Number Systems - Introduction Quant Notes | EduRev

Quantitative Aptitude (Quant)

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INTRODUCTION

Numbers form an integral part of our lives. In this lesson we will learn about the different types of numbers and the different categories under which they fall. The concepts discussed in this lecture will be your first step towards a general understanding of the mathematics requirements to clear MBA entrance exams. As we proceed with this lecture, you will realize that you have already learnt many of the concepts, included in this lesson, in school. This would further help build confidence in you. Although Number theory is important in the context of all the MBA entrance exams, it gains all the more importance for the students aiming for success in the CAT.


Understanding Numbers
A measurement carried out, of any quantity, leads to a meaningful value called the Number. This value may be positive or negative depending on the direction of the measurement and can be represented on the number line.


Natural Numbers (N)
The numbers 1, 2, 3, 4, 5…are known as natural numbers. The set of natural numbers is denoted by N.
Hence, N = {1, 2, 3, 4…}. The natural numbers are further divided as even, odd, prime etc.

 

Whole Numbers (W)
All natural numbers together with ‘0’ are collectively called whole numbers. The set of whole numbers is denoted by W, and W = {0, 1, 2, 3, ……}

Integers (Z)
The set including all whole numbers and their negatives is called a set of integers. It is denoted by Z, and Z = {- ∞, … - 3, - 2, - 1, 0, 1, 2, 3, ……. ∞}. They are further classified into Negative integers, Neutral integers and positive integers.

Negative integers (Z-)
All integers lesser than Zero are called negative integers.
Z = {- 1, - 2, - 3…- ∞ }

Neutral integers (Z0)
Zero is the only integer which is neither negative nor positive and it is called a neutral integer.

Positive integers (Z+)
All integers greater than Zero are called positive integers.
Z+ = {1, 2, 3, …….., ∞ }.


Classification of Numbers
(i) Even Numbers: All numbers divisible by 2 are called even numbers. E.g., 2, 4, 6, 8, 10 …Even numbers can be expressed in the form 2n, where n is an integer. Thus 0, - 2, − 6, etc. are also even numbers.
(ii) Odd Numbers: All numbers not divisible by 2 are called odd numbers. e.g. 1, 3, 5, 7, 9…Odd numbers can be expressed in the form (2n + 1) where n is any
integer. Thus - 1, − 3, − 9 etc. are all odd numbers.
(iii) Prime Numbers: A natural number that has no other factors besides itself and unity is a prime number.
Examples: 2, 3, 5, 7, 11, 13, 17, 19 …


Important Observation about prime numbers:
A prime number greater than 3, when divided by 6 leaves either 1 or 5 as
the remainder. Hence, a prime number can be expressed in the form of 6K
± 1. But the converse of this observation is not true, that a number leaving a
remainder of 1 or 5 when divided by 6 is not necessarily a prime number. For eg: 25, 35 etc

 TIP: If a number has no prime factor equal to or less than its square root, then the number is a prime number.

Must Know
1 to 25 ⇒ 9 prime
1 to 50 ⇒ 15 prime
1 to 100 ⇒ 25 prime
1 to 200 ⇒ 45 prime

Example 1. If a, a + 2, a + 4 are consecutive prime numbers. Then how many solutions ‘a’ can have?
(1) One 

(2) Two 

(3) Three 

(4) More than three
Ans. (1)
Solution. 
No even value of ‘a’ satisfies this. So ‘a’ should be odd. But out of three consecutive odd numbers,
atleast one number is a multiple of 3.
So, only possibility is a = 3 and the numbers are 3, 5, 7.


(iv) Composite Numbers: A composite number has other factors besides itself
and unity. e.g. 8, 72, 39, etc. On the basis of this fact that a number with
more than two factors is a composite we have only 34 composite from 1 to
50 and 40 composite from 51 to 100.

TIP: 1 is nether a prime number nor a composite number

(v) Perfect Numbers: A number is said to be a perfect number if the sum of ALL its factors excluding itself (but including 1) is equal to the number itself.

Or

The sum of all the possible factors of the number is equal to twice the number.

FUNDA: If the factors of any perfect number other than 1 are written and reciprocal of them are added together then result is always unity.

Example:
6 is a perfect number because the factors of 6, i.e., 1, 2 and 3 add up to the number 6 itself.
Also   1/6 + 1/3 + 1/2 = (1+ 2+3/6) = 6/8 = 1 (Unity)
Other examples of perfect numbers are 28, 496, 8128, etc. There are 27 perfect numbers discovered so far.


