Placement of factors:
We will learn this theory through examples. Let us take factors of 3-4 different numbers.
Factors of 12: 1, 2, 3, 4, 6, 12 (6 factors).
Factors of 35: 1, 5, 7, 35 (4 factors).
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 (9 factors).
Factors of 45: 1, 3, 5, 9, 15, 45 (6 factors).
After observing these, we should get an idea that number of factors can be either odd or even.
When number of factors is even, where will the centre position be if all the factors are written in ascending order?
Answer should be that centre position will always lie between two factors. For e.g. if there are 4 factors, centre position will always be between 2nd and 3rd factor. If a number has 8 factors, centre position will always be between 4th and 5th factor. But what if the number of factors is odd, where will the centre position lie? Answer should be exactly at one of the factors. We already saw 36 has odd number of factors and centre will lie exactly at 6 which is at 5th position. If a number has 19 factors, the centre position will be 10th factor.
By observing these points, we can ascertain that a perfect square which has odd number of factors, the centre position will lie exactly at the square root of that perfect square.
For e.g. : 100 = 22 × 52 = 9 factors.
Factors of 100 : 1, 2, 4, 5, 10, 20, 25, 50, 100. Since 100 has 9 factors, the centre position will be 5th position and it is occupied by 10 which happens to be the square root of 100.
So, now we can conclude one point:
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 144 : 1, 2,3, 4, 6, 8,9, 12, 16, 18, 24, 36,48, 72, 144.
If we denote 36 by N, then 6 will be denoted by √N. Centre position will lie at √N. There are three factors between 1 and √N, and 3 factors are also present between √N and N.
In case of 144, it has 15 factors. If we denote 144 by N, then 12 will be denoted by √N. There are 6 factors between 1 and √N, and 6 factors are also present between √N and N.
In case of perfect square, this logic is working fine. But, what if the number is not a perfect square, will this logic hold true?
Factors of 12: 1, 2, 3, 4, 6, 12 (6 factors).
Factors of 24: 1,2, 3, 4, 6, 8, 12, 24 (8 factors).
If we take 12 as N, √N will be lie between 3 and 4 which happens to be 3rd and 4th factor of 12. There are two factors between 1 and √N, and 2 factors also lie between √N and N.
If we 24 as N which has 8 factors, the centre will lie between 4th and 5th factor which happens to be 4 and 6 in the increasing order sequence. There are 3 factors between 1 and √N, and there are exactly 3 factors between √N and N. (√N = √24 = 4.4 approx.)
After observing all these examples, we can conclude that the logic is applicable for all natural numbers. We just have to be careful in case of perfect squares. In case of perfect squares, square-root of that number is also a part of factors, while in case of number not being a perfect square, square-root of that number is not a part of factors.
Note: If there are ‘x’ number of factors between 1 and √N, then there will exist ‘x’ number of factors between √N and N, for N being a natural number.
E.g. 1: If there are 20 factors between 1 and √N, find the total number of factors of N. If it is given that N is a natural number but √N is not a natural number.
On reading √N is not a natural number, it should strike our mind that N is not a perfect square. Had N been a perfect square, √N must have been a natural number. Just use the logic learnt and get the answer. If there are 20 factors between 1 and √N, there would be exactly 20 factors between √N and N. We also know that 1 and the number itself are factors of any natural number. So, total number of factors of N will be = 20 + 20 + 2 = 42 factors.
What would have been the number the number of factors of N had N been a perfect square? Answer will be 43 as now √N would also have been part of the group of factors.
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But why the same number of factors exist between 1 and √N, and between √N and N?
Answer to this question is that factors of all natural numbers exist in pairs.
Look at this example, you will get the picture. 12 can be written as product of two natural numbers in following way: 12 = 1 × 12, 2 × 6, 3 × 4.
These are the three ways in which 12 can be written as product of two natural numbers. But what are 1, 12, 3, 4, 2 and 6. They are factors of 12. And why 1 × 12 gives us 12. Because (1, 12) are factors of 12 which are situated at equal distance from the centre when all the factors of 12 are written in ascending order.
Factors of 12: 1 2 3 4 6 12.
Centre will lie between 3 and 4. If we take one factor to the left of the centre and one factor to the right of centre, we will get factors 3 and 4. When we multiply them we will get 12. Similarly when we move 2 places to the left of the centre and 2 places to the right, we get factors 2 and 6, when we multiply them we get 12. And finally when we move 3 places to the left of the centre and 3 places to the right of centre, we get 1 and 12. When we multiply them, we will again get 12. This pattern will be found in all the natural numbers.
Now, you should understand the logic of why the factors exist in pairs. Because a factor completely divides a number, so when that factor divides a number, we will get a quotient and not a remainder. And that quotient will also be a factor of that number. Those two factors will be located equidistant from the centre.
But in case of a perfect square, we have already seen that square root of perfect square is also a factor, so that square root multiplied with itself will also result in original number.
