find out number of factors of 16. 16 can be broken down into 24, so it has 5 factors.
So, total number of factors of 16N = factors of 16 multiplied with factors of N = 5 × 15 = 75 factors.
b) Second part is slightly tricky since we do not know whether N is an even number or odd number. It can be either of them, so we will have to take both the scenarios.
So, N has 15 factors. We will first find the factorised form of N, after that we will get the answer easily.
15 = 1 × 15 = P14.
15 = 3 × 5 = P12 × P24.
If we take the first factorised form i.e. N = P14, we will have to take both the scenario that N can be either odd or even. If N is even, then we can say that N = 214. Then number of factors of 16N = 24 × 214 = 218 = 19 factors. But if N is odd, then we will get the same answer as we got the answer for the first part. As N is odd, no prime factor can be 2, answer for 16N will be 75 factors.
If we take the second factorised form which is = P12 × P24, again we will have to take two conditions. Either of P1 or P2 can be 2 which will give us different answer.
If we take P1 as 2, N becomes 22 × P24, then number of factors of 16N = 24 × 22 × P24 = 26 × P24 = 7 × 5 = 35 factors.
If we take P2 as 2, N becomes 24 ×P12, then number of factors of 16N = 24 × 24 ×P12 =28 × P12 = 27 factors.
So, answer for the number of factors of 16N is 19, 75, 35 and 27. You must be wondering that why didn’t we take both the prime factors as odd because if we take the both the prime factors as odd and multiply it by 16, it becomes the same problem of finding out the number of factors of two co-prime numbers. And also we cannot take both the prime factors as even since 2 is the only even prime factor.
So, in exam you might be asked the question that we solved right now in a different way.
Which of the following cannot be the number of factors of 16N, if N has 15 factors?
a) 75 b) 30 c) 27 d) 35 e) 19
Answer is obviously option (b).
E.g. 6: How many two digit numbers have 6 factors?
This a pretty standard question. We will first find out the factorised form of the number having 6 factors and after that we will proceed.
The factorised form having 6 factors can be of two forms: 1 × 6 = P5 or 2 × 3 = P11 × P22.
If we take the first factorised form i.e. P5, now we can start plug-in values as P can take all the prime values. If P = 2, 25 = 32, it is valid as 32 is a two-digit number. If we take P as 3, we get 35 =243 which is not a two-digit number. If 35 is a three-digit number, then fifth power of all the remaining prime numbers will be greater than 100. Only one possibility is there which is 32.
If we take the second factorised form which is P11 × P22, we can get many possibilities. If we start plug-in values randomly, we will definitely miss one or two of numbers. So, we need to follow a simple algorithm which will give us the required answer.
Factorised form is P11 × P22. The logic is to give the highest power to smallest prime number and fix it. During this process, keep changing the remaining prime number until we get two-digit numbers.
So, we will start with 2 and 3, we will fix 2.
22 × 31 = 12, 22 × 51 = 20, 22 × 71 = 28, 22 × 111 = 44, 22 × 131 = 52, 22 × 171 = 68, 22 × 191 = 76, 22 × 231 = 92. If we take next prime number which is 29, we will exceed two digit number. So, all the possibilities of 2 is exhausted. So, now change it to 3 and fix it until we keep getting two-digit numbers.
32 × 21 = 18, 32 × 51 = 45, 32 × 71 = 63, 32 × 111 = 99. If we take the next prime number, we will get three-digit number. So, now change it to 5.
52 × 21 = 50, 52 × 31 = 75. If we take the next prime number, we will get three digit number. So, now change it to 7.
72 × 21 = 98. After this we will not get any two-digit number. Even if we change it to 11, 112 is only greater than 100, so we don’t need to go any further. All the possibilities are exhausted. So, just count all of them.
Answer: 16 two-digit numbers have 6 factors.
You must be thinking that solving this question will take 5-10 minutes. But, its not like that. If the fundamental logic is clear, you will hardly take 1 minute to solve it correctly using the logarithm.
E.g. 7: How many two digit numbers have 15 factors?
Again it’s the same type of question; we will use the same logic. We will first find the factorised form after that we will see.
15 = 1 × 15 = P14 and 3 × 5 = P12 × P24. Only two possibilities are there and if we take the first factorised form which is P14 and start substituting values starting from 2, we will get 214 which is a very large number which is greater than 100. So, we can say that we would not get any two-digit number from the first factorised form.
If we take the second factorised form and give the highest power to smallest prime number 2, we will get 24 × 32 = 16 × 9 = 144. This number only is greater than 100, we don’t need to go any further.
So, answer is no such number exists.
E.g. 8: If a number N is represented as 2a × 3b. It is also given that N has 10 factors, 2N has 12 factors, 3N has 15 factors, so find the number of factors of 12N.
This is also a standard question and can be solved pretty easily.
If N = 2a × 3b,
Then 2N = 2a+1 × 3b,
And 3N = 2a × 3b+1,
Now we can equate the ratio of number of factors of N and 2N with the given number of factors of N and 2N in the question.