(vi) Co-Prime numbers:
Two numbers are (prime or composite) said to be co-prime to one another, if
they do not have any common factor other than 1. e.g. 35 & 12, since they
both don’t have a common factor among them other than 1.

 TIP: The HCF of Co-prime number is always 1.

(vii) Fractions
A fraction denotes part or parts of a unit. Several types are:

  • Common Fraction: Fractions whose denominator is not 10 or a multiple of it. e.g. 2/3, 17/18
    Decimal Fraction: Fractions whose denominator is 10 or a multiple of 10.
  • Proper Fraction: In this the numerator < denominator e.g. 2/10, 6/7, 8/9 etc. Hence its value < 1.
  • Improper Fraction: In these the numerator > denominator e.g. 10/2, 7/6, 8/7  etc. Hence its value > 1.
  • Mixed Fractions: When a improper fraction is written as a whole number and proper fraction it is called mixed fraction. e.g. 7/3 can be written as 2 + 1/3 = 2(1/3)

Rational Numbers
Rational Number is defined as the ratio of two integers i.e. a number that can be represented by a fraction of the form p/q where p and q are integers and q ≠ 0.They also can be defined as the non-terminating recurring decimal numbers. Such as 3.3333…., 16.123123….. are all rational numbers as they can be expressed in the
form p/q.
Examples: Finite decimal numbers, whole numbers, integers, fractions i.e.
3/5, 16/9, 0.666...∞, 2/3, 7, 0 etc.

Irrational Numbers
Any number which can not be represented in the form p/q where p and q are integers and q ≠ 0 is an irrational number. On the basis of non-terminating decimals, irrational numbers are non-terminating non recurring decimals. Such as 3.4324546345……. is a non-terminating, non-repeating number.
Examples: π, √5, √7, e

Non-Terminating Decimal Numbers: 

When we divide any number by other number, either we get a terminating number or a non - terminating number. A non - terminating number on the basis of occurrence of digits after decimal can classified as following:

  • Pure Recurring Decimals: A decimal in which all the figures after the decimal point repeat, is called a pure recurring decimal.
    Examples: 0.6 , 0Number Systems - Introduction Quant Notes | EduRev are examples of pure recurring decimals. (0.6 = 0Number Systems - Introduction Quant Notes | EduRev = 0.666 ……….)
  • Mixed Recurring Decimals: A decimal in which some figures do not repeat and some of them are repeated is called a mixed recurring decimal.
    Examples:Number Systems - Introduction Quant Notes | EduRev            Number Systems - Introduction Quant Notes | EduRev
  • Non - Recurring Decimals: A decimal number in which the figure don’t repeat themselves in any pattern are called non-terminating non- recurring decimals and are termed as irrational numbers.
  • Converting Recurring Decimal as Fraction: All recurring decimals can be converted into fractions. Some of the common types can be 0.33….. ,0.1232323…, 5.33…., 14.23636363…. etc.

Pure Recurring to Fractions
FUNDA 1. If a number is of the form of 0.ababab……. then divide the repeating digits with as many 9’s as we have repeated digits.

Example: 0.363636...= 36/99 = 4/11

Mixed Recurring to Fractions:

FUNDA 2. If N = 0.abcbcbc…. Then

N = abc - a/990 = Repeated & non Repeated digits - Non Repeated digits/ As many 9's as repeated digits followed by as many zero as non - repeated digits

eg. 0.25757...= 257 - 2/990 = 255/990 = 17/66
FUNDA 3. If N = a.bcbc…. Then
Write N = a + 0.bcbc….
Proceed as Funda 1
5.3636… = 5 + 0.3636… = 5 + 36/99 = 59/11
Divisibility Test

Number Systems - Introduction Quant Notes | EduRev


FUNDA

How to calculate remainder, when a number is divided by 11, without
division?

Step 1: Add all the odd place numbers (O) and even place numbers (E)
counted from right to left.

Step 2: If O - E is positive, remainder will be the difference less than 11.

Step 3: If O - E is negative, remainder should be (11 - difference).