Factors of 36: 1 2 3 4 6 9 12 18 36.
This will hold true for all the perfect squares. 12 × 12 = 144, 7 × 7 = 49, 10 × 10 = 100.
Note: Factors of any number which are equidistant from the centre, when multiplied with each other will always result in that particular number.
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All the perfect squares have odd factors and if a number has odd factors then it has to be a perfect square.
Why all the perfect squares have odd number of factors?
Powers of prime factors in a perfect square is even, and when we are finding out number of factors, we add 1 to the power and then multiply. 1 added to an even number results in an odd number and when multiplied with another odd number will remain an odd number only.
For e.g. N = P12 × P24 is a perfect square. (P1 and P2 are prime numbers). When we are finding out number of factors of N, we will add 1 to 2 and 4, then multiply them. We will always get an odd number as two odd numbers are being multiplied.
E.g. 2: If the product of factors of a natural number N at 12th position and 25th position results in N, then find the total number of factors of N.
The question is based on the logic of relative placement of factors. If factor at 12th position multiplied with factor at 25th position results in N, then factor at 11th position multiplied with 26th position will also result in N. That should be sufficient to find the answer. There would be 11 factors on the left hand side of 12th factors, and since factors exist in pairs and factors equidistant from centre always result in number. There would be 11 factors on the right of factor at 25th position, so N has total 36 factors. Once you got hold of logic, these problems can be solved orally.
We can observe one more thing. Add up the positions and see, you will get a pattern.
12th position + 25th position = 37. (While multiplying factors at 12th & 25th position gives us N)
11th position + 26th position = 37. (While multiplying factors at 11th & 26th position gives us N)
10th position + 27th position = 37. (While multiplying factors at 10th & 27th position gives us N)
9th position + 28th position = 37. And so on.(While multiplying factors at 9th & 28th position gives us N)
All the factors which are at these position when multiplied with each other will result in original number and the sum will always be constant which is 1 greater than total number of factors.
So, we could have solved the above problem by just adding 12 + 25 = 37 and subtracting 1 from it giving us 36 total factors.
E.g. 3: Which factor will occupy the 68th position if all the factors of 62 × 52 × 142 are written in ascending order?
We just need to find the number of factors of given number and then we can think about applying logic of placement of factors.
62 × 52 × 142 = 22 × 32 × 52 × 22 × 72 = 24 × 32 × 52 × 72 = 5 × 3 × 3 × 3 = 135 factors. If the number of factors of a number is odd, then that number has to be a perfect square. It means that the centre position when all the factors are written in ascending order will be (135 + 1)/ 2 = 68th position. That is what the question is asking. And we know, in case of perfect square, centre position always belongs to the square-root of that number.
So, answer is 6 × 5 × 14 = 420.
E.g. 4: If all the factors of 800 are written in ascending order, find which factor will occupy the 13th position?
First of all, we need to find how many factors do 800 has? 800 = 25 × 52 = 6 × 3 = 18 factors.
Now, we can use the placement of factors learnt earlier.
Product of factor at 1st position with factor at 18th position will give us 800. (Sum of positions = 1 + 18 = 19)
Product of factor at 2nd position with factor at 17th position will give us 800. (Sum of positions = 2 + 17 = 19)
And so on.
So, we can say that sum of positions will always be one more than number of factors of 800.
Which position added with 13th position = 19, answer should be factor at 6th position.
So, product of factor at 6th position with factor at 13th position will give us 800.
Now we can find factor at 6th position of 800 easily by just checking which numbers starting from 1 will divide 800. 1, 2, 4, 5, 8, 10 and others are factors of 800 in ascending order. So, 10 is at 6th position.
Now, 10 × ? = 800, answer = 80 is the factor which will occupy the 13th position.
E.g. 5: Find the product of all the factors of 240.
This sum is also an application of placement of factors. Just find out the number of factors of 240, you will get the answer instantly.
240 = 24 × 10 = 23 × 31 × 21 × 51 = 24 × 31 × 51 = 5 × 2 × 2 = 20 factors. If 240 has 20 factors, then 10 pairs of factors will exist which when multiplied with each other will give 240.
So, answer should be 24010 or (24 × 31 × 51)10.
[1 × 240 = 240, 2 × 120 = 240, 3 × 80 = 240 and so on]. 1, 2, 3, 240, 120 and 80 are factors of 240, when these factors are multiplied with factors placed equidistant from the centre they always result in original number.
To write a natural number as a product of two integers or natural numbers:
We will learn this logic through the help of examples which is directly related to number of factors.
E.g. 6: Find the number of ways in which 24 can be written as the product of two natural numbers.
We have already seen that any natural number can be written as the product of two factors which are equidistant from the centre.
24 = 1 × 24 or 2 × 12 or 3 × 8 or 4 × 6. But one point that is worth noting that we would not count 24 × 1 as a different way. 1 × 24 is same as 24 × 1, so, answer is 4 ways.
