CBSE Class 10 Real Numbers Sure Shot Questions Set E

System of numbers

• Numbers
  • Real
  • Imaginary
• Real
  • Rational
    • Integer
      • Whole Number (0, 1, 2, ......)
        • Natural (1, 2, 3, 4, 5,.....)
          • Prime (2, 3, 5, 7, ...)
          • Composite (4, 6, 8, ...)
          • Perfect (6, 28 ...)
      • Negative
      • odd (\( \pm 1, \pm 3, \pm 5 \dots \))
      • Even (multiples of 2)
    • Fractions
      • Proper (1/3, 2/5, 3/8)
      • Improper (4/3, 5/2, 6/5 ect.)
      • Mixed (\( 1\frac{1}{4}, 3\frac{1}{15} \),.....etc).
  • Irrational (\( \sqrt{2}, \sqrt{3} \) etc)

Natural numbers

Counting numbers 1, 2, 3, 4, 5,.....are known as natural numbers. The set of all natural numbers can be represented by \( N = \{1, 2, 3, 4, 5 \dots \} \)

Whole numbers

If we include 0 among the natural numbers, then the numbers 0, 1, 2, 3, 4, 5.....are called whole numbers. The set of whole numbers can be represented by \( W = \{0, 1, 2, 3, 4, 5, \dots \} \)

Clearly, every natural number is a whole number but 0 is a whole number which is not a natural number.

Integers

All counting numbers and their negatives including zero are known as integers. The set of integers can be represented by \( Z \) or \( \{ \dots -4, -3, -2, -1, 0, 1, 2, 3, 4, \dots \} \)

Positive integers

The set \( I^+ = \{1, 2, 3, 4, \dots \} \) is the set of all positive integers. Clearly, positive integers and natural numbers are synonyms.

Negative integers

The set \( I^- = \{ -1, -2, -3, \dots \} \) is the set of all negative integers. 0 is neither positive nor negative.

Non-negative integers

The set \( \{ 0, 1, 2, 3, \dots \} \) is the set of all non-negative integers.

Rational numbers

The numbers of the form \( \frac{p}{q} \), where p and q are integers and \( q \neq 0 \), are known as rational numbers, e.g. \( \frac{4}{7}, \frac{3}{2}, \frac{5}{8}, \frac{0}{1}, -\frac{2}{3} \), etc. The set of all rational numbers is denoted by Q.

i.e. \( Q = \{ x : x = \frac{p}{q}; p, q \in I, q \neq 0 \} \)

Since every natural number a can be written as \( \frac{a}{1} \), so is a rational number. Since 0 can be written as \( \frac{0}{1} \) and every non-zero integer 'a' can be written as \( \frac{a}{1} \), so it is also a rational number.

Every rational number has a peculiar characteristic that when expressed in decimal form is expressible either in terminating decimals or non-terminating repeating decimals.

For example, \( \frac{1}{5} = 0.2, \frac{1}{3} = 0.333\dots, \frac{22}{7} = 3.1428714287\dots, \frac{8}{44} = 0.18188\dots \), etc.

The recurring decimals have been given a short notation as \( 0.333\dots = 0.\bar{3} \), \( 4.1555\dots = 4.1\bar{5} \), \( 0.323232\dots = 0.\overline{32} \).

Irrational numbers

Those numbers which when expressed in decimal form are neither terminating nor repeating decimals are known as irrational numbers, e.g. \( \sqrt{2}, \sqrt{3}, \sqrt{5}, \pi \), etc.

• Note, that the exact value of \( \pi \) is not \( \frac{22}{7} \). \( \frac{22}{7} \) is rational while \( \pi \) is irrational numbers. \( \frac{22}{7} \) is approximate value of \( \pi \). Similarly, 3.14 is not an exact value of it.

Real numbers

The rational and irrational numbers combined together are called real numbers, e.g. \( \frac{13}{21}, \frac{2}{5}, -\frac{3}{7}, \sqrt{3}, 4 + \sqrt{2} \), etc. are real numbers. The set of real numbers is denoted by R.

• Note, that the sum, difference or product of a rational and irrational number is irrational, e.g. \( 3 + \sqrt{2}, 4 - \sqrt{3}, \frac{2}{3} - \sqrt{5}, 4\sqrt{3}, -7\sqrt{5} \) are all irrational.

Even numbers

All those numbers which are exactly divisible by 2 are called even numbers, e.g. 2, 6, 8, 10 etc. are even numbers.

Odd numbers

All those numbers which are not exactly divisible by 2 are called odd numbers, e.g. 1, 3, 5, 7, etc, are odd numbers.

Prime numbers

Except 1 each natural number which is divisible by only 1 and itself is called as prime number e.g., 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,.....etc.

• There are total 25 prime numbers upto 100
• There are total 46 prime numbers upto 200
• 2 is the only even prime number and the least prime number.
• 1 is neither prime nor composite number.
• There are infinite prime numbers.

Table of prime Numbers (1-100):

2 11 23 31 41 53 61 71 83 97
3 13 29 37 43 59 67 73 89
5 17 47 79
7 19

Test to find whether a given number is a prime

• Step 1 Select a least positive integer n such that \( n^2 > \) given number.
• Step 2 Test the divisibility of given number by every prime number less than n.
• Step 3 The given number is prime only if it is not divisible by any of these primes.