PROPERTIES OF NUMBER

  • Every number has the same unit’s digit at its fifth power as it has at its first power, thus the standard method that can be followed is to divide the power given by 4, find the remaining power and check the unit’s digit in that number. This short cut can be applied because you will always get the same unit’s digit as otherwise.
  • For calculating number of zeroes at the end of factorial of a number, you should divide the number by 5, the quotient obtained is again divided by 5 and so on till the last quotient obtained is smaller than 5. The sum of all the quotients is the number of 5s, which then becomes the number of zeroes in the given number.
  • The digital root of a number is the sum of the digits, over and over again, till it becomes a single digit number. For example, the digital root of 87984 will be 8 + 7 + 9 + 8 + 4 ⇒ 36 = 3 + 6 ⇒ 9.
  • When the concept of Euler number is used and the dividend and divisor happen to be co-prime, the remainder questions become very easy.
  • The product of 3 consecutive natural numbers is divisible by 6.
  • The product of 3 consecutive natural numbers, the first of which is an even number is divisible by 24.
  • The sum of a two-digit number and a number formed by reversing its digits is divisible by 11. E.g. 28 + 82 = 110, this is divisible by 11. At the same time, the difference between those numbers will be divisible by 9. e.g. 82 – 28 = 54, this is divisible by 9.
  • ∑n = n(n+1)/2, ∑n is the sum of first n natural numbers.
  • ∑n2 = n(n+1)(n+2)/6, ∑n2 is the sum of first n perfect squares.
  • ∑n3 = n2(n+1)2/4 = (∑n)2, ∑n3 is the sum of first n perfect cubes.
  • xn + yn = (x + y) (xn-1 - xn-2.y + xn-3.y2 - ... +yn-1) when n is odd. Therefore, when n is odd, xn + yn is divisible by x + y. e.g. 33 + 23 = 35 and is divisible by 5, which is (3 + 2).
  • xn - yn = (x + y) (xn-1 - xn-2.y + ... yn-1) when n is even. Therefore, when n is even, xn - yn is divisible by x + y. e.g. 72 - 32 = 40, is divisible by 10, which is (7+3).
  • xn - yn = (x - y) (xn-1 + xn-2.y + .... + yn-1) for both odd and even n. Therefore, xn - yn is divisible by x - y. For example: 94 - 24 = 6545, is divisible by 7, which is (9 – 2).

CAT NUMBER SYSTEM QUESTIONS
Example 1. N = (18n2 + 9n + 8)/n; where N belongs to integer. How many integral solutions of N are possible?
Solution. The given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n. Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8. The integers, which will satisfy this condition are ±1, ±2, ±8, ±4. Thus, in total,n can take 8 values.

Example 2. N = 960. Find total number of factors of N.
Solution. The number of divisors of a composite number: If D is a composite number in the form D = ap × bq × cr, where a, b, c are primes, then the no. of divisors of D, represented by n is given by n = (p+1)(q+1)(r +1). Following the same, after dividing 960 into prime factors: 26 × 31 × 51, we can calculate the total number of factors as: (6+1).(1+1).(1+1) = 28.

Example 3. Find the unit digit of the following expression: (123)34 × (876)456 × (45)86.
Solution. Whenever an even unit digit and a 5 at the unit digit are present, they will always give a 0 at the unit digit, no matter if any other number is present or not. Hence, with this approach, this question proves to be a sitter.

Example 4. What will be the number of zeroes at the end of the product of the first 100 natural numbers?
Solution. In this kind of questions, you need to find greatest power of 5, which can divide the product of the first 100 natural numbers. Remember, a multiple of 5 multiplied by any even number, gives you a zero. Now divide 100 by 5 and take 20 as quotient. Then divide 20 (the quotient) by 5 and get the new quotient 4, which further cannot be divided by 5. The sum of all such quotient gives you the greatest power of 5, which can divide that number. The sum is 24 and this is the number of zeroes at the end of the product of the first 100 natural numbers.

Example 5. Which letter should replace the $ in the number 2347$98, so that it becomes a multiple of 9?
Solution. As you know that if the sum of all the digits is divisible by 9, then the number is divisible by 9. Now sum of the given digits is 2 + 3 + 4 + 7 + 9 + 8 = 33 + $. Now think the next multiple of 9 after 33 i.e. 36. This means you add 3 in this. The value of $ is 3.

