The number theory or number systems happens to be the back bone for CAT preparation. Number systems not only form the basis of most calculations and other systems in mathematics, but also it forms a major percentage of the CAT quantitative section. The reason for that is the ability of examiner to formulate tough conceptual questions and puzzles from this section. In number systems there are hundreds of concepts and variations, along with various logics attached to them, which makes this seemingly easy looking topic most complex in preparation for the CAT examination. The students while going through these topics should be careful in capturing the concept correctly, as it’s not the speed but the concept that will solve the question here. The correct understanding of concept is the only way to solve complex questions based on this section.
Here is the way numbers are categorized.
Real numbers: The numbers that can represent physical quantities in a complete manner. All real numbers can be measured and can be represented on a number line. They are of two types:
Rational numbers: A number that can be represented in the form p/q where p and q are integers and q is not zero. Example: 2/3, 1/10, 8/3 etc. They can be finite decimal numbers, whole numbers, integers, fractions.
Irrational numbers: A number that cannot be represented in the form p/q where p and q are integers and q is not zero. An infinite non recurring decimal is an irrational number. Example: √2 , √5 , √7 and Π(pie)=3.1416.
The rational numbers are classified into Integers and fractions
Integers: The set of numbers on the number line, with the natural numbers, zero and the negative numbers are called integers, I = {…..-3, -2, -1, 0, 1, 2, 3…….}
Fractions:
A fraction denotes part or parts of an integer. For example 1/6, which can represent 1/6th part of the whole, the type of fractions are:
1. Common fractions: The fractions where the denominator is not 10 or a multiple of it. Example: 2/3, 4/5 etc.
2. Decimal fractions: The fractions where the denominator is 10 or a multiple of 10. Example 7/10, 9/100 etc.
3. Proper fractions: The fractions where the numerator is less than the denominator. Example ¾, 2/5 etc. its value is always less than 1.
4. Improper fractions: The fractions where the numerator is greater than or equal to the denominator. Example 4/3, 5/3 etc. Its value is always greater than or equal to 1.
5. Compound fraction: A fraction of a fraction is called a compound fraction
Example 3/5 of 7/9 = 3/5 x 7/9 = 21/45
6. Complex fractions: The combination of fractions is called a complex fraction.
Example (3/5)/ (2/9)
7. Mixed fractions: A fraction which consists of two parts, an integer and a fraction. Example 3 ½, 6 ¾
The integers are classified into negative numbers and whole numbers
Negative numbers: All the negative numbers on the number line, {…..-3, -2, -1}
Whole numbers: The set of all positive numbers and 0 are called whole numbers, W = {0, 1, 2, 3, 4…….}.
Natural numbers: The counting numbers 1, 2, 3, 4, 5……. are known as natural numbers, N = {1,2,3,4,5….. }. The natural numbers along with zero make the set of the whole numbers
Even numbers: The numbers divisible by 2 are even numbers. e.g. 2, 4,6,8,10 etc . Even numbers can be expressed in the form 2n where n is an integer other than 0.
Odd numbers: The numbers not divisible by 2 are odd numbers. e.g. 1,3,5,7,9 etc . Odd numbers are expressible in the form ( 2n + 1 ) where n is an integer other than 0.
Composite numbers: A composite number has other factors besides itself and unity .e.g. 8 , 72 , 39 etc . A real natural number that is not prime is a composite number.
Prime numbers: The numbers that has no other factors besides itself and unity is a prime number. Example 2 , 23,5,7,11,13 etc. Here are some properties of prime numbers:
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1. The only even prime number is 2
2. 1 is neither a prime nor a composite number
3. If p is a prime number then for any whole number a , ap – a is divisible by p.
4. 2,3,5,7,11,13,17,19,23,29 are first ten prime numbers (should be remembered)
5. Two numbers are supposed to be co-prime of their HCF is 1, e.g. 3 & 5, 14 & 29 etc.
6. A number is divisible by ab only when that number is divisible by each one of a and b, where a and b are co prime.
7. To find a prime number, check the rough square root of the given number and divide the number by all the prime number lower than the estimated square root
8. All prime numbers can be expressed in the form 6n-1 or 6n+1, but all numbers that can be expressed in this form are not prime
Prime Factors: The composite numbers express in factors, wherein all the factors are prime. To get prime factors we divide number by prime numbers till the remainder is a prime number. All composite numbers can be expressed as prime factors, for example prime factors of 150 are 2,3,5,5.
