Table of contents  
What is Number System?  
Important Formulas in Algebra  
Visual Representation of Formulas in Algebra  
Important Formulas in Number System  
Types of Numbers  
Divisibility Rules  
HCF and LCM 
Understanding the number system is crucial for competitive exams like CAT, as it forms the foundation for various quantitative topics. The topic is known for intriguing conceptual problems that test the best brains. Instead of just solving problems, focus on grasping the concepts thoroughly. Explore alternative approaches to solve questions, as it enhances your problemsolving skills and prepares you comprehensively for the exam.
Number system is a method of representing a number on the number line. Number system is a system of writing or expressing numbers. This Page from here on contains Formulas and definition of Number System.
Here is a list of Important Formulas that candidates need to master to crack the exam successfully.
Algebra is a branch of mathematics that substitutes letters for numbers. An algebraic equation depicts a scale, what is done on one side of the scale with a number is also done to either side of the scale.
Solution: We know that there are 25 odd numbers between 1 to 50. Thus, by using the sum of n odd numbers formula which is n^{2}, we get, S_{25 }= 25^{2} = 625.
We can alternatively show this using the formula S_{n} = n/2 × [a + l]. We know that the sum of odd numbers 1 to 50 is represented as S_{n} = 1 + 3 + ... + 49.
Thus, a = 1, l = 49, and n = 25.
S_{25} = (25/2) × [1 + 49]
= (25/2) × 50
= 25 × 25 = 625
Thus, the sum of odd numbers 1 to 50 is equal to 625.
Example: What is the sum of even numbers from 1 to 50?
Solution: We know that, from 1 to 50, there are 25 even numbers.
Thus, n = 25
By the formula of sum of even numbers we know;
Sn = n(n+1)
Sn = 25(25+1) = 25 x 26 = 650
Highest power of n in m! is [m/n] + [m/n^{2}] + [m/n^{3}] +….. where, [x] is the greatest integer less than or equal to x.
Example: Find the highest power of 7 in 100!.
Highest power of 7 in 100! = [100/7] + [100/49] = 16
[23/5] = 4. It is less than 5, so we stop here. The answer is 4.
If all possible permutations of n distinct digits are added together with the sum
= (n1)! * (sum of n digits) * (11111… n times)
Example: What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated?
Solution: The sum of the numbers formed by taking all the given n digits is (sum of all the n digits) * (n  1)! * (111…..n times).
Here n = 4, and Sum of 4 digits = 16
The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7= (16) * (4 – 1)! * ( 1111) = 16 * 3! * 1111
If the number can be represented as N = a^{p}∗ b^{q }∗ c^{r}. Then, the number of factors of N is (p+1) * (q+1) * (r+1)
Sum of all factors:
Example: Find the number of factors of 98 and also find the sum and product of all factors.
Solution: First, write the number 98 into prime factorization.
i.e. 98 = 2 × 49 = 2 × 7 × 7 = 2^{1} x 7^{2 }
Here A = 2, B = 7, p = 1, q = 2
Number of factors for 98 = (p + 1)(q +1) = 2 × 3 = 6
Sum of all factors of 98= 3 × 57 = 171
Product of all factors of number 98 = (98)^{6/2} = 941192
If the number of factors is odd, then N is a perfect square.
Example: Find the number of odd, even, perfect square, perfect cube factors of 4500.
4500 = 45 × 100 = 9 × 5 × 10 × 10 = 3 × 3 × 5 × 5 × 2 × 5 × 2 = 2^{2} × 3^{2} × 5^{3}
Here, consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3
Here identifying that odd number are 3 and 5.
Numbers of odd factors of number 4500 = (q + 1) (r + 1) = 3 × 4 = 12
∴ Total number of factors = (p + 1)(q +1)(r +1) = 3 × 3 × 4 =36
Numbers of even factors of number = (Total number of factors – Numbers of odd factors) = 36 – 12 = 24
Number of perfect square factors of number 4500 = 2 x 2 x 2 = 8
Number of perfect cube factors of number 4500 = 1 x 1 x 2 = 2
If there are n factors, then the number of pairs of factors would be n/2. If N is a perfect square, then the number of pairs (including the square root) is (n+1)/2.
Example: In how many ways you can express 36 as the product of two of its factors?
Prime factorization of 36 i.e. we write 36 = 2^{2} × 3^{2}
Number of factors of 36 will be (2+1)(2+1) = 9
(i.e. factors are 1, 2,3, 4, 6, 9, 12, 18, 36).
Since we are asked the total number of ways hence we include the square root of 36 i.e. 6 as well.
Thus the number of ways you can express 36 as the product of two of its factors is (9+1)/2 = 5.
If the number can be expressed as N = 2^{p} ∗ a^{q} ∗ b^{r} . . . where the power of 2 is p and a, b are prime numbers then:
(i) The number of even factors of N = p (1 + q) (1 + r) . . .
(ii) The number of odd factors of N = (1 + q) (1 + r)…
When we are asked to calculate how many positive integral solutions are possible for the equation X^{2}– Y^{2}= N, there can be 4 cases.