Example 6. In a party there are 20 persons present. If each of them shakes hand with all the other persons,in total how many handshakes will take place?
Solution. Out of 20 persons, the first person will shake hand with 19 persons. The second person will shake hand with 18 persons (because he has already shaken hand with first person). The third person will shake hand with 17 persons and so on. The second last person shakes hand with only one person. And last will shake hand with none (because he has already shaken hand with all persons). In order to find the total number of handshakes you have to add all the natural numbers from 1 to 19 i.e. ∑ 19. ∑19 = 19 x 20/2 = 190 handshakes.

Example 8. What is the remainder when 2354789341 is divided by 11?
Solution. Odd place digit sum (O) = 1 + 3 + 8 + 4 + 3 = 19.
Even place digits sum (E) = 4 + 9 + 7 + 5 + 2 = 27.
Difference (D) = 19 - 27 = - 8
Remainder = 11 - 8 = 3.

TIP: ⇒ When any number with even number of digits is added to its reverse, the sum is always divisible by 11. e.g. 2341 + 1432 = 3773, which is divisible by 11.
⇒ Any number written 6 times consecutively will be divisible by 7 and 13.

Example 9. If 567P55Q is divisible by 88; Find the value of P + Q.
(a) 11 

(b) 12 

(c) 5 

(d) 6 

(e) 10
Ans. (e)
Solution. 
The number is divisible by 8 means; the number formed by the last 3 digits should be divisible by 8 which are 55 Q. Only Q = 2 satisfies this. From the divisibility rule of 11, (2 + 5 + 7 + 5) - (5 + P + 6) is divisible by 11. So 8-P is divisible by 11. if P= 8, then only it is possible. So P = 8 and Q = 2.
So P + Q = 10.


Example 10. If the first 100 natural numbers are written side by side to form a big number and it is divided by
 8. What will be the remainder?
(a) 1 

(b) 2 

(c) 4
(d) 7 

(e) Cannot be determined
Ans. (c)

Solution. 
The number is 1234…..9899100
According to the divisibility rule of 8, we will check only the last 3 digits.
If 100 is divided by 8, the remainder is 4.

Example 11. What will be the remainder when 4444……..44 times is divided by 7?
(a) 1 

(b) 2 

(c) 5 

(d) 6 

(e) 0
Ans. (b)

Solution. 
If 4 is divided by 7, the remainder is 4.
If 44 is divided by 7, the remainder is 2.
If 444 is divided by 7, the remainder is 3
By checking like this, we come to know that 444444 is exactly divisible by 7.
So if we take six 4’s, it is exactly divisible by 7. Similarly twelve 4’s is also exactly divisible by 4 and 42 4’s will be exactly divisible by 7. So out of 44, the remaining two 4,s will give a remainder of 2.


Absolute value of a number:
The modulus of a number is the absolute value of the number or we can say the distance of the number from the origin. The absolute value of a number a is defined as
|a| = a, if a ≥ 0
= - a, if a ≤ 0
|a| is read as MODULUS of a pr Mod a

Example: |79| = 79 & | - 45| = - (- 45) = 45
Also, | x - 3 | = x - 3, if x ≥ 3
= 3 - x, if x < 3.


Always Keep in Mind
The number 1 is neither prime nor composite.

  • 2 is the only even number which is prime.
  • (xn + yn) is divisible by (x + y), when n is an odd number.
  • (xn - yn) is divisible by (x + y), when n is an even number.
  • (xn - yn) is divisible by (x - y), when n is an odd or an even number.
  • The difference between 2 numbers (xy) - (yx) will always be divisible by 9.
  • The square of an odd number when divided by 8 will always give 1 as a remainder.
  • Every square number is a multiple of 3 or exceeds a multiple of 3 by unity.
  • Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.
  • If a square number ends in 9, the preceding digit is even.
  • If m and n are two integers, then (m + n)! is divisible by m! n!
  • (a)n / (a + 1) leaves a remainder of   Number Systems - Introduction Quant Notes | EduRev
  • Product of n consecutive numbers is always divisible by n!.

Example 1. If ‘X’ is an even number; Y is an odd number, then which of the following is even?
(a) X+ Y 

(b) X + Y2 

(c) X2 + Y2 

(d) X2Y2 

(e) None of these
Ans. (d)
Solution. 
Since X is even, X2 is even.
Y is odd, Y2 is odd
So options (1), (2), (3) are even + odd = odd.
Option (4) is (even) (odd) = Even.