So, the logic is to find the number of factors of the number asked in question, in this case number is 24. Number of factors is 24 is 8, we have to write 24 as a product of two natural numbers. So, just divide 8 by 2 as only 4 pairs can be formed.
E.g. 7: Find the number of solutions to a × b = 24, given that a and b are natural numbers.
You must be thinking that both the questions are same and answer still should be same. No, there is difference between number of ways and solutions.
Logic is still the same, but there is just one point that should be kept in mind. We will find out factors of 24 which is 8. We can write 24 as product of two natural numbers in 4 ways. But now each way will give us two distinct solutions.
For e.g., if a takes values 1, then b takes the value 24. But it can be also other way around. Now, a takes 24 while b takes the value of 1 and both of them are distinct solutions. So, every way gives us 2 solutions, and we had total of4 ways, so answer should be 4 × 2 = 8 solutions.
So, in case of finding solutions, after finding out the ways, we will have to multiply it by 2.
E.g. 8: Find the number of ways in which 6! can be written as the product of two integers.
This sum is also the same, just we have to take care of integers. In case of integers, we will also have to take the negative numbers.
First of all, we need to factorise 6! and find out its number of factors. (6! can be factorised easily using the highest power of prime learnt earlier). 6! = 24 × 32 × 51 = 5 × 3 × 2 = 30 factors.
If 6! has 30 factors, then we can write 6! as the product of two natural numbers in 15 ways as 15 pairs will be formed. But the question is asking about the product of two integers.
It means that every pair will now give us two ways. 1 × 720 is the first way, but we can also write 720 as( – 1 × – 720). (Value of 6! is 720).
So, final answer is 15 × 2 ways = 30 ways.
E.g. 8: Find the number of solutions in which 120 can be written as the product of two integers.
Again the same type of question, only difference is instead of natural numbers we have to find solutions for integers. We factorise 120 and find out number of factors. 120 = 15 × 8 = 23 × 31 × 51 = 4 × 2 × 2 = 16 factors. If a number has 16 factors, that number can be written as the product of two natural numbers in 8 ways. But, now each way will give us 4 integral solutions, so answer is 32 solutions.
If a = 1, b = 120 (1st solution).
If a = 120, b = 1 (2nd solution).
If a = –1, b = –120 (3rd solution).
If a = –120, b = –1 (4th solution).
Similarly, other ways will also give 4 solutions each.
Until now, we have not dealt with perfect squares. Will the logic remain same for perfect squares?
E.g. 9: In how many ways can 144 be written as product of two natural numbers?
144 is a perfect square, so number of factors of 144 must be an odd number. 144 = 24 × 32 = 15 factors. In case of odd number of factors, we cannot make exact number of pairs, but we know in case of a perfect square, the factor which comes at the centre multiplies with itself to give the original number itself. In this case 12 will be at centre position and 12 × 12 results in 144.
So, to avoid confusion, best way in finding out number of ways for a perfect square is to ignore the square root of number initially and deal with it later on. 144 has 15 factors, so ignore 1 factor (i.e. 12 in this case), now remaining factors are 14. If a number has 14 factors, number of ways will be 7 and one factor ignored will give us one more way.
Total number of ways = 7 + 1 = 8 ways.
E.g. 10: In how many ways can 100 be written as the product of two integers?
Again 100 is a perfect square, but only difference in this question to the previous question is the product of two integers and not natural numbers. 100 = 22 × 52 = 9 factors.
Again ignore 1 factor (i.e. 10 in this case), now remaining factors are 8. We can write these 8 factors in 4 ways. But every way will give us two distinct ways. And one factor ignored will also give us two ways. (10 × 10 and –10 × –10).
So, total number of ways is 4 × 2 + 2 = 10 ways.
E.g. 11: Find the number of solutions for a × b = 144, where a and b are integers.
Number of factors of 144 = 15. Again, we ignore one factor, then remaining 14 factors can be written as product of two natural numbers in 7 ways. But since question is asking for integral solutions, every way will give us 4 integral solutions. (a = 1, b = 144),(a = 144, b = 1), (a = –1, b = –144), (a = –144, b = –1). So, 7 ways will result in 28 integral solutions and one factor ignored will give us just 2 integral solutions. (i.e. a = 12, b = 12 and a = –12, b = –12).
This is the error-prone area, so be careful here.
So, answer is 30 integral solutions.
E.g. 12: Find the number of solutions to a × b = 196, where a and b are distinct integers.
Same type of question, only difference is that in this case question is asking about product of distinct integers. 196 = 142 = 22 × 72 = 3 × 3 = 9 factors.
In case of distinct integers, the factor which occupies the centre position will be ruled out as factor at centre position multiplied with itself results in the number. So, 14 is ruled out (as 14 × 14 and –14 × –14 are not distinct). Remaining factors are 8 and they can be written as product of two natural numbers in 4 ways. And every way will give us 4 integral solutions. So, answer is 16 integral solutions.