Question. Investigate whether 571 is a prime number.
Answer: Sol. Since \( (23)^2 = 529 < 571 \) and \( (24)^2 = 576 > 571 \). Therefore \( n = 24 \). Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. Since 571 is not divisible by any of these primes, 571 is a prime number.

Co-prime

A pair of two natural numbers having no common factor, other than 1, is called a pair of co-primes. For example: (3, 5), (4, 5), (5, 6), (7, 9), (6, 7) etc., are co-primes.

Twin primes

Prime numbers differing by two are called twin primes, e.g. (3, 5), (5, 7), (11, 13) etc, are called twin primes.

Prime triplet

A set of three consecutive primes differing by 2, such as (3, 5, 7) is called a prime triplet. "every prime number except 2 is odd but every odd number need not be prime."

Fractions

(a) Common fraction : Fractions whose denominator is not 10.

(b) Decimal fraction : Fractions whose denominator is 10 or any power of 10.

(c) Proper fraction : Numerator < Denominator i.e. \( \frac{3}{5} \).

(d) Improper fraction : Numerator > Denominator i.e. \( \frac{5}{3} \).

(e) Mixed fraction : Consists of integral as well as fractional part i.e. \( 3\frac{2}{7} \).

(f) Compound fraction : Fraction whose number and denominator themselves are fractions i.e. \( \frac{2/3}{5/7} \).

Composite numbers

All natural numbers, which are not prime are composite numbers. If C is the set of composite number than \( C = \{4, 6, 8, 9, 10, 12 \dots \} \)

Imaginary numbers

All the numbers whose square is negative are called imaginary numbers. e.g. \( 3i, -4i, \dots \); where \( i = \sqrt{-1} \).

Complex numbers

The combined form of real and imaginary numbers is known as complex number. It is denote by \( Z = A + iB \), where A is real and B is imaginary part of Z and \( A, B \in R \).

Square & Square roots

The second power of number is called the square of that number. In other words the square of a number is the product of the number with the number itself. A given number is a perfect square, if it is expressed as a product of pairs of equal factors.

Important properties

(i) A natural number having 2, 3, 7 or 8 in the unit's place is never a perfect square (or squared number). 17, 23, 118, 222 are not perfect squares.

(ii) The square of an even number is always an even number. \( 2^2=4, 6^2=36, 10^2=100, 12^2=144 \).

(iii) The square of an odd number is always an odd number. \( 3^2=9, 7^2=49, 13^2=169, 15^2=225 \).

(iv) The number of zeroes at the end of a perfect square is never odd. 100, 400, 3600, 640000 are perfect squares and 1000, 4000, 6400000 are not perfect squares.

(v) The square of a natural number n is equal to the sum of the first n odd numbers.

\( 1^2=1= \) sum of the first 1 odd number.

\( 2^2=1+3 = \) sum of the first 2 odd numbers.

\( 3^2=1+3+5= \) sum of the first 3 odd numbers.

(vi) For every natural numbers n, \( (n+1)^2 – n^2 = (n+1+n) (n+1–n) = (n+1)+n \).

\( 4^2–3^2 = (3+1)+3=7 \).

\( 16^2–15^2= (15+1)+15=31 \).

(vii) A perfect square (other than 1) is either a multiple of 3 exceeds a multiple of 3 by 1. \( 49 = (7)^2= 3\times 16+1, 169 = (13)^2 = 3 \times 56 + 1 \).

(viii) A perfect square (other than 1) is either a multiple of 4 or exceeds a multiple of 4 by 4. \( 441 = (21)^2 = 4 \times 110+1 \).

Some other properties

(i). If the unit digit of the number is zero then the unit digit of the square of this number will also be zero and the number of zeros will be double in the square than that of its root. e.g., \( (60)^2= 3600, (130)^2 = 16900 \)

(ii) If the unit digit of the number is 5 then the unit digit of its square is also 5 and the number formed by last two digits is 25. e.g., \( (35)^2= 1225, (45)^2= 2025, (55)^2= 3025 \) etc.

(iii) If the unit digit of any number is 1 or 9 then the unit digit of the square of its number is always 1. e.g., \( (71)^2 = 5041, (31)^2 = 961, (19)^2 = 361 \)

(iv) If the unit digit of any number is 2 or 8 then the unit digit of the square of its number is always 4.

(v) If the unit digit of any number is 3 or 7 then the unit digit of its square is always 9. e.g., \( (23)^2 = 529, (27)^2 = 729 \)

(vi) If the unit digit of any number is 4 or 6 then the unit digit of its square is always 6. e.g., \( (26)^2 = 676, (24)^2 = 576, (14)^2 = 196, (16)^2 = 256 \) etc.

(vii) The square of any number is always positive irrespective of the nature of the given number.

(viii) Non square numbers between the squares of two consecutive natural numbers n & n +1 \( \to (n + 1)^2 – n^2 – 1 = n^2 + 1 + 2n – n^2 – 1 = 2n \).

(ix) If a natural number cannot be expressed as a sum of successive odd natural numbers starting with 1, then it is not a perfect square.

(x) Square root of a negative integer is an imaginary number.

(xi) Square roots of integers that are not perfect squares are always irrational numbers.