Example 2. What is the difference between 0.343434....…and 0.2343434…… in fraction form?
(a) 6/55
(b) 6/11
(c) 9/55
(d) 9/13
(e) 5/11

Ans. (a)

Solution.
0.343434.....= 34/99 and 0.23434.... = 234 - 2/990 = 232/990
∴ Difference = 34/99 - 232/990 = 108/990 = 6/55

Example 3. How many of the following numbers are divisible by at least 3 distinct prime numbers 231, 750, 288 and 1372?
(a) 0 

(b) 1 

(c) 2 

(d) 3 

(e) 4
Ans. (c)

Solution. 
231 = 3 × 7 × 11 (3 prime factors)
750 = 2 × 3 × 53 (3 prime factors)
288 = 25 × 32 (only 2 prime factors)
1372 = 22 × 73 (only 2 prime factors)
So, only 231 & 750 has 3 prime factors.


Example 4. n3 + 6n2 + 11n + 6 (where n is a whole no) is always divisible by
(a) 4 

(b) 5 

(c) 6 

(d) 8 

(e) 12
Ans. (c)

Solution. 
n3 + 6n2 + 11n + 6 = (n + 1) (n + 2) (n + 3).
Product of 3 consecutive numbers is always divisible by 3! = 6.

(or)

Take n = 0, 1, 2, 3 and check it is always divisible by 6.


Example 5. What is the remainder, if 351 × 352 × 353 × - - - - - - - × 356 is divided by 360?
(a) 0 

(b) 1 

(c) 2 

(d) 3 

(e) 359
Ans. (a)

Solution. 
Since the given is the product of 6 consecutive numbers, it is always divisible by 6! = 720.
⇒ It is divisible by 360 also. So, the remainder will be 0.

CAT NUMBER SYSTEM QUESTIONS
Example 1. N = (18n2 + 9n + 8)/n; where N belongs to integer. How many integral solutions of N are possible?
Solution. The given expression can be broken as: 18n2/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n. Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n only ⇒ n can have any integral value, which is a factor of 8. The integers, which will satisfy this condition are ±1, ±2, ±8, ±4. Thus, in total,n can take 8 values.

Example 2. N = 960. Find total number of factors of N.
Solution. The number of divisors of a composite number: If D is a composite number in the form D = ap × bq × cr, where a, b, c are primes, then the no. of divisors of D, represented by n is given by n = (p + 1)(q + 1)(r + 1). Following the same, after dividing 960 into prime factors: 26 × 31 × 51, we can calculate the total number of factors as: (6 + 1).(1 + 1).(1 + 1) = 28.

Example 3. Find the unit digit of the following expression: (123)34 × (876)456 × (45)86.

Solution. Whenever an even unit digit and a 5 at the unit digit are present, they will always give a 0 at the unit digit, no matter if any other number is present or not. Hence, with this approach, this question proves to be a sitter.

Example 4. What will be the number of zeroes at the end of the product of the first 100 natural numbers?
Solution. In this kind of questions, you need to find greatest power of 5, which can divide the product of the first 100 natural numbers. Remember, a multiple of 5 multiplied by any even number, gives you a zero. Now divide 100 by 5 and take 20 as quotient. Then divide 20 (the quotient) by 5 and get the new quotient 4, which further cannot be divided by 5. The sum of all such quotient gives you the greatest power of 5, which can divide that number. The sum is 24 and this is the number of zeroes at the end of the product of the first 100 natural numbers.

Example 5. Which letter should replace the $ in the number 2347$98, so that it becomes a multiple of 9?
Solution. As you know that if the sum of all the digits is divisible by 9, then the number is divisible by 9. Now sum of the given digits is 2 + 3 + 4 + 7 + 9 + 8 = 33 + $. Now think the next multiple of 9 after 33 i.e. 36. This means you add 3 in this. The value of $ is 3.

Example 6. In a party there are 20 persons present. If each of them shakes hand with all the other persons,in total how many handshakes will take place?
Solution. Out of 20 persons, the first person will shake hand with 19 persons. The second person will shake hand with 18 persons (because he has already shaken hand with first person). The third person will shake hand with 17 persons and so on. The second last person shakes hand with only one person. And last will shake hand with none (because he has already shaken hand with all persons). In order to find the total number of handshakes you have to add all the natural numbers from 1 to 19 i.e. ∑ 19. ∑19 = 19 × 20/2 = 190 handshakes.

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