Writing a natural number as difference of squares:
(i) When the natural number is odd:
E.g. 13: In how many ways can 45 be written as the difference of squares of two natural numbers?
We can frame the equation as x2 – y2 = 45, where x and y are natural numbers. The equation can be further broken down into (x + y) (x – y) = 45, where (x + y) will always be greater than (x – y), since x and y are natural numbers.
Now, (x + y) (x – y) = 45; = 1 × 45; = 3 × 15; = 5 × 9.
(Since, we are dealing with number of ways, we would not go any further since we will get the same ways).
Now since, (x + y) is greater than (x – y), (x + y) will always correspond to the larger number which is 45 and (x – y) will correspond to the smaller number which is 1.
If we solve x + y = 45 and x – y =1, will we get natural number solutions for x and y. Answer is yes, we will get x = 23 and y = 22 which are natural numbers.
If we take x + y = 15 and x – y = 3, on solving we get x = 9 and y = 6.
And finally when we take x + y = 9 and x – y = 5, on solving we get x = 7 and y = 2.
So, all the three ways are working since we are getting natural number values for x and y. Now, you should think why all the factors are working?
Because 45 is an odd number and all the factors of an ODD NUMBER are always odd and when we add up two odd numbers, we always get an even number and that even number divided by two always results in a natural number.
So, in case of any odd number, all the ways will give us answer. So, we just need to find out number of factors of that odd number and divide it by 2 to get the required number of ways.
So, we could have just find out the number of factors of 45 and divided it by 2. Number of factors of 45 = 32 × 51 = 6 factors that leads to 3 ways.
E.g. 14: In how many ways can 315 be written as the difference of squares of two natural numbers?
Again 315 is an odd number, so we will find out number of factors of 315 = 32 × 51 × 71 = 12 factors.
If 315 has 12 factors, that means we can write 315 as difference of squares of two natural numbers in 6 ways.
(ii) When the natural number is even:
E.g. 15: In how many ways can 60 be written as the difference of squares of two natural numbers?
Question is asking x2 – y2 = 60 in how many ways.
(x + y) (x – y) = 1 × 60; = 2 × 30; = 3 × 20; = 4 × 15; = 5 × 12; = 6 × 10.
First way, (x + y) = 60 and (x – y) = 1 is not valid because when we solve them simultaneously, values of x and y will not be natural numbers.
But second way, (x + y) = 30 and (x – y) = 2 is valid since summation of 30 and 2 is even and on solving we will get x = 16 and y = 14. We just need to check whether the sum is even or odd. If the sum is even, then even number divided by 2 will result in a natural number. But, if the sum is odd, then odd number divided by 2 will result in a fractional number.
So, we just need to check which pair gives us even sum, those pairs will only result in natural numbers. (1 + 60), (3 + 20), (4 + 15) and (5 + 12) give us odd sums, so these pairs will not work. Only two pairs (2 + 30) and (6 + 10) give us even sums, only these pairs will give us required solutions.
So, answer is 2 ways.
60 was a small even number, so we could factorise it and write it as a product of factors. What if the number is large even number? For e.g. let’s say 960. Will we write all ways? Don’t you think it will be a cumbersome process? Then what should be the efficient way of solving.
Just observe in case of 60 which all ways are working. Only (2, 30) and (6,10) give us the required solutions, and both the numbers in two ways are even. So, in case of even numbers only those pairs will work in which both the numbers are even. In case, even if one of the numbers is odd, that pair will not be valid since odd + even = odd and a odd number divided by 2 always results into a fractional number.
So, we can write (x + y) (x – y) = even × even. This means that both (x + y) and (x –y) has to be even. The standard form of representing an even number is 2n where n is a whole number.
So, we can assume (x + y) as 2n and (x – y) as 2m where n > m as (x + y) will always be greater than (x – y), since x and y are natural numbers.
So, our equation can be rewritten as (2n) (2m) = 60. On solving, it reduces to nm = 15.
We have already seen such problems where a natural number is written as the product of two natural numbers. Here, the question becomes the number of ways in which 15 can be written as the product of natural numbers.
Answer should be 2 ways as number of factors of 15 are 4 and since we are finding out number of ways of writing 15 as the product of two natural number, we will be needed to divide the number of factors by 2, i.e. the answer is 2 ways. We got the same answer through the first method also, but this is a faster way of solving such problems.
So, in case of any even number which has to be expressed as a difference of squares of two natural numbers, we first need to divide the number by 4 and then find out the number of factors of that number, and then divide it by 2 to get the required ways.
E.g. 16: In how many ways can 80 be written as the difference of squares of two natural numbers?
We will solve this problem by both the methods to verify whether we are getting the same answer or not.