(xii) Every composite number can be uniquely factored as a product of prime numbers only.

(xiii)If a perfect square is of n-digits, then its square root will have \( n/2 \) digits if n is even or \( \frac{(n + 1)}{2} \) if n is odd.

Cube & cube roots

Cube of a number is obtained by multiplying the number itself thrice. For example, 27 is the cube of 3 as \( 27 = 3 \times 3 \times 3 \).

Cube root

Cube root of a given number is that number which when raised to the third power produces the given numbers, that is the cube root of a number x is the number whose cube is x. The cube root of x is written as \( \sqrt[3]{x} \). For example, cube root of 64 is 4 as \( 4 \times 4 \times 4 = 64 \).

Short-cut method of finding cube roots of exact cubes consisting of up to 6 digits:

Before we discuss the method to find the cube roots of exact cubes, the following two remarks are very useful

(i) \( 1^3= 1; 2^3= 8; 3^3 = 27; 4^3= 64; 5^3= 125; 6^3=216; 7^3= 343; 8^3= 512; 9^3 = 729; 10^3=1000 \).

(ii)

If the cube ends in 1, then its cube root ends in 1

If the cube ends in 2, then its cube root ends in 8

If the cube ends in 3, then its cube root ends in 7

If the cube ends in 4, then its cube root ends in 4

If the cube ends in 5, then its cube root ends in 5

If the cube ends in 6, then its cube root ends in 6

If the cube ends in 7, then its cube root ends in 3

If the cube ends in 8, then its cube root ends in 2

If the cube ends in 9, then its cube root ends in 9

If the cube ends in 0, then its cube root ends in 0

Some other properties

(i) Cubes of all odd natural numbers are odd.

(ii) Cubes of all even natural numbers are even.

(iii) The cube of a natural number which is a multiple of 3 is a multiple of 27.

(iv) The cube of a natural number which of the form 3n+1 (e.g., 4, 7, 10.........) is also a number of the form 3n+1.

(v) The cube of natural number which is of the form 3n+2 (e.g., 5, 8, 11,.....) is also a number of the form 3n+2.

Squares (short-cut methods)

Question. To square any number ending with 5.
Answer: Method : \( (A5)^2 = A(A+1)/25 \)

Question. To square a number in which every digit is one.
Answer: Method: Count the number of digits in the given number and start writing numbers in ascending order from one to this number and then in descending order up to one.

Question. To square number which is nearer to 10x.
Answer: Method: Use the formula. \( x^2 = (x^2–y^2) + y^2 = (x+y)(x–y)+y^2 \)

Question. Find the squares of following:
(i) \( (25)^2 \)
(ii) \( (45)^2 \)
(iii) \( (85)^2 \)
(iv) \( 11^2 \)
(v) \( 111^2 \)
(vi) \( 1111^2 \)
(vii) \( 222^2 \)
(viii) \( 3333^2 \)
(ix) \( (97)^2 \)
(x) \( (102)^2 \)
(xi) \( (994)^2 \)
(xii) \( (1005)^2 \)
Answer: Sol. (i) \( (25)^2 = 2 (2+1)/25=6/25 = 625 \)
(ii) \( (45^2) = 4 (4+1)/25 = 20/25 = 2025 \)
(iii) \( (85)^2 = 8 (8+1)/25 = 72/25 = 7225 \)
(iv) \( 11^2 = 121 \)
(v) \( 111^2 = 12321 \)
(vi) \( 1111^2 = 1234321 \)
(vii) \( 222^2 = 2^2(111)^2 = 4 (12321) = 49284 \)
(viii) \( 3333^2 = (1111)^2 \times 9 = 9 (1234321) = 11108889 \)
(ix) \( (97)^2 = (97+3) (97–3)+3^2 = 9400+9=9409 \)
(x) \( (102)^2 = (102–2) (102+2)+2^2 = 10400+4=10404 \)
(xi) \( (994)^2 = (994+6) (994–6) + 6^2 = 988000 + 36 = 988036 \)
(xii) \( (1005)^2 = (1005–5) (1005+5) + 5^2 = 1010000 + 25 = 1010025 \)

Multiplication (short-cut methods)

Question. Multiplication of a given number by 9, 99, 999, etc., that is by \( 10^n–1 \).
Answer: Method: Put as many zeros to the right of the multiplication as there are nines in the multiplier and from the result subtract the multiplicant and get the answer.

Question. Multiply (i) 3893 by 99 (ii) 4327 by 999 (iii) 5863 by 9999
Answer: Sol. (i) \( 3893 \times 99 = 389300 – 3893 = 385407 \).
(ii) \( 4327 \times 999 = 4327000 – 4327 = 4322673 \)
(iii) \( 5863 \times 9999 = 58630000 – 5863 = 58624137 \)

Question. Multiplication of a given number by 11, 101, 1001, etc., that is by \( 10^n+1 \).
Answer: Method: Place n zeros to the right of the multipliciant and then add the multiplicant to the number so obtained.

Question. Multiply (i) \( 4782 \times 11 \) (ii) \( 9836 \times 101 \) (iii) \( 6538 \times 1001 \)
Answer: Sol. (i) \( 4782 \times 11 = 47820 + 4782 = 52602 \)
(ii) \( 9836 \times 101 = 983600 + 9836 = 993436 \)
(iii) \( 6538 \times 1001 = 6538000 + 6538 = 6544538 \)

Question. Multiplication of a given number by 15, 25, 35, etc.
Answer: Method: Double the multiplier and then multiply the multiplicant by this new number and finally divide the product by 2.