A composite number can be uniquely expressed as a product of prime factors.
Ex. 12 = 2 x 6 = 2 x 2 x 3 = 22 x 31
20 = 4 x 5 = 2 x 2 x 5 = 22 x 51 etc
Note : The number of divisors of a given number N ( including one and the number itself ) where N = am x bn x cp ……. Where a , b , c are prime numbers
are = ( 1 + m ) ( 1 + n ) ( 1 + p ) …………..
e.g. 90 = 2 x 3 x 3 x 3 x 5 = 21 x 32 x 51
Hence here a = 2 b = 3 c = 5
m = 1 n = 2 p = 1
then the number of divisors are = ( 1 + m ) ( 1 + n ) ( 1 + p ) = 2 x 3 x 2 = 12
the factors of 90 = 1 , 2 , 3 , 5 , 6 , 9 , 10 , 15 , 18 , 30 , 45 , 90 = 12
the sum of divisors of given number N is ( am+1 – 1 ) ( bn+1 - 1 ) ( cp+1 – 1 ) ……..
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( a – 1 ) ( b – 1 ) ( c – 1 ) ……….
Perfect number: If the sum of the divisor of N excluding N itself is equal to N , then N is called a perfect number. E.g. 6, 28 , 496
Finding a perfect number through Euclid’s method
Euclid's method makes use of the powers of 2, which are numbers obtained by multiplying by 2 by itself over and over again, which are 1, 2, 4, 8, 16, 32, 64, 128….
Note that the sum of the two numbers in this series (in ascending order) is equal to the third number minus 1:
1+2 = 3 = 4 - 1,
1+2+4 = 7 = 8 - 1,
STARTING FROM THE NUMBER 1, IF YOU ADD THE POWERS OF 2 AND IF THE SUM IS A PRIME NUMBER, THEN YOU GET A PERFECT NUMBER BY MULTIPLYING THIS SUM TO THE LAST POWER OF 2.
If you add 1+2, the sum is 3, which is a prime number. Therefore 3 x 2 = 6 is a perfect number.
If you add 1+2+4, the sum is 7, a prime number. Therefore 7 x 4 = 28 is a perfect number.
If you add 1+2+4+8, the sum 15 is not a prime number, so you can't use Euclid's method here.
If you add 1+2+4+8+16, the sum is 31, a prime number. Therefore 31 x 16 = 496 is another perfect number.
Absolute value of a number:
The absolute value of a number a is | a | and is always positive.
Fibonacci numbers: The Fibonacci numbers is a sequence where
X (n+2) = X (n+1) + X (n), X(1) = 1, X(2) = 1
Example1,1,2,3,5,8,13,21,34,55,89,144.., it can be clearly seen that any number in the series is the addition of the last two numbers, other than the first two numbers
Example 1: The price of pens has increased over the years. Each year for the last 7 years the price has increased, and the new price is the sum of the prices for the two previous years. Last year a pen cost 60 rupees. How much does a pen cost today? How much did a pen cost 7 years ago?