80 is an even number, so divide it by 4. We get 20 and then find out the number of factors of 20. 20 has 6 factors. If a number has 6 factors, it can be written in 3 ways. So, answer is 3 ways.
Now, we will verify it.
(x + y) (x – y) = 1 × 80; = 2 × 40; = 4 × 20; = 5 × 16; =8 × 10
Only those methods will work where both (x + y) and (x – y) are even. There are only three instances namely (2, 40), (4, 20) and (8, 10) where both (x + y) and (x – y) are even. So, only these three ways will give us the natural number ways on solving them simultaneously.
So, both the methods are yielding the same answer, but we can calculate the answer orally with the help of the first method, but the second one will take time to write 80 as the product of two natural numbers.
E.g. 17: In how many ways can 960 be written as the difference of squares of two natural numbers?
960 is an even number, so we need to divide 960 by 4. We get 240 and now we need to factorize 240 and find out its factors. 240 = 24 × 31 × 51 = 5 × 2 × 2 = 20 factors. If a number has 20 factors, we can write them in 10 ways. So, answer is 10 ways.
If we had followed the second method, it would have taken us a long time to write 960 as the product of two natural numbers. Hopefully, after doing so many examples you would have understood the logic and the efficiency of the first method.
What if the number is even but not a multiple of 4? Will the same logic work?
E.g. 18: In how many ways can 50 be written as the difference of squares of two natural numbers?
We will use the long-cut method, after that we will see how we could streamline the method of working.
Question is asking x2 – y2 = 50 in how many ways.
(x + y) (x – y) = 1 × 50; = 2 × 25; = 5 × 10.
Only those pairs will work where both of them are even. In this case none of the pairs are even, in all the pairs one of them is even and another one is odd. So, answer should be 0 ways or no way. And remember in case of an even number, we will not get any pair in which both the numbers are odd, since odd × odd = an odd number. All the factors of an EVEN NUMBER are either even or odd. Only those pairs will work where (even + even) is found.
So, answer is no way. Will this hold true for any even number which is not a multiple of 4?
Answer is yes but why?
Because if an even number which is not a multiple of 4 will have just one 2, for e.g., (30 = 2 × 3 × 5) OR (70 = 2 × 5 × 7). So, if a number has just one 2 and we are writing that number as a product of two natural numbers, only one number of that pair will have 2 (means it will be even), while the other one will be odd. When we add up (odd + even), we get odd and odd number divided by 2 results in a fractional number. This will be true for all the pairs of that number. Of all the pairs, only one number will have 2 and another one will be odd which will result into no solution.
SUMMARY:
So, whenever a sum comes “In how many ways can (a number) be written as difference of squares of two natural numbers?”
We will see whether its an odd number or an even number or even number which is not a multiple of 4. If its an odd number, we have already seen how to deal with it. Just find out number of factors and divide by 2 to get the number of ways.
If the number is even, divide then number by 4 first, find the number of factor of the number left behind after division and then divide number of factor obtained in the previous stage by 2 to get the number of ways.
If the number is even but not a multiple of 4, we can write the answer directly as no ways.
If the number is a odd perfect square, then we will have to see whether the question is asking in how many ways can 25 be written as the difference of squares of two natural numbers or whole numbers.
x2 – y2 = 25 = 1 × 25 or 5 × 5. The first way will work since the sum of 1 + 25 is even, but when we solve for the second way, we will get x = 5 and y =0. So, we will have to see if the question is asking to find ways to write 25 as difference of squares of two natural numbers or of whole numbers. If asking for natural numbers, answer will be 1 and if asking for whole numbers, answer will be 2.
Same thing has to be observed for even perfect square.
Integral Solutions:
This problem looks like Algebra problem, but it is also an application of number of factors. We will learn the standard way, after that you can learn the short-cut through observation.
On simplifying the equation given, we get 10a + 10b = ab.
On further solving, we get ab – 10a – 10b = 0. We will add 100 on both sides so that the expression gets factorized easily. (How did we get 100?.Multiply coefficient of a which is (– 10) with that of b which also is (– 10 )).
Now the expression on adding becomes ab – 10a – 10b + 100 = 100.
On factorizing, it becomes (a – 10) (b – 10) = 100.
Now , on seeing this you should become familiar with pattern that whenever two numbers (natural or integers) are multiplied, it becomes an application of number of factors.
How to solve this? We have already learnt about making pairs.
(a – 10) (b – 10) = 100, find out factors of 100 which comes to 9. If a number has 9 factors, 5 pairs can be made.
(a – 10) (b – 10) = 1 × 100; = 2 × 50 ; = 4 × 25 ; = 5 × 20 ; = 10 × 10.
Since we are talking about natural number solutions, every pair will give us two natural number solutions except the last pair. We have already seen in case of a perfect square, the square root of that number multiplied with itself results in the number. Since both numbers are identical, we will just get one solution and not two.