Question. Multiply (i) \( 7054 \times 15 \) (ii) \( 3897 \times 25 \) (iii) \( 4563 \times 35 \)
Answer: Sol. (i) \( 7054 \times 15 = \frac{1}{2} (7054 \times 30) = \frac{1}{2} (211620) = 105810 \).
(ii) \( 3897 \times 25 = \frac{1}{2} (3897 \times 50) = \frac{1}{2} (194850) = 97425 \)
(iii) \( 4563 \times 35= \frac{1}{2} (4563 \times 70)= \frac{1}{2} (319410)= 159705 \)

Question. Multiplication of a given number by 5, 25, 125, 625, etc., that is, by a number which is some power of 5.
Answer: Method: Place as many zeros to the right of the multplicant as in the power of 5 in the multiplier, then divide the number so obtained by 2 raised to the same power as is the power of 5.

Question. Multiply (i) \( 3982 \times 5 \) (ii) \( 4739 \times 25 \) (iii) \( 7894 \times 125 \) (iv) \( 4863 \times 625 \)
Answer: Sol. (i) \( 3982 \times 5 = \frac{39820}{2} = 19910 \)
(ii) \( 4739 \times 25 = \frac{473900}{2^2} = \frac{473900}{4} = 118475 \)
(iii) \( 7894 \times 125 = \frac{7894000}{2^3} = \frac{7894000}{8} = 986750 \)
(iv) \( 4863 \times 625 = \frac{48630000}{2^4} = \frac{48630000}{16} = 3039375 \)

Test of divisibility

• Divisibility by 2: A number is divisible by 2 if the unit's digit is zero or divisible by 2. For example, 4, 12, 30, 18, 102, etc, are all divisible by 2.
• Divisibility by 3: A number is divisible by 3 if the sum of digits in the number is divisible by 3. For example, the number 3792 is divisible by 3, since 3 + 7 + 9 + 2 = 21, which is divisible by 3.
• Divisibility by 4: A number is divisible by 4 if the number formed by the last two digits (ten's digit and unit's digit) is divisible by 4 or are both zero. For example, the number 2616 is divisible by 4, since 16 is divisible by 4.
• Divisibility by 5: A number divisible by 5 if the unit's digit in the number is 0 or 5. For example 13520, 7805, 640, 745, ect. are all divisible by 5.
• Divisibility by 6: A number is divisible by 6 if the number is even and sum of its digits is divisible by 3. For example, the number 4518 is divisible by 6 since it is even and sum of its digits 4+5+1+8=18 is divisible by 3.
• Divisibility by 7: The unit digit of the given number is doubled and then it is subtracted from the number obtained after omitting the unit digit. If the remainder is divisible by 7, then the given number is also divisible by 7. For example, consider the number 448. On doubling the unit digit 8 of 448 we get 16. Then, 44–16 = 28. Since 28 is divisible by 7, 448 is divisible by 7.
• Divisibility by 8: A number is divisible by 8, if the number formed by the last 3 digits is divisible by 8. For example, the number 41784 is divisible by 8 as the number formed by last three digits i.e. 784 is divisible by 8.
• Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. For example, the number 19044 is divisible by 9 as the sum of its digits 1 + 9 + 0 + 4 + 4 = 18 is divisible by 9.
• Divisibility by 10: A number is divisible by 10, if ends in zero. For example, the last digit of 580 is zero, therefore 580 is divisible by 10.
• Divisibility by 11: A number is divisible by 11, if the difference of the sum of the digits at odd places and sum of the digits at even places is either zero or divisible by 11. For example, in the number 38797, the sum of the digits at odd places is 3 + 7 + 7 = 17 and the sum of the digits at even places is 8 + 9 = 17. The difference is 17 – 17=0, so the number is divisible by 11.
• Divisibility by 12: A number is divisible by 12 if it is divisible by 3 and 4.
• Divisibility by 25: A number is divisible by 25 if the number formed by the last two digits is divisible by 25 or last two digits are zero. For example, the number 13675 is divisible by 25 as the number formed by the last two digits 75 which is divisible by 25.
• Divisibility by 125: A number is divisible by 125 if the number formed by the last three digits is divisible by 125 or the last three digits are zero. For example, the number 5250 is divisible by 125 as 250 is divisible by 125.
• Divisibility by 18: An even number satisfying the divisibility test of 9 is divisible by 18.
• Divisibility by 88: A number is divisible by 88 if it is divisible by 11 and 8.

Decimal fractions

Decimal fraction : Fractions in which the denominators are the powers of 10 are called decimal fractions. In general, the decimal fractions are of the following types

• Recurring decimals : If in a decimal fraction, a figure or a set of figures is repeated continually, then such a number is called a recurring decimal. If a single figure is repeated, it is shown by putting a dot on it. But if a set of a figures is repeated, we express if either by putting one dot at the starting digit and one dot at the last digit of the repeating digits or by placing a bar on the repeating digit (s).
  (i) \( 2/3 = 0.6666....= 0.\dot{6} = 0.\bar{6} \)
  (ii) \( 22/7 = 3.142857142857 = 3.\dot{1}4285\dot{7} = 3.\overline{142857} \)
  (iii) \( 95/6 = 15.83333...= 15.8\dot{3} = 15.8\bar{3} \)
• Pure recurring decimals : A decimal in which some figures after the decimal point repeat is called a pure recurring decimal. Ex. \( 0.\dot{6}, 3.\overline{142857} \) etc.