Let this year price be x. Last year it was 60, so the previous year it must have been x-60, continuing this process backwards gives us a sequence of expressions:
x, 60, x-60, 120-x, 2x-180, 300-3x, 5x-480, 780-8x, 13x-1260
All of these increases must be positive as every year prize has gone up. That gives us a sequence of inequalities, each of which can be solved to find a range for x:
x > 0
60 > 0
x-60 > 0 x > 60
120-x > 0 x 120
2x-180 > 0 x > 90
300-3x > 0 x 100
5x-480 > 0 x > 96
780-8x > 0 x 97.5
13x-1260 > 0 x > 96.92
Looking at this, we can say
96.92 x 97.5
The whole number value x can have is 97, with which we get
x = 97
60 = 60
x-60 = 37
120-x = 23
2x-180 = 14
300-3x = 9
5x-480 = 5
780-8x = 4
13x-1260 = 1
Seven years ago, the price was 4 rupees
In the CAT/MCQ format, where you have the four answers, you can check it by working forward and seeing if the results are correct. You can try putting the given answers for original price and see which one fits in the equation.
Golden ratio: The golden ratio is a special number approximately equal to:
1.6180339887498948482...
Golden ratio = (1 + √ 5)/2
To find the golden ratio, we define the golden ratio as the ratio between x and y if
x y
--- = -----
y x+y
Let's say x is 1. Then we have 1/y = y/(y+1). If we solve this equation
to find y, we'll find that it is the value given above, about 1.618
A golden rectangle is a rectangle in which the ratio of the length to the width is the golden ratio.
The concepts like Fibonacci and golden ratio are reference concepts, students are advised not to cram them but just understand the concepts as they are.
BASIC ARITHMETIC OPERATIONS
Addition, subtraction, multiplication and division are the four basic mathematical operations. We have not gone into details of these concepts as they are very basic; we have added some formulae wherever required. Students preparing for CAT are expected to know the basic arithmetic.
Addition: Addition is used to find the total as a single number of two or more given numbers. The number obtained is called the sum of two numbers.
Subtraction: Subtraction is the quantity left when a smaller number is taken from a greater one. The number obtained is called the difference of two numbers. If a smaller number is subtracted from a greater number, the difference is positive; if a greater numbers is subtracted from a smaller number the result is negative.
Multiplication: Multiplication is the short method of finding the sum of given number of repetitions of the same number. The resultant sum of the repetition is called the product. If one factor is zero than the product is zero. If same factors are multiplied, they can be represented as power or the exponent for example 3 x 3 x 3 = 33
Some short methods in multiplication :
Multiplication by 11 , 101 , 1001 etc
Rule : add 1 , 2 , 3 zeroes resp to the multiplicand and add the multiplicand to the resulting number .
Ex 5023 x 11 = 50230 + 5023 = 55253
i. 5023 x 1001 = 5023000 + 5023 + 5028023
Multiplication by 5
Rule : annex a zero to the right of the multiplicand and then divide it by 2
Ex 89356 x 5 = 893560/2 = 446780
Multiplication by 25
Rule : annex two zeroes the right of the multiplicand and then divide it by 4
Ex 890023 x 25 = 89002300/4 = 22250575
Multiplication by 125
Rule : annex 3 zeroes to the right of the multiplicand and then divide it by 8
Multiplication by a number wholly made of nines , i.e. 9 , 99 , 999 etc
Rule : place as many zeroes to the right of the multiplicand as there are nines in the multiplier and from the result subtract the multiplicand.
Ex 895023 x 999 = 895023000 - 895023 = 894127977.
- Power Patterns: see the table below and notice the pattern of last digits of powers:
If you notice, the follow is the pattern of last digits:
Pattern of 2: 2, 4, 6, 8 – repeat every four powers
Pattern of 3: 3, 9, 7, 1 – repeat every four powers
Pattern of 4: 4,6 – repeat every two powers
Pattern of 7: 7, 9, 3, 1 – repeat every four powers
Pattern of 8: 8, 4, 2, 6 – repeat every four powers
Pattern of 9: 9,1 – repeat every two powers
Application:
Since we have seen the cyclicity of 2,3,7,8 is 4, if we want to find the last digit of any power of these numbers of numbers with last digit as 2,3,7,8(like 12, 13, 27) can be calculated by finding out remainder of the power divided by four. The last digit of the remainder power will be the last digit of given number.