Still if there are some doubts, then pay attention here. If (a – 10) = 1, then (b – 10) = 100. On solving them a and b will be natural numbers. But, it could be other way round also that (a – 10 ) takes 100 and ( b – 10 ) = 1. Similarly, we will get 2 natural number solutions each for remaining three pairs. But what about the last pair? If (a – 10 ) = 10, then (b – 10 ) = 10. Even if we take the value other way around, we would not get anything new.
So, answer is 9 natural number solutions.
What if the question asked to find the integral solutions to the same question?
Still the logic was same, but now every pair will give us 4 integral solutions except the last pair. (Concept discussed in the beginning of this topic). If (a – 10) = 1, then (b – 10) = 100. Also, if (a – 10) takes 100 and (b – 10) = 1. And If (a – 10) = – 1 , then (b – 10) = –100. On solving this also, we will get integral values for a and b. And the last one will be If (a – 10) = –100, then (b – 10) = –1.
So, rest of the three pairs will give us 4 integral solutions each. So, we got 4 × 4 integral solutions up till now. What about the last pair, will it give us two integral solutions or just one?
Error-prone area: It will just give one integral solution. If (a – 10) = 10, then (b – 10) = 10. This will be the first one. If we take the second part which is If (a – 10) = –10, then (b – 10) = –10. On solving this, we get we get a = 0 and b = 0. This value is not acceptable as despite 0 being an integer, any number divided by zero is not defined in Maths. And our question was 
.
And if we substitute value of a and b as 0, the expression will become undefined. So, that solution is rejected. So, final answer is 17 integral solutions.
In case of Integral solutions, we have to be careful, as many of us make mistakes here. In all such problems asking for integral solutions, there will be one pair which on solving give us value of a and b as 0, so, we have to be cautious there. No such problem in case of natural number solution.
We can simplify it immediately as it of the same pattern as (a – 12) (b – 12) = 144 as 12 will be multiplied with 12 to make 144.
Again if we are looking for natural number solution, we will find out number of factors of 144 which is 15. If a number has 15 factors, we can make 8 pairs and every pair will give us two natural number solutions except the last pair which will only result into one natural number solution. So, answer is 7 × 2 + 1 = 15 natural number solutions.
For Integral solutions: We have to be careful, since one pair will result in values of a and b as 0 which will not be acceptable. And every pair will give us 4 integral solutions and the last pair which will be 12 × 12 will just one integral solution. So, answer is 4 × 7 + 1 =29 integral solutions.
So, if logic is clear we can solve this problem orally by just finding out number of factors of the number asked.
This is also of the same type, so follow the same logic.
On simplifying it we get, 10b + 15a = ab; or ab – 15a – 10b = 0. The coefficient of a and b are (– 15) and (–10) respectively. Multiply them and add on both sides, we get
ab – 15a – 10b + 150 = 150. On factorizing, we get (a – 10) (b – 15) = 150. Now, if we are looking for natural number solutions, we will find out number of factors of 150 which is 12. If a number has 12 factors, we can form 6 pairs or write them into 6 ways. All the pair will give us 2 natural number solutions each since 150 is not a perfect square.
So, answer is 6 × 2 = 12 natural number solutions.
If we are solving for integral solutions, as we have discussed before we will have to tread carefully since one pair will be there which will result in values of a and b as 0 which will not be acceptable. So, 6 pairs are there, every pair will give us 4 integral solutions, except one pair which will give us just 3 integral solutions.
Which pair will give us 3 integral solutions? The pair which contains 10 and 15.
If (a – 10) = 10, then (b – 15) = 15; will give integral solutions for a and b.
If (a – 10) = 15, then (b – 15) = 10; will give integral solutions for a and b.
If (a – 10) = – 10, then (b – 15) = – 15; not acceptable since a= 0 and b = 0.
And if (a – 10) = –15, then (b – 15) = –10; will give integral solutions for a and b.
So, answer is 6 × 4 – 1 = 23 integral solutions.
Same type of problem, but just negative sign comes in for positive.
On rearranging, we get 8b – 12a = ab; On further solving we get ab + 12a – 8b = 0. (Now add – 96 to both sides since coefficients of a and b are (12) and (–8) respectively).
So, our equation becomes ab + 12a – 8b – 96 = –96.
On factorizing, we get (a – 8 ) (b + 12) = –96.
Do not worry about – 96, just find out the factors of 96. 96 has 12 factors. If a number has 12 factors, we can form 6 pairs and every pair will result in 4 integral solutions except one pair which will give us just 3 integral solutions. And that pair will be (12, 8).
So, final answer is 6 × 4 – 1 = 23 integral solutions.
Exercise:
1. Find the product of number of factors of 480.
2. If a number N has 30 factors between 1 and √N, then find the total number of factors of N (both cases).
3. If all the factors of N are written in ascending order and the product of factors at the 20th and 32nd position results in N only, then find the total number of factors of N.