So, after finding out the values of a and b, we can substitute their values in N.
So, N becomes 24 × 31.
So, 12N = 22 × 24 × 31 × 31 = 26 × 32. So, number of factors of 12N is 7 × 3 = 21 factors.
Now, to practice whatever you have learnt, solve the exercise.
Exercise:
1. Which greatest two digit number has 6 factors?
2. How many two-digit numbers have 8 factors?
3. Which greatest two-digit number has 4 factors?
4. If N has 15 factors, find the number of factors of 16N.
5. If P and Q are two co-prime numbers having 10 and 12 factors respectively, find the number of factors of P × Q.
6. If A2 has 35 factors, then find the number of factors of A and A3 respectively.
7. How many three-digit numbers have 7 factors?
8. If N has 20 factors, 2N has 24 factors and 3N has 30 factors, then find out the number of factors of 36N.
9. If the number A3 has 16 factors, find the number of factors of A2.
10. Which greatest three-digit number has 10 factors?
Answer Key:
1. 99 2. 10 3. 95 4. 19 or 75 or 27 or 35 factors
5. 120 6. A has 12 or 18 factors while A3 has 70 or 52 factors.
7. one number(729) 8. 56 factors 9. A2 has 9 or 11 factors. 10. 976
Explanations:
Question 1:
There are two ways of dealing with this question. First one is using hit or trial method.
We start from the largest two-digit number which is 99 and find out its number of factors. And Bingo, we got the number which satisfies our need.
Theoretical method: Any number having 6 factors can be represented in the form of either P5 or P12 × P21, where P, P1, P2 are prime factors.
If we take the first case, number of the form P5, we start substituting values of prime numbers starting from 2 onwards. If we take 2, number which we get is 25 = 32, if we take 3, we get 35 = 243. 243 is not a two-digit number, so there is no point in plug-in values in the first case as prime number greater than 3 raised to fifth power will be greater than 100.
So, the greatest two-digit number which we got from the first case was 32. Now, let us see the second case.
The number is of the format P12 × P21, now if we want to get greatest two digit number then we should start assuming values of prime factors and see which one of them reaches close to 90. If we take P1 as 2 and P2 as 11, we will get the number = 22 × 111 = 44, if we take prime numbers as 3 and 11, we will get number = 32 ×111 = 99. So, that should be our answer.
Question 2:
Two-digit numbers having 8 factors: First of all we will frame the possibilities of the number having 8 factors.
First case: 1 × 8 : P7
Second case: 4 × 2 : P13 × P21
Third case: 2 × 2 × 2: P11 ×P21 × P31.
If we take the first case: We start plug –in –values for prime numbers starting from 2 onwards. When we plug 2, we get 27, which is greater than 100. So, we don’t need to go ahead as no number will satisfy the condition given in the question.
If we take the second case which is P13 × P21, we will start with prime numbers 2 and 3. We will follow a simple algorithm to find out how many two-digit numbers will satisfy this format. If we do it randomly, we will surely miss out on some of the numbers.
First is 23 × 31 = 24; 23 × 51 = 40; 23 × 71 = 56; 23 × 111 = 88; 33 × 21 = 54. After this all the number will be of three-digit. So, we got 5 two-digit numbers.
Third case: P11 ×P21 × P31, now start substituting prime number values in this format.
2 × 3 × 5 = 30; 2 × 3 × 7 = 42; 2 × 3 × 11 = 66; 2 × 3 × 13 = 78; 2 × 5 × 7 = 70. After this no number will satisfy the required condition.
So, in total 10 two-digit numbers satisfy the condition of having 8 factors.
Question 3:
Greatest two-digit number having 4 factors can also be solved using hit-and-trial as we did in question 1. We can start from 99 and keep calculating factors of each number. We will get the answer as 95.
Alternative method is to find the format of number and use common sense logic to reach the answer. The format of number having 4 factors is either P3 or P11 × P21.
If we take the first case, greatest two digit number which we will get is 33 i.e. 27 since 53 = 125 which is a three-digit number.
If we take the second case, we should understand that number will be product of two prime numbers. If we take P1 as 2, then take another prime number which multiplied with 2 results in a number close to 99. So, that second prime number can be 47 and then (P11 × P21) will be 94. After reaching 94, we just need to check remaining five numbers (i.e. 95, 96, 97, 98 and 99). We just need to verify whether any one of these 5 numbers is product of two prime numbers or not.
Only 95 is the product of two prime numbers that is of 5 and 19. 96, 98 and 99 are not product of two prime numbers while 97 is a prime number.
So, answer should be 95.
Question 4:
If N has 15 factors, format of the number N can be either P14 or P14 × P22.
If we take first case which is P14, there can be two possibilities. N can be an odd number or an even number as nothing has been stated in the question.
If we take N as an even number, then N takes the value 214. Then number of factors of 16N will be 24 × 214 = 218 is 19 factors. And if N is odd, then P can take any odd prime number. Then number of factors of 16N = 24 × P14 = 5 × 15 = 75 factors.