Examples
Last digit of 232, since 2 has cyclicity of 4, 32/4 has remainder = 0, so the last digit will be same as of 20 or 24, which is 6
Last digit of 325, since 3 has cyclicity of 4, 25/4 has remainder = 1, so the last digit will be same as 31, which is 3
Division: Division is the method of finding how many times one number called the divisor is contained in another number called dividend. The number of times is called the quotient. The number left after the operation is called the remainder.
(Divisor * quotient) + Remainder = dividend
The number of divisors (including 1 and itself) of a given number N where
N = Am * Bn * Co … where A,B,C are prime numbers are (1+m)(1+n)(1+o)…
Example 2: 90 = 2 * 32 * 5, Here a,b,c are 2,3,5 and m,n,o are 1,2,1. So number of divisors are 2*3*2 = 12, which actually are 1,2,3,5,6,9,10,15,18,30,45,90
Here the sum of the divisors is given by
(a(m+1) – 1)/(a -1) * (b(n+1) – 1)/(b -1) * (c(o+1) – 1)/(c -1) * ….
Taking values from the previous example
(22 – 1)/1 * (33 – 1)/2 * (52 – 1)/4 = 234
Tests for divisibility:
1. A number is divisible by 2 if its unit’s digit is even or zero
2. A number is divisible by 3 if the sum of its digit is divisible by 3.
3. A number is divisible by 4 when the number formed by last two right hand digits is divisible by 4.
4. A number is divisible by 5 if its unit’s digit is five or zero
5. A number is divisible by 6 if its divisible by 2 and 3 both.
6. Divisibility by 7 has two ways:
Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also. This method uses the fact that 7 divides 2*10 + 1 = 21. Start with the numeral for the number you want to test. Chop off the last digit, double it, and subtract that from the rest of the number. Continue this until you get a one-digit number. The result is 7, 0, or -7, if and only if the original number is a multiple of 7.
Example 3:
123471023473
12347102347 - 2*3 = 12347102341
1234710234 - 2*1 = 1234710232
123471023 - 2*2 = 123471019
12347101 - 2*9 = 12347083
1234708 - 2*3 = 1234702
123470 - 2*2 = 123466
12346 - 2*6 = 12334
1233 - 2*4 = 1225
122 - 2*5 = 112
11 - 2*2 = 7.
This rule holds good for numbers with more than 3 digits is as follows:
Group the numbers in three from unit digit.
add the odd groups and even groups separately
the difference of the odd and even should be divisible by 7
e.g. 85437954 the groups are 85 , 437 , 954
sum of odd groups = 954 + 85 = 1039
Sum of even groups = 437
difference = 602 which is divisible by 7
7. A number is divisible by 8 if the number formed by the last three right hand digits is divisible by 8.
8. A number is divisible by 9 if the sum of its digits is divisible by 9.
9. A number is divisible by 10 if its unit’s digit is zero.
10. To check the divisibility by 11, take the test, Alternately add and subtract the digits from left to right. If the result (including 0) is divisible by 11, the number is also. Example: to see whether 365167484 is divisible by 11, start by subtracting: 3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11
11. A number is divisible by 12 if its divisible by 3 and 4 both.
12. A number is divisible by 13 if it fits the following rule:
Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.
Example: 676, 67 – 6*9 = 13, which is divisible by 13 and so is 676
13. A number is divisible by 15 when it is divisible by 3 and 5 both. E.g. 930
14. A number is divisible by 25 if the number formed by the last two right hand digits is divisible by 25. e.g. 1025 , 3475 , 55550 etc.
15. A number is divisible by 125 if the number formed by the last three right hand digits is divisible by 125. e.g. 2125 , 4250 , 6375 etc.
Some common properties of numbers:
1. The numbers, which give a perfect square on adding as well as subtracting its reverse, are rare and hence termed as Rare Numbers.
If X is a positive integer and X1 is the integer obtained from X by writing its decimal digits in reverse order, then X + X1 and X - X1 both are perfect square then X is termed as Rare Number. For example:
For X=65, X1=56
X+X1 = 65+56 = 121 = 112
X-X1 = 65 - 56 = 9 = 32
So 65 is a Rare Number.