4. Find the integral solutions to a × b = 300.
5. Find the number of ways in which 420 can be written as the product of two natural numbers.
6. Find the integral solutions to a × b = 900, if a and b are distinct integers.
7. If all the factors of 153 × 60 are written in ascending order, then find which factor will occupy the 38th position.
8. Find the number of ways in 440 can be written as the difference of squares of two natural numbers.
9. Find the number of ways in which 370 can be written as the difference of squares of two natural numbers.
10. Find the number of ways in which 475 can be written as the difference of squares of two natural numbers.
14. How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?
a. 748 b. 300 c. 750 d. 250
15. Let N = 210 × 39. How many factors of N2 are less than N but do not divide N completely?
Answer Key
1. (260 × 312 × 512) or (480)12. 2. a) 63 factors, if N is a perfect square; b) 62 factors, if N is not a perfect square.
3. 51 factors 4. 36 solutions 5. 12 ways 6. 52 solutions
7. 450 8. 4 ways 9. 0 ways 10. 3 ways
11. 17 solutions 12. 20 solutions 13. 23 solutions 14. C 15. 90 factors
Explanations:
Question 1:
We have to find the product of all the factors of 480. For that, we need to find total factors of 480 first. 480 = (25 × 31 × 51) = 24 factors.
Now, we have to multiply all the 24 factors of 480. We have learnt about “RELATIVE PLACEMENT OF FACTORS”, the same logic will be applied here. We know that all the factors which are equidistant from the centre, when multiplied result in the number itself. The factor at the 1st position multiplied with factor at the 24th position will give us the number itself i.e. 480. Similarly, the factor at the 2nd position multiplied with factor at 23rd position will result in 480. So, if there are 24 factors, we will get 12 pairs which multiplied with each other will result in 480.
So, answer is (25 × 31 × 51)12 = (260 × 312 × 512) or 48012.
Question 2:
If there are 30 factors between 1 and √N, then there would be 30 factors between √N and N. But, in case of number N being a perfect square, √N is also a part of factor.
So, if the number is perfect square, total number of factors of N will be 63 (30 + 30 + 2 + 1).
And if the number is not a perfect square, total number of factors will be 62 as √N will not a part of the factor.
Question 3:
This problem is also based on relative placement of factors.
If the factor at 20th position multiplied with factor at 32nd position results in N.
Then factor at 19th position multiplied with factor at 33rd position will also result in N.
Similarly the factor at 18th position multiplied with factor at 34th position will also result in N.
So, we can conclude that there must be 19 factors before the factor at the 20th position, and to accommodate them there must be 19 factors after 32nd factor. Since, number of factors exist in pairs and when multiplied result in N, so total number of factors of N = 32 + 19 = 51 factors.
Question 4:
To find the integral solutions to a × b = 300, we need to find factors of 300.
300 has 18 factors and we can make 9 pairs of numbers multiplied to give us 300. Every pair will give us 4 integral solutions, so in total 36 integral solutions.
Question 5:
To find the number of ways in which 420 can be written as the product of two natural numbers, we need to find number of factors of 420. 420 has 24 factors and 12 pairs can be made. Since number of ways is asked, 12 pairs will give us 12 ways in which 420 can be written as the product of two natural numbers.
Question 6:
To find integral solutions to a × b = 900, we will find how many factors does 900 has. And since 900 is a perfect square, the number of factors will be odd i.e. 27 factors. We have learnt earlier how to deal with perfect square.
If 900 has 27 factors, we will ignore one factor (i.e. √900 = 30), then remaining factors of 900 are 26. If a number has 26 factors, we can have 13 pairs and every pair will give us 4 integral solutions. So, in total 52 integral solutions of 300 are there. And the pair which we ignored was (30 × 30) will not be considered since the question is asking for distinct integral solutions.
Question 7:
Number of factors of 153 × 60 = (33 × 53 × 51 × 31 × 22) = 22 × 34 × 54. So, this number is a perfect square and it will have odd factors i.e. 75. If there are 75 factors of a number the centre position will be the 38th position. And in case of perfect square, centre position is always occupied by the square-root of that perfect square.
So, factor which will occupy the 38th position is (21 × 32 × 52) or 450.
Question 8:
We have already understood the logic how to solve for an even number in terms of difference of squares of two natural numbers. We need to divide the even number by 4, in this case number given is 440 and divided by 4 will give us 110.
Now, we need to find number of factors of 110 which 8 factors. If 8 factors are there, then we can write the number in 4 ways.
Question 9:
370 cannot be written as the difference of squares of two natural numbers since 370 is not a multiple of 4. Because when we write 370 as the product of two natural numbers, only one number will be even while other will be odd. And odd + even = odd and odd number divided by 2 results in fraction. [Already discussed in detail in theory].
So, answer is 0 ways.