Conversion of a pure recurring decimal into fraction

Rule : Write the recurring figures only once in the numerator and take as many nines in the denominator as the number of repeating figures.

(1) \( 0.\bar{6} = 6/9 = 2/3 \).

(2) \( 16.\bar{6} = 16 + 0.\bar{6} = 16+6/9 = 16 + 2/3 = 50/3 \).

To convert a mixed recurring decimal into fraction

\( \frac{p}{q} \text{ form} = \frac{(\text{Complete numbers}) – (\text{number formed by Non–repeating digit})}{\text{No. of 9 as no. of repeating digits after that write no. of 0 as no. of non repeating digits.}} \)

e.g. (i) \( 0.\overline{35} = \frac{35 – 0}{99} = \frac{35}{99} \)
(ii) \( 0.4\overline{35} = \frac{435 – 4}{990} = \frac{431}{990} \)

H.C.F & L.C.M of numbers

The least number which when divided by \( d_1, d_2 \text{ and } d_3 \) leaves the remainders \( r_1, r_2 \text{ and } r_3 \) respectively, such that \( (d_1–r_1) = (d_2–r_2) = (d_3–r_3) \) is \( (L.C.M. \text{ of } d_1, d_2 \text{ and } d_3) – (d_1–r_1) \).

Question. Find the least number which when divided by 9, 10 and 15 leaves the remainders 4, 5 and 10, respectively.
Answer: Sol. Here, \( 9–4 = 10–5=15–10=5 \). Also, L.C.M. (9, 10, 15)= 90. \ the required least number = 90–5=85.
A number on being divided by \( d_1 \text{ and } d_2 \) successively leaves the remainders \( r_1 \text{ and } r_2 \), respectively. If the number is divided by \( d_1 \times d_2 \), then the remainder is \( (d_1 \times r_2 + r_1) \).

Question. A number on being divided by 10 and 11 successively leaves the reminders 5 and 7, respectively. Find the remainder when the same number is divided by 110.
Answer: Sol. The required remainder \( = d_1 \times r_2 + r_1 = 10 \times 7 + 5 = 75 \).
To find the number of numbers divisible by a certain integer.

Question. (i) How many numbers up to 532 are divisible by 15 ?
Answer: Sol. We divide 532 by 15. \( 532 = 35 \times 15 + 7 \). The quotient obtained is the required number of numbers. Thus there are 35 such numbers.

Question. (ii) How many numbers up to 300 are divisible by 5 and 7 together ?
Answer: Sol. L.C.M. of 5 and 7 = 35. We divide 300 by 35. \( 300 = 8 \times 35 + 20 \). Thus there are 8 such numbers.
Two numbers when divided by a certain divisor give remainders \( r_1 \text{ and } r_2 \). When their sum is divided by the same divisor, the remainder is \( r_3 \). The divisor is given by \( r_1 + r_2 – r_3 \).

Question. Two numbers when divided by a certain divisor give remainders 473 and 298, respectively. When their sum is divided by the same divisor, the remainder is 236. Find the divisor.
Answer: Sol. The required divisor \( = 437 + 298 – 236 = 499 \).
Product of two numbers. = L.C.M. of the numbers \( \times \) H.C.F. of the numbers.

Question. The H.C.F. and the L.C.M. of any two numbers are 63 and 1260, respectively. If one of the two numbers is 315, find the other number.
Answer: Sol. The required number \( = \frac{L.C.M. \times H.C.F.}{\text{First number}} = \frac{1260 \times 63}{315} = 252 \).
To find the greatest number that will exactly divide x, y and z. Required number = H.C.F. of x, y, and z.

Question. Find the greatest number that will exactly divide 200 and 320.
Answer: Sol. The required greatest number \( = H.C.F. \text{ of } 200 \text{ and } 320 = 40 \).
To find the greatest number that will divide x, y, and z leaving remainders a, b, and c, respectviely. Required number = H.C.F. fo (x–a), (y–b) and (z–c).

Question. Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11, respectively.
Answer: Sol. The required greatest number \( = H.C.F. \text{ of } (148–4), (246–6) \text{ and } (623–11) \), i.e. H.C.F. of 144, 240 and 612 = 12.
To find the least number which is exactly divisible by x, y and z. Required number = L.C.M. of x, y and z.

Question. What is the smallest number which is exactly divisible by 36, 45, 63 and 80?
Answer: Sol. The required smallest number \( = L.C.M. \text{ of } 36, 45, 63 \text{ and } 80 = 5040 \).
To find the least number which when divided by x,y, and z leaves the remainders a,b, and c, respectively. It is a always observed that (x–a) = (y–b)=(z–c) = k(say). \ Required number = (L.C.M. of x, y, and z)–k.