2. When n is odd , n( n2 – 1 ) is divisible by 2 For e.g. n = 9 then n(n2-1) = 9(92 – 1) = 720 is divisible by 24
3. If n is odd, 2n + 1 is divisible by 3, e.g. n=5, 25+1 =33, which is divisible by 3
And if n is even, 2n – 1 is divisible by 3, e.g. n=6, 26-1 =63, which is divisible by 3
4. If n is prime, then n (n4-1) is divisible by 30, e.g. n=3, 3(34-1) = 240, which is divisible by 30
5. If n is odd, 22n + 1 is divisible by 5, e.g. n=5, 22*5+1 =1025, which is divisible by 5
And if n is even, 22n – 1 is divisible by 5, e.g. n=6, 22*6-1 = 4095, which is divisible by 5
6. If n is odd, 52n + 1 is divisible by 13, e.g. n=3, 52*3+1 =15626, which is divisible by 13
And if n is even, 52n – 1 is divisible by 13, e.g. n=4, 52*4-1 =390624, which is divisible by 13
7. The number of divisors of a given number N ( including 1 and the number itself ) where N = ambncp where a,b,c are prime numbers , are ( 1 + m ) ( 1 + n ) ( 1 + p ).
8. xn + yn = ( x + y ) ( xn-1 – xn-2 y + …. + yn-1 ), xn + yn is divisible by x + y when n is odd.
9. xn - yn = ( x + y ) ( xn-1 – xn-2 y + …. - yn-1 ) when n is even, so xn - yn is divisible by (x+y)
10. xn - yn = ( x - y ) ( xn-1 + xn-2 y + …. + yn-1 ) when n is either odd or even, so so xn - yn is divisible by (x-y)
SURDS
Surds are irrational roots of a rational number.
e.g. √6 = a surd = it can’t be exactly found . Similarly e , √7 , √8 , √9 , √27 etc are all surds .
Pure surd : the surds which are made up of only an irrational number e.g. √6 , √7 , √8 etc
Mixed surd : surds which are made up of partly rational and partly irrational numbers . e.g. 3 √3 , 6 4√27
Example 6: Convert √96 to a mixed surd
Solution: √96 = √16 x 6 = 4 √6
Example 7: Convert 5√8 to a pure surd
Solution: 5√ 8 = √8 x 25 = √200
Laws of surds:
1. ( n√a )n = a
2. n√a x n√b = n√ab
3. n√a / n√b = n√a/b
4. m√ n√a = mn√a = n√ m√a
5. ( n√a )m = n √am
Rationalization of surds: In order to rationalize a given surd, multiply and divide by the rationalizing factor (conjugate). Rationalizing factor of (x + √y) is (x - √y).
Indices
The indices word comes from index, which is the power or the exponent given to a number, it is the number of times number is going to be multiplied wit itself. In case of xn, n is called the exponent or index.
Example: 24 = two multiplied four times = 2 × 2 × 2 × 2 = 8
LAWS :
1. am x an = am+n
2. am / an = am-n
3. (am)n = amn
4. a-m = 1/am
5. a0 = 1
6. (ab)m = am bm
7. m √a = a1/m
8. ap/q = q √ap
Square of a number: If a number is multiplied by itself, it is called the square of that number. Example 4 × 4 = 16 is square of 4.
Important Properties:
1. A square cannot end in odd number of zeroes.
2. The square of an odd number is odd and that of an even number is even.
3. Every square number is a multiple of 3 or exceeds a multiple of 3 by unity.
4. Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.
5. If a square number ends in 9, the preceding digit is even.
Square root: The square root of a number is the number, whose square is the given number.
Example √16 is 4, as 4 × 4 = 16
Methods for finding square roots:
1. Factorization: Resolve the number into prime factors and deduce if there are numbers which are repeating themselves (square of numbers).