Question 10:
To find the number of ways in which 475 can be written as the difference of squares of two natural numbers, we just need to find number of factors of 475 and divide it by 2 as 475 is an odd number.
475 has 6 factors, so the answer is 3 ways.
Question 11:
We can rearrange the equation like 14a + 14b = ab.
Further, ab – 14a – 14b = 0, [Add 196 on both sides as coefficients of a and b are (– 14) each].
Now, we get ab – 14a – 14b + 196 = 196. This equation can be factorized as (a – 14) (b – 14) = 196.
Now, we know how to solve it as the expression is a product of two numbers. We need to find number of factors of 196 which is 9 factors.
In case of perfect square, we need to ignore one factor (i.e. as 14 = √196), then remaining factors are 8. We can make 4 pairs and every pair will give us 4 integral solutions. So, we will get in total 16 integral solutions. And the one factor that we ignored will give us just one integral solution.
[(a – 14) = 14 and (b – 14) = 14 while (a – 14) = –14 & (b – 14) = –14 will be rejected since (a =0 and b = 0 is not acceptable).
So, answer is 17 integral solutions.
Question 12:
We can simplify the expression as 6 (3b + 4a) = ab. Take all the terms on R.H.S.
We get ab – 24a – 18b = 0, (Multiply 24 × 18 on both the sides).
Now, ab – 24a – 18b + 24 × 18 = 24 × 18.
And after factorizing, we get (a – 18) (b – 24) = 24 × 18.
Now, we can find out factors of 24 × 18 which will be 20 factors. If number has 20 factors, we will get 10 pairs. And every pair will give us two natural number solutions.
So, answer is 20 natural number solutions.
Question 13:
On simplifying, we get 5(3b – 2a) = ab;
ab – 15b + 10a = 0. [Add (–150) on both the sides since the coefficients of a and b are (–15) and 10 respectively.
We get, ab – 15b + 10a – 150 = –150.
On factorizing we get (a – 15) (b + 10) = –150.
Since we are looking for integral solutions and we have discussed this earlier that in case of integral solutions, there will one pair that will make values of (a =0 & b =0) which is not acceptable.
150 has 12 factors, so we can make 6 pairs. Every pair will give us 4 integral solutions, so a total of 24 integral solutions. But one of the pair will yield value of a & b as 0 which will be rejected.
(a – 15) = 15, then (b + 10) = –10; we will get integral solution.
(a – 15) = –10, then (b + 10 ) = 15; we will get integral solution.
(a – 15) = –15, then (b + 10) = 10; will be rejected as values of a & b becomes 0.
And (a – 15) = 10, then (b + 10) = –15; we will get integral solution.
So, we should remember that in case of integral solutions, there will be one pair which will give value of a & b as 0 which will have to be rejected.
So, final answer is 6 × 4 – 1 = 23 integral solutions.
Question 14:
We have discussed this type of question earlier, only difference is that in this question we have to find how many numbers from 1 to 1000 can be represented as the difference of squares of two whole numbers (non-negative integers).
We have learnt that all the even numbers which are multiples of 4 can be represented as the difference of squares of two natural numbers. So, all those numbers like 4, 8, 12, 16 and so on can also be represented as the difference of squares of two whole numbers also. In total there are 250 multiples of 4 in the range 1 to 1000.
All the even numbers which are multiples of 2 but not multiples of 4 (2, 6, 10, 14 and so on) cannot be represented in the form of difference of squares of two whole numbers.
We had seen that in case of odd numbers, there is no such problem as all the odd numbers satisfy the condition. There are total of 500 odd numbers in the range 1 to 1000, so all of them will work. You must be wondering about whether a2 – b2 = 1 will work or not.
It will work as on simplifying we get (a + b) (a – b) = 1 × 1;
On solving we get (a + b) = 1 and (a – b) = 1. When we solve them simultaneously, we get a = 1 and b = 0. And 0 being a whole number will work. And all the remaining odd numbers will certainly be represented as the difference of squares of two natural numbers as well as two whole numbers.
So, in total 500 + 250 = 750 numbers will satisfy the required condition.
Question 15:
This is a good question based on relative placement of factors. Let us see how to solve it.
N = 210 × 39, so N has (11 × 10) = 110 factors.
N2 = 220 × 318, so N2 has (21 × 19) = 399 factors.
If we write all the factors of N2 in ascending order, N will occupy the centre position i.e. 200th position since N2 is a perfect square.
1……………………………………………….N………………………………………………….N2.
N2 will be at 399th position, N will be at 200th position.
Between 1 and N, there are 109 factors of N including 1.
Between 1 and N, there are 199 factors of N2 including 1 which are less than N.
Of these 199 factors, 109 factors are of N which are less than N and will divide N completely since they are factors of N. Remaining 90 factors of N2 which are less than N (199 – 109 = 90) will not divide N since they are factors of N2 and not N.
So, answer is 90 factors.