Let us look at them one by one by solving examples:
Positive Integral Solutions for the equation X^{2}  Y^{2} = N?
When we are asked to calculate how many positive integral solutions are possible for the equation X^{2}  Y^{2} = N, there can be 4 cases.
Case 1: N is an odd number and not a perfect square
Case 2: N is an odd number and a perfect square
Case 3: N is an even number and not a perfect square
Case 4: N is an even number and a perfect square
Case 1: N is an odd number and not a perfect square
The total number of positive integral solutions will be = (Total number of factors of N)/2
Example: How many positive integral solutions are possible for the equation X^{2 }– Y^{2 }= 135?The total number of factors of 135 = 1, 3, 5, 9, 15, 27, 45 and 135 is 8.
So, total number of positive integral solutions = 8/2 = 4.
Case 2: N is an odd number and a perfect square
The total number of positive integral solutions will be = [(Total number of factors of N) – 1]/ 2
Example: How many positive integral solutions are possible for the equation X^{2}– Y^{2 }= 121?
The total number of factors of 121 = 1, 11 and 121 is 3.
So, the total number of positive integral solutions = (31)/2 = 1
Case 3: N is an even number and not a perfect square
The total number of positive integral solutions will be = [Total number of factors of (N/4)] / 2
Example: How many positive integral solutions are possible for the equation X^{2}– Y^{2 }= 160?Total number of factors of 40 = 1, 2, 4, 5, 8, 10, 20 and 40 is 8 (as N = 160 and N/4 = 40)
So, total number of positive integral solutions = 8/2 = 4
Case 4: N is an even number and a perfect square
The total number of positive integral solutions will be = {[Total number of factors of (N/4)] – 1 } / 2
Example: How many positive integral solutions are possible for the equation X^{2}– Y^{2}= 256?
The total number of factors of 64 = 1, 2, 4, 8, 16, 32 and 64 is 7 (as N = 256 and N/4 = 64)
So, (71)/2 = 3 positive integral solutions.
EduRev Tip: A number can be written as the sum of two squares while including prime factors of the form 4k + 3 as long as those prime factors are raised to an even power.
The number of digits in a^{b} = [b log_{m}(a)] + 1; where m is the base of the number and [.] denotes the greatest integer function.
Even a number that is not a multiple of 4, can never be expressed as a difference of 2 perfect squares.
Basic summation properties:
(i) Sum of first n odd numbers is n^{2}
(ii) Sum of first n even numbers is n(n + 1)
(iii) 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2
(iv) (1^{²} + 2^{²} + 3^{²} + ….. + n^{²}) = n ( n + 1 ) (2n + 1) / 6
(v) (1^{³} + 2^{³} + 3^{³} + ….. + n^{³}) = (n(n + 1)/ 2)²
The product of the factors of N is given by N^{a/2}, where 'a' is the number of factors.
The last two digits of a^{2}, (50 – a)^{2}, (50 + a)^{2}, (100 – a)^{2} . . . . . are same. Every number can be written as (50n ± x), where x is a number from 0 to 25. Hereby 50n, it just means any multiple of 50 i.e. 0, 50, 100, 150, ……
0 to 25 = (0 to 25) itself.
25 to 50 = 50 – (25 to 0)
50 to 75 = 50 + (0 to 25)
75 to 100 = 100 – (25 to 0)
100 to 125 = 100 + (0 to 25)
125 to 150 = 150 – (25 to 0) and so on.
But how does this help us in finding the last two digits of any square?
(50n ± x)^{2} = 2500n^{2} ± 100nx + x^{2}
The last two digits of each of 2500n^{2} and 100nx will be 00. Thus the last two digits of the RHS, and hence of the LHS, will be the last two digits of x^{2}.
268 = 50 × 5 + 18.
Thus, the last two digits of 268^{2} will be same as the last two digits of 18^{2} i.e. 24.
While you can consider 278 = 50 × 5 + 28, it will consider
278 = 50 × 6 – 22, to be x in the range 0 to 25.
Thus, the last two digits of 278^{2} will be the same as the last two digits of 22^{2} i.e. 84.
If the number is written as 2^{10}^{n}:
(i) When n is odd, the last 2 digits are 24.
(ii) When n is even, the last 2 digits are 76.
Q.1. Is 7248 is divisible (i) by 4, (ii) by 2, and (iii) by 8?
Answer:
(i) The number 7248 has 48 on its extreme right side which is exactly divisible by 4. When we divide 48 by 4 we get 12.
Therefore, 7248 is divisible by 4.
(ii) The number 7248 has 8 on its unit place which is an even number so, 7248 is divisible by 2.
(iii) 7248 is divisible by 8 as 7248 has 248 at its hundred place, tens place and unit place which is exactly divisible by 8.
Q.2. A number is divisible by 4 and 12. Is it necessary that it will be divisible by 48? Give another example in support of your answer.
Answer: 48 = 4 × 12 but 4 and 12 are not coprime.
Therefore, it is not necessary that the number will be divisible by 48.
Let us consider the number 72 for an example
72 ÷ 4 = 18, so 72 is divisible by 4.
72 ÷ 12 = 6, so 72 is divisible by 12.