Example: Find √2601
Here 2601 = 32 × 172 = √2601 = 3 × 17 = 68
2. Approximation: The approximation method is the simplest method to find the square root of a number, but as the name suggests it is an approximate method.
This method is best explained with an example. Suppose you want to find the square root of -[if !vml]
, you know the square root of 100 is 10 and 121 is 11, now 104 lies between 100 and 121. Difference is 21, and number is 4 more than the lower number which is 100. Therefore we can say the square root is
10 (of 100) + 4/21 = 10.19
Cube of a number: when a number is multiplied three times with the same number, it is called the cube of a number.
Example: 4 × 4 × 4 = 64
Cube root: The cube root of a number is the number, which if multiplied three times by same number gives the given number.
Example: 
= 4. It is represented by
or with the power of 1/3, example (64)1/3 = (43)1/3 = 4.
To find the cube root of a number you have to find prime factors of the numbers, and deduce if in those numbers if a number is repeated thrice.
Example 5: 
= 
= 3 × 17 = 51
Complex numbers
Complex numbers are numbers with square root of a negative number. They were created as there is no root of a negative number, by assuming i (called iota) = √-1 , it was possible to do arithmetic operations on these numbers. A complex number is represented by (a + bi), where a and b are real numbers.
Since i = -[if !vml]
, i2 = –1, i3 = –1 × i = –i
and i4 = (i2)2 = (–1)2 = 1
Just like surds, to rationalize complex numbers, the rationalizing factor or conjugates are used like (a + ib) and (a – ib) are relative conjugates.
HCF AND LCM
HCF: HCF is the Highest common factor or greatest common divisor (GCD). Actually GCD explains it well, that is the greatest division that divides given set of numbers. Example: HCF of 10, 15 and 30 = 5 and HCF of 15, 30 and 45 is 15. It is obvious to see in each case 5 and 15 are the highest numbers which can divide the three numbers.
To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the common term from them and multiplying them will give you the HCF. The HCF is 1 when no common prime factors are there, as 1 is the only number which divides the two and is the highest.
Example 6: Find the HCF of 24, 48, 102
Prime factors 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2
Common numbers = 2 × 3 = 6, therefore 6 is the HCF
LCM: LCM is Least common multiple. It represents the smallest number which is divisible by all of the given numbers.
Example: LCM of 3, 4 and 5 = 60, as it is smallest number divisible by them.
To find the LCM, resolve all the numbers into their prime factors, take the ones which are common and the ones which are left (uncommon) and multiplying them will give you the LCM of the number.
Example 7: Find the LCM of 24, 48, 102
Prime factors 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2
Common numbers = 2 × 3
Numbers left = 2 × 2 × 2 × 2 × 2 × 17
LCM = common numbers x numbers left
= 2 × 3 × 2 × 2 × 2 × 2 × 2 × 17 = 3264
Important Note:
1. For two numbers, LCM × HCF = product of two numbers
Example: LCM of 4, 5 = 20 and HCF is 1, 20 × 1 = 4 × 5, 20 = 20
2. HCF of fractions = HCF of numerators/ LCM of denominators
Example: Find HCF of 3/5 and 4/10 .
Here HCF of 3,4 is 1
And LCM of 5, 10 is 10
HCF of 3/5and 4/10 = 1/10
3. LCM of fractions = LCM of numerators/ HCF of denominators
Example: Find LCM of 3/5 and 4/10.
Here LCM of 3,4 is 12
And HCF of 5, 10 is 5
LCM of 3/5 and 4/10 = 12/5
4. HCF and LCM of decimals: To calculate the HCF and LCM of decimals, remove the decimals and convert them into non-decimals, by multiplying with 10 or 100 or…. Post that calculate the HCF and LCM in regular fashion and once you have the regular HCF and LCM, convert that number into decimal by dividing it with power of 10 with which you multiplied earlier.