Question. Find the least number which when divided by 36, 48 and 64 leaves the remainders 25, 37 and 53, respectively.
Answer: Sol. Since (36–25) = (48–37) = (64–53) = 11, therefore the required smallest number \( = (L.C.M. \text{ of } 36, 48 \text{ and } 64) –11 = 576–11 = 565 \).
To find the least number which when divided by x, y and z leaves the same remainder r in each case. Required number = (L.C.M. of x, y, and z) +r.

Question. Find the least number which when divided by 12, 16 and 18, will leave in each case a remainder 5.
Answer: Sol. The required smallest number \( = (L.C.M. \text{ of } 12, 16 \text{ and } 18) + 15 \text{ (OCR says 15, but then } 144+5=149 \dots \text{ likely error, } +5) = 144 + 5 = 149 \).

To find the greatest number that will divide x, y, and z leaving the same remainder in each case.
(a) When the value of remainder r is given : Required number = H.C.F. of (x–r), (y–r) and (z–r).
(b) When the value of reminder is not given: Required number = H.C.F. of |(x–y) |,|(y–z)| and |(z–x)|

Question. Find the greatest number which will divide 772 and 2778 so as to leave the remainder 5 in each case.
Answer: The required greatest number
= H.C.F. of \( (772 - 5) \) and \( (2778 - 5) \)
= H.C.F. of 767 and 2773
= 59.

Question. Find the greatest number which on dividing 152, 277 and 427 leaves remainder.
Answer: The required greatest number.
= H.C.F. of \( |(x - y)|, |(y - z)| \) and \( |(z - x)| \)
= H.C.F. of \( |(152 - 277)|, |(277 - 427)| \) and \( |(427 - 152)| \)
= H.C.F. of 125, 275 and 150
= 25.

To find the n-digit greatest number which, when divided by x, y and z.

• (a) leaves no remainder (i.e., exactly divisible)
  • Step 1 – L.C.M. of x, y and z = L
  • Step 2 – \( \frac{\text{n-digit greatest number}}{L} \), Remainder = R
  • Step 3 – Required number = n-digit greatest number – R
• (b) leaves reminder K in each case
  • Required number = (n-digit greatest number – R) + K.

Question. Find the greatest number of 4 digits which, when divided by 12, 18, 21 and 28, leaves 3 as a remainder in each case.
Answer: L.C.M. of 12, 18, 21 and 28 = 252.
Divide 9999 by 252, we get remainder 171.
The required number = \( (9999 - 171) + 3 = 9828 + 3 = 9831 \).

To find the n-digit smallest number which when divided by x, y and z

• (a) leaves no remainder (i.e. exactly divisible)
  • Step 1 – L.C.M. of x, y and z = L
  • Step 2 – \( \frac{\text{n-digit smallest number}}{L} \), Remainder = R
  • Step 3 – Required number = n-digit smallest number + (L – R).
• (b) leaves remainder K in each case.
  • Required number = n-digit smallest number + (L – R) + k.

Question. Find the least number of four digits which is divisible by 4, 6, 8 and 10.
Answer: L.C.M. of 4, 6, 8, and 10 = 120.
Divide 1000 by 120, we get remainder 40.
The required number = \( 1000 + (120 - 40) = 1080 \).

Question. Find the smallest 4-digit number, such that when divided by 12, 18, 21 and 28, it leaves remainder 3 in each case.
Answer: L.C.M. of 12, 18, 21 and 28 = 252.
Divide 1000 by 252, we get remainder 244.
The required number = \( 1000 + (252 - 244) + 3 = 1000 + 8 + 3 = 1011 \).

Number of factors of a given number

Number of factors (or divisors) of a given number (composite number)
Lets us assume a composite number say 24 then find the number of factors.
\( 24 = 1 \times 24 \)
\( 2 \times 12 \)
\( 3 \times 8 \)
\( 4 \times 6 \)
We see that there are total 8 factors namely, 1, 2, 3, 4, 6, 8, 12 and 24.
Let there be a composite number N and its prime factors be a, b, c, d,....... etc and p, q, r, s...etc. be the indices (or powers) of the a, b, c, d..... etc. respectively i.e., if N can be expressed as \( N = a^p \cdot b^q \cdot c^r \cdot d^s \dots \), then, the number of total divisors or factors of N is \( (p+1)(q+1)(r+1)(s+1)\dots \)

Question. Find the total number of factors of 540:
(a) 24
(b) 20
(c) 30
(d) None of these
Answer: (a)
Answer: \( 540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5^1 \).
Therefore total number of factors of 540 is \( (2+1)(3+1)(1+1) = 3 \times 4 \times 2 = 24 \).

Sum of factors of given number

Once again if you want to find the sum of smaller composite numbers, then you can do it manually, but for larger numbers its a problem.
e.g. Sum of factors of 24 = \( 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60 \)
Let N be the composite number and a, b, c, d.. be its prime factors and p, q, r, s be the indices (or powers) of a, b, c, d i.e., if N can be expressed as \( N = a^p \cdot b^q \cdot c^r \cdot d^s \dots \), then the sum of all the divisors (or factors) of N = \( \frac{(a^{p+1}-1)(b^{q+1}-1)(c^{r+1}-1)(d^{s+1}-1)}{(a-1)(b-1)(c-1)(d-1)} \)

Question. Find the sum of factors of 270.
Answer: \( 270 = 2 \times 3^3 \times 5^1 \).
Sum of factors of 270 = \( \frac{(2^{1+1}-1)(3^{3+1}-1)(5^{1+1}-1)}{(2-1)(3-1)(5-1)} = \frac{(4-1)(81-1)(25-1)}{1 \times 2 \times 4} = \frac{3 \times 80 \times 24}{8} = 3 \times 80 \times 3 = 720 \).