If we take the second case which is P14 × P22, then again there can be two sub-cases. First sub-case can be when P1 is 2 and second sub-case when P2 is 2.
If P1 is 2, then other prime factor P2 can be any odd prime factor because both prime factors cannot be even as there is only one even prime number. So, N = 24 × P22, then 16N = 24 × 24 × P22 = 28 × P22 = 27 factors.
If P2 is 2, then other prime factor P1 can be any odd prime factor. So, N = 22 × P14, then 16N = 24 × 22 × P14 = 26 × P14 = 35 factors.
So, there are four possibilities of factors of 16N if N has 15 factors. 16N can have 19, 75, 27 and 35 factors.
Question 5:
We have already dealt with such question earlier. The number of factors of (P × Q) will be 10 × 12 = 120 since P and Q are co-prime numbers.
Question 6:
A2 has 35 factors, then first of all we will frame the number A2 first.
A2 can be either P34 or P16 × P24.
If we take the first case: A2 = P34, then A = P17 which gives18 factors of A.
If we take the second case: A2 = P16 × P24, then A = P13 × P22 which gives 12 factors of A.
If we found out the format of number A, we can find out number of factors of A3.
If A = P17, then A3 = P51, which gives 52 factors of A3.
If A = P13 × P22, then A3 = P19 × P26, which gives 70 factors of A3.
So, finally A can have 18 or 12 factors. And A3 can have 52 factors or 70 factors.
Question 7:
Numbers having 7 factors will be of the form P6. Now, since we want three-digit numbers, we will start substituting values of P = 2, 3, 5, 7 and so on.
If P =2, we will get 26 = 64 which is not a three-digit number.
If P =3, we will get 36 = 729.
If P = 5, we will get 56 = a five-digit number which does not satisfy our requirement. We don’t need to go any further since on taking values for P greater than 5 will result into large numbers.
Question 8:
We have discussed the logic of this type of problem earlier. We will start off with assuming N = 2a × 3b × Pc.
We have assumed two prime factors as 2 and 3 because when we multiply N by 2 and 3 according to the question, the extra 2 and extra 3 will go and add up to the power of 2 and 3 in N. So, it is just manipulation of data. If the question had data related with 5N and 7N, we would have definitely taken two prime factors as 5 and 7. So, we are just trying to assume such values of N which makes our job easy.
If N = 2a × 3b × Pc, (where P can any prime factor except 2 & 3).
Then 2N = 2a + 1 × 3b × Pc;
And 3N = 2a × 3b + 1× Pc;
Now, we can equate the ratio of number of factors of N and 2N with the number of factors of N and 2N given in the question.
If we know the value of (a, b), we can plug their value in N.
Now, N becomes 24 × 31 × Pc; Now, we can also calculate value of ‘c’. As number of factors of N are [5 × 2 × (c + 1)] = 20 or (c + 1) = 2, which gives the value of ‘c’ as 1.
Now, N becomes 24 × 31 × P1. We need to find factors of 36N.
36N = 22 × 32 × 24 × 31 × P1 = 26 × 33 × P1 = 7 × 4 × 2 = 56 factors.
Question 9:
If the number of factors of A3 is given, we can frame the possibility of format of A3. And after finding A3, we can find format of A. Then we can find the format of A2.
The different formats of A3 can be P15 or P11 × P27 or P13 × P23 or P11 × P23 × P31 or P11 × P21 × P31 × P41.
Among these formats, only P15 or P13 × P23 are applicable since A3 is a perfect cube and the requirement for a number to be perfect cube is that power of all the prime factors must be multiple of 3. Other formats are rejected because of that reason.
Now, we can solve for both the possibilities of A3.
If A3 = P15, then A = P5. If A = P5, then A2 = P10 which results in 11 factors of A2.
If A3 = P13 × P23, then A = P11 × P21, so A2 = P12 × P22 which results in 9 factors of A2.
Question 10:
The number having 10 factors will be of the form of P9 or P14 × P21.
If we take the first case which is P9 and start substituting values of prime numbers. We will get 29 = 512, 39 = a five-digit number. So, we got just one possibility of a three-digit number with the first case.
If we take the second case which is P14 × P21, we can get various numbers. But, we want the greatest 3-digit number in the vicinity of 999.
We cannot take P1 as prime number greater than 3 because if we take P1 as 5, then P14 = 625 and multiplied with P2 (let’s say P2 = 2) will result into a 4-digit number.
So, P1 can take either 2 or 3. Let us start with P1 as 2. Then 24 × (a prime number) = close to 999. 16 × 611 = 976. If we take a prime number greater than 61 and multiply it with 16, we will get 4-digit number.
If we take P1 as 3, then 34 × (a prime number) = close to 999. 81 × 111 = 891. If we take 81 × 13, we will get a 4-digit number.
So, the greatest 3-digit number having 10 factors is 976.