Example 8: Find the HCF and LCM of 0.6, 0.9, 1.5
Convert the numbers by multiplying them with 10, therefore numbers are 6, 9, 15
HCF of 6, 9, 15 is 3, dividing by 10 it is 0.3
LCM of 6, 9, 15 is 90, dividing by 10 it is 9
For large numbers the way to find the HCF is using Euclid’s Algorithm, but it works for two numbers only. From the larger number, subtract the biggest multiple of the smaller number without getting a negative answer. Replace the larger number with the answer and repeat this until the last number is zero, and the HCF is the next-to-last number computed.
Example 9: What is the HCF of 347236 and 297228?
347236 – 1 × 297228 = 50008
297228 – 5 × 50008 = 47188
50008 – 1 × 47188 = 2820
47188 – 16 × 2820 = 2068
2820 – 1 × 2068 = 752
2068 – 2 × 752 = 564
752 – 1 × 564 = 188
564 – 3 × 188 = 0
The HCF here is 188. Once we have the HCF, the LCM is the product of the two numbers divided by the HCF as per the properties mention on LCM and HCF. Therefore LCM of 347236 and 297228 = 347236 × 297228/188 = 549090936.
Example 10: Find the LCM of 32 and 40 if their HCF is 8.
LCM × HCF = Product of two numbers
LCM = 32 × -[if !vml] = 160
Example 11: One ice cream truck visits Rahul’s neighborhood every 4 days and another ice cream truck visits her neighborhood every 5 days. If both trucks visited today, when is the next time both trucks will visit on the same day?
Since both the trucks visited today, next they will visit together on the LCM of 4 and 5, which is 20, so on the 20th day from now
BASE SYSTEM
The base is the number of distinct symbols used in a number system; it can also be defined as the place value of a symbol in a system.
The normal number system is called the decimal system which has 10 digits, 0 to 9. For example if we take a number 385 in this system, it is given by
385 = 3 × 102 + 8 × 101 + 5 × 100
Here 10 is the termed as the base for the decimal system. There could be other systems with other bases; some other systems are binary, septenuary, octal, decimal and hexadecimal.
Binary system has two digits: 0, 1
Septenuary has seven digits: 0, 1, 2, 3, 4, 5, 6
Octal has 8 digits: 0, 1, 2, 3, 4, 5, 6, 7
Decimal obviously have 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Hexadecimal has 16 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.
In the hexadecimal system, A has value 10; B has value 11 and so on.
Conversions of numbers from one system to other: The counting sequences in each of the systems would be different though they follow the same principle. For instance, the sequence of the first few numbers on the number line starting with 0 are:
Decimal Binary Octal Hexadecimal
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 10111 13 B
15 11110 17 F
16 100001 20 10
17 100010 21 11
18 10010 22 12
Decimal into other systems
The number in decimal is consecutively divided by the number of the base to which we are converting the decimal number. Then list down all the remainders in the reverse sequence to get the number in that base.
Example 12: Convert (142)10 to base 8 system.
Looking at the remainder is the reverse order 216
So here (122)10 = (216)8
Example 13: Convert (14.625)10 to binary system
First take the non-decimal part
So it comes as 1110, now lets take the decimal part. In converting the decimal, the parts (0.625, 0.25, 0.5. 0.0) have to be multiplied by the base and the integers of the products have to be taken from first to last, here is how it will be done of the current case:
0.625 × 2 = 1.25
0.25 × 2 = 0.50
0.50 × 2 = 1.0
0.0 × 2 = 0.0
Going from first to last for the integer part of the products, 1010, so this is the decimal conversion.
Total conversion (14.625)10 = (1110.1010)2
Example 14: Convert (270)10 to hexadecimal system
The solution is not 1014, it is 10E, where E = 14 in the hexadecimal system.
Binary to decimal
To convert any system into decimal, the base with the power place value has to multiplied with the number on that place value and added. So if in a binary system, which is a base system, there is a number for example 101, then it will be represented in decimal system the following manner:
101 = 1 × 22 + 0 × 21 + 1 × 20
(101)2 = (10)10
Octal to decimal
Example 15: Convert (231)8 into decimal system.