Product of factors

Let us assume a very small number 24 and see the factors
24 = \( 1 \times 24 \)
= \( 2 \times 12 \)
= \( 3 \times 8 \)
= \( 4 \times 6 \)
Note, it is obvious from the above explanation that the product of factors of 24 = \( (1 \times 24) \times (2 \times 12) \times (3 \times 8) \times (4 \times 6) = 24 \times 24 \times 24 \times 24 = (24)^4 \).
Thus, the product of factors of composite number \( N = N^{n/2} \), where n is the total number of factors of N.

Question. Product of divisors of 7056 is :
(a) \( (84)^{48} \)
(b) \( (84)^{44} \)
(c) \( (84)^{45} \)
(d) None of these
Answer: (c)
Answer: \( 7056 = 2^4 \times 3^2 \times 7^2 \).
Number of factors /divisors of 7056 = \( (4+1)(2+1)(2+1) = 5 \times 3 \times 3 = 45 \).
Product of factors = \( (7056)^{45/2} = ((84)^2)^{45/2} = (84)^{45} \).

Number of odd factors of given number

Let us assume a smaller number e.g., 24
24 = \( 1 \times 24 \); also \( 24 = 2^3 \times 3^1 \)
= \( 2 \times 12 \)
= \( 3 \times 8 \)
We can see that there are total 2 odd factors namely 1 and 3.
Further assume another number say 36
36 = \( 1 \times 36 \); also \( 36 = 2^2 \times 3^2 \)
= \( 2 \times 18 \)
= \( 3 \times 12 \)
= \( 4 \times 9 \)
= \( 6 \times 6 \)
So, we can see that there are only 3 odd factors viz., 1, 3 and 9.
Once again we assume another number say 90
then 90 = \( 1 \times 90 \); also \( 90 = 2 \times 3^2 \times 5 \)
= \( 2 \times 45 \)
= \( 3 \times 30 \)
= \( 5 \times 18 \)
= \( 6 \times 15 \)
= \( 9 \times 10 \)
Thus there are only 6 odd factors namely 1, 3, 5, 9, 15, 45.
To get the number of odd factors of a numbers N first of all express the number N as \( N = (p_1^a \times p_2^b \times p_3^c \dots) \times (e^x) \)
(where, \( p_1, p_2, p_3, \dots \) are the odd prime factors and e is the even prime factor)
Then the total number of odd factors = \( (a+1)(b+1)(c+1)\dots \)

Question. The number of odd factors of 36 is ....
Answer: \( 36 = 2^2 \times 3^2 \).
Number of odd factors = \( (2 + 1) = 3 \).

Number of even factors a composite number

Number of even factors of a number = (Total number of factors of the given number) – (Total number of odd factors)

Question. Find the number of even factors (or divisors) of (a) 24, (b) 36, (c) 90.
Answer:
(a) number of even factors = 8 – 2 = 6
(b) number of even factors = 9 – 3 = 6
(c) number of even factors = 12 – 6 = 6

Number of ways of expressing composite number as a product of two factors

Let us consider an example of small composite number say, 24
then 24 = \( 1 \times 24 \)
\( 2 \times 12 \)
\( 3 \times 8 \)
\( 4 \times 6 \)
So it is clear that the number of ways of expressing a composite no. as a product of two factors = \( \frac{1}{2} \times \) the no. of total factors.

Question. Find the number of ways of expressing 180 as a product of two factors.
Answer: \( 180 = 2^2 \times 3^3 \times 5^1 \). (Note: Actual prime factorization is \( 180 = 2^2 \times 3^2 \times 5^1 \), but following OCR verbatim)
Number of factors = \( (2+1)(2+1)(1+1) = 18 \).
Hence, there are total \( \frac{18}{2} = 9 \) ways in which 180 can be expressed as a product of two factors.

Note - As you know when you express any perfect square number 'N' as a product of two factors namely \( \sqrt{N} \) and \( \sqrt{N} \), and also know that since in this case \( \sqrt{N} \) appears two times but it is considered only once while calculating the no. of factors so we get always an odd number as number of factors so we can not divide the odd number exactly by 2 as in the above formula. So if we have to consider these two same factors then we find the number of ways of expressing N as a product of two factors = \( \frac{\text{Number of factors } + 1}{2} \).
Again if it is asked that find no. of ways of expressing N as a product of two distinct factors then we do not consider 1 way (i.e. \( N = \sqrt{N} \cdot \sqrt{N} \)) then no. of ways = \( \frac{\text{Number of factors } - 1}{2} \).

Question. Find the number of ways expressing 36 as a product of two factors.
Answer: \( 36 = 2^2 \times 3^2 \).
Number of factors = \( (2+1)(2+1) = 9 \).
Hence the no. of ways of expressing 36 as a product of two factors = \( \frac{9 + 1}{2} = 5 \), as \( 36 = 1 \times 36, 2 \times 18, 3 \times 12, 4 \times 9 \) and \( 6 \times 6 \).