But 72 is not divisible by 48.
Q.3. Without actual division, find if 235932 is divisible (i) by 4 and (ii) 8.
Answer:
(i) The number formed by the last two digits on the extreme right side of 235932 is 32
32 ÷ 4 = 8, i.e. 32 is divisible by 4.
Therefore, 235932 is divisible by 4.
(ii) The number formed by the last three digits on the extreme right side of 235932 is 932
But 932 is not divisible by 8.
Therefore, 235932 is not divisible by 8.
Q.4. Check whether 998 is divisible by 9.
Answer:
According to the rule, if the sum of the digits in a number is a multiple of 9, then it is divisible by 9.
Sum of the digits in 998: 9 + 9 + 8 = 26
26 is not a multiple of 9
So, 998 is not divisible by 9.
Q.5. Check whether 1782 is divisible by 11.
Answer:
According to the rule, in a number, if the sum of the digits in odd places and sum of the digits in even places are equal or they differ by a number divisible by 11, then the number is divisible by 11.
In 1782, the sum of the digits in odd places:
1 + 8 = 9
In 1782, the sum of the digits in even places:
7 + 2 = 9
In 1782, the sum of the digits in odd places and sum of the digits in even places are equal.
So, 1782 is divisible by 11.
Example 1: Find the HCF of 96, 36, and 18.
96 = 2 × 3 × 2 × 2 × 2 × 2
36 = 2 × 3 × 2 × 3
18 = 2 × 3 × 3
Therefore, the HCF of 96, 36 and 18 is the product of the highest number of common factors in the given numbers i.e., 2 × 3 = 6.
In other words, 6 is the largest possible integer, which can divide 96, 36 and 18 without leaving any remainder.
Example 2: Find the HCF of 42 and 70.
42 = 3 × 2 × 7
70 = 5 × 2 × 7
Hence, HCF is 2 x 7 = 14.
Example 3: Find the HCF of numbers 144, 630, and 756.
144 = 2^{4} × 3^{2}
630 =2 × 3^{2 }× 5 × 7
756 = 2^{2} x 3^{3} x 7
Hence, HCF of 144, 630, 756 = 2 × 3^{2} = 18.
Example 4: Find the LCM of 96, 36, and 18.
96 = 2 × 2 × 2 × 2 × 2 × 3 = 2^{5 }× 3^{1}
36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}
18 = 2 x 3 x 3 = 2^{1} x 3^{2}
Therefore, LCM of 96, 36 and 18 is the product of the highest powers of all the prime factors, i.e. 2^{5} x 3^{2 }= 32 x 9 = 288.
That is, 288 is the smallest integer which is divisible by 96, 36 and 18 without leaving any remainder.
Example 5: Find the LCM of 42 and 70.
42 = 3 × 2 × 7
70 = 5 × 2 × 7
Hence, LCM is 2 × 3 × 5 × 7 = 210.
Apart from the method of prime factorization, there is another method of finding the LCM of given numbers, and the method is known as the long division method. This method is quite helpful in getting LCM quickly if there are three or more than three numbers.
Example 6: Find out the LCM of 48 and 300.
Here, we need to find out the LCM of 48 and 300.
48 = 2 × 2 × 2 × 2 × 3 = 2^{4} × 3^{1}
300 = 2 × 2 × 3 × 5 × 5 = 2^{2} × 3^{1} × 5^{2}
List all numbers found, as many times as they occur, most often any one given number, and multiply them together to find the LCM.
Thus, 2^{4 }× 3^{1} × 5^{2} = 1200
Hence, the LCM of 48 and 300 is 1200.
Example 7: Find the LCM of 12, 18, 30.
Here, we need to find the LCM of 12, 18, 30.
Now, let us find the prime factors of the above three numbers.
12 = 2 × 2 × 3 = 2^{2} × 3^{1}
18 = 2 × 3 × 3 = 2^{1} × 3^{2}
30 = 2 × 3 × 5 = 2^{1} × 3^{1 }× 5^{1}
Now, list all the prime numbers found as many times as they occur most often for anyone given number and multiply them together to find the LCM.
So, 2 × 2 × 3 × 3 × 5 = 180
Using exponents instead, multiply together each of the prime numbers with the highest power
Thus, 2^{2} × 3^{2} × 5^{1} = 180
Hence, the LCM of 12, 18, 30 is 180.
EduRev Tip:
HCF of two prime numbers is always 1.
HCF of coprime numbers is always 1.
Links for other topics of Number System:
Introduction: Number System
Important Concepts: Number System  1
Important Concepts: Number System  2
Important Concepts: Number System  3
Base Systems
Overview: Number System
Divisibility Rules
Introduction & Concept: HCF & LCM
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1. What are the types of numbers discussed in the article on Number System Important Formulas? 
2. What are some important divisibility rules mentioned in the article on Number System Important Formulas? 
3. Can you explain the concept of HCF and LCM as discussed in the article on Number System Important Formulas? 
4. How are the important formulas in Algebra visually represented in the article? 
5. What are some frequently asked questions related to the article on Number System Important Formulas in Bank Exams? 
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