Here 231 = 2 × 82 + 3 × 81 + 1 × 80
= 1 + 24 + 128 = (153)10
Hexadecimal to decimal
Example 16: Convert (1AB )16 into decimal system
(1AB)16 = 1 × 162 + A × 161 + B × 160
= 256 + 160 + 11 = 427
Hence (1AB)16 = (427)10
Binary to Octal
To convert binary into octal, make sets from left to eight of three in the given binary number, eg for 101111, the sets will be 101 and 111, for numbers not multiple of 3, add leading zeros from starting digit for make it multiple of three, e.g. for 1111, make it 001111, and then the sets are 001 and 111. Once the sets are made, get the decimal equivalent of each group and multiply the equivalent.
Example 17: (101111)2
= (101)2(111)2
= (1 × 22 + 0 × 21 + 1 × 20)(1 × 22 + 1 × 21 + 1 × 20)
= (5)(7) = (35)8
Binary into Hexadecimal
In this conversion, the process is absolutely same as in case of “Binary to Octal”, the difference is that here we need to make sets of 4 digits, eg for 11001111, the sets will be 1100 and 1111.
Example 18: (11001111)2 = (1100)(1111)
= (1 × 23 + 1 × 22 + 0 × 21 + 0 × 20)
(1 × 23 + 1 × 22 + 1 × 21 + 1 × 20)
= (12)(7)
= (C7)16
ARITHMETIC IN THE BASE SYSTEMS
For any arithmetic operation of numbers in base systems, follow the steps:
1. Convert the digit into decimal system
2. Do the arithmetic operation
3. Convert the result back into the original base system.
For addition you can also do it directly as in the example below:
Example 19: Add (330)8 and (355)8
330
355
705
Addition steps
Step 1: Add in decimal system 0 + 5 = 5, converting in Octal
5 = 8(0) + 5, so you place 5
tep 2: Again add in decimal system 5 + 3 = 8, converting in Octal
8 = 8(1) + 0, so you place 0, and carry forward 1
Step 3: Again add in decimal system 3 + 3 + 1 (carried forward) = 7
7 = 8(0) + 7, so you place 7
So (330)8 + (355)8 = (705)8
The Roman system of notation
In this system, the symbols I, V, X, L, C, D and M are used to denote resp 1, 5, 10, 50, 100, 500 and 1000. a bar is placed over a symbol multiplies its value by one thousand.
Thus D = 500,000
The following rules should be noted:
The symbols I, X, C, M may be repeated twice or thrice, i.e. II = 2, III = 3, XX = 20, XXX = 30, CC = 200, CCC = 300, MM = 2000, MMM = 3000
When a smaller number is placed to the right of the greater number, it is added to the greater.
Thus VI = 6, XV = 15, CX = 110
When any one of the numbers I, X, C is placed on the left of the greater number , it is subtracted from the greater.
Thus IV = 4, XL = 40, XC = 90
Some common numbers are:
I 1 XV=15
II 2 XVI=16
III 3 XVII=17
IV 4 XVIII=18
V 5 XIX=19
VI 6 XX=20
VII 7 XXX=30
VIII 8 XL=40
IX 9 L=50
X 10 LX=60
XI 11 LXX=70
XII 12 LXXX=80
XIII 13 XC=90
XIV 14 C=100
Here are some basic formulae used in solving various mathematical problems:
1. (a + b)2 = a2 + 2ab + b2
2. (a + b)2 – (a – b)2 = 4ab
3. (a + b)2 + (a – b)2 = 2 (a2 + b2)
4. a2 – b2 = (a + b) (a – b)
5.(a + b)3 = a3 + b3 + 3ab (a + b)
6. (a – b)3 = a3 – b3 – 3ab (a – b)
7. a3 + b3 = (a + b) (a2 – ab + b2)
8. a3 – b3 = (a – b) (a2 + ab + b2 )
9. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – ac – bc)
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