Question. In how many ways can 576 be expressed as the product of two distinct factors?
Answer: \( 576 = 2^6 \times 3^2 \).
Total number of factors = \( (6 + 1)(2 + 1) = 21 \).
So the number of ways of expressing 576 as a product of two distinct factors = \( \frac{21 - 1}{2} = 10 \).
Note – Since the word 'distinct' has been used therefore we do not include \( 576 = 24 \times 24 \).

Cyclicity

We are having 10 digits in our number systems and some of them shows special characteristics like they, repeat their unit digit after a cycle, for example 1 repeat its unit digit after every consecutive power. So its cyclicity is 1 on the other hand digit 2 repeat its unit digit after every four power, hence the cyclicity of 2 is four. The cyclicity of digits are as follows:
Digit | Cyclicity
0, 1, 5 and 6 | 1
4 and 9 | 2
2, 3, 7 and 8 | 4
So, if we want to find last digit of \( 2^{45} \), divide 45 by 4. The remainder is 1 so the last digit of \( 2^{45} \) would be same as the last digit of \( 2^1 \) which is 2.

To Find the Unit Digit in Exponential Expression:

(i) Where there is 2 in unit's place of any number. Since, in \( 2^1 \) unit digit is 2, in \( 2^2 \) unit digit is 4, in \( 2^3 \) unit digit is 8, in \( 2^4 \) unit digit is 6, after that the unit's digit repeats e.g. unit digit \( (12)^{12} \) is equal to the unit digit of \( 2^4 \) i.e. 6.
Ex. In \( (32)^{33} \) unit digit is equal to the unit digit of \( 2^1 \) i.e. 2.

(ii) When there is 3 in unit's place of any number. Since, in \( 3^1 \) unit digit is 3, in \( 3^2 \) unit digit is 9, in \( 3^3 \) unit digit is 7, in \( 3^4 \) unit digit is 1, after that the unit's digit repeats.
Ex. In \( (43)^{46} \) unit digit be 9.

(iii) When there is 4 in units place of any number. Since, in \( 4^1 \) unit digit is 4, in \( 4^2 \) unit digit is 6, after that the unit's digit repeats.
Ex. In \( (34)^{14} \) unit digit is 6.
Ex. In \( (34)^{33} \) unit digit is 4.

(iv) When there is 5 in unit's place of any number. Since, in \( 5^1 \) unit digit is 5, in \( 5^2 \) unit digit is 5 and so on.
Ex. In \( (25)^{15} \) unit digit is 5.

(v) When there is 6 in unit's place of any number. Since, in \( 6^1 \) unit digit is 6, in \( 6^2 \) unit digit is 6 & so on.
Ex. In \( (46)^{13} \) unit digit is 6.

(vi) When there is 7 in unit's place of any number. Since, in \( 7^1 \) unit digit is 7, in \( 7^2 \) unit digit is 9, in \( 7^3 \) unit digit is 3, in \( 7^4 \) unit digit is 1, after that the unit's digit repeats.
Ex. In \( (57)^9 \) unit digit is 7.
Ex. In \( (97)^9 \) unit digit is 3. (OCR says 3, actual is 7)

(vii) When there is 8 is unit's place of any number. Since in \( 8^1 \) unit digit is 8, in \( 8^2 \) unit digit is 4, in \( 8^3 \) unit digit is 2, in \( 8^4 \) unit is 6, after that unit's digit repeats after a group of 4.

(viii) When there is 9 in unit's place of any number. Since, in \( 9^1 \) unit's digit is 9, in \( 9^2 \) unit's digit is 1, after that unit's digit repeats after a group of 2.

(ix) When there is zero in unit's place of any number. There will always be zero in unit's place.

Question. (a) Find the last digit of (i) \( 3^{57} \) (ii) \( 13^{59} \)
Answer: (i) The cyclicity of 3 is 4. Hence, \( \frac{57}{4} \) gives the remainder 1. So the last digit of \( 3^{57} \) is same as the last digit of \( 3^1 \), i.e. 3.
(ii) The number of digits in the base will not make a difference to the last digit. It is last digit of the base which decides the last digit of the number itself. For \( 13^{59} \), we find \( \frac{59}{4} \) which gives a remainder 3. So the last digit of \( 13^{59} \) is same as the last digit of \( 3^3 \), i.e 7.

Question. (b) Find the last digit of the product \( 7^{23} \times 8^{13} \).
Answer: Both 7 and 8 exhibit a cyclicity of 4. the last digit are:
\( 7^1 = 7, 8^1 = 8 \)
\( 7^2 = 9, 8^2 = 4 \)
\( 7^3 = 3, 8^3 = 2 \)
\( 7^4 = 1, 8^4 = 6 \)
\( 7^5 = 7, 8^5 = 8 \)
The cycle would repeat itself for higher powers. \( 7^{23} \) ends with the same last digit as \( 7^3 \), i.e. 3. \( 8^{13} \) ends with the same last digit as \( 8^1 \), i.e. 8. Hence, the product of the two numbers would end with the same last digit as that of \( 3 \times 8 \), i.e. 4.

Question. (c) Find unit's digit in \( y = 7^{17} + 7^{34} \)
Answer: \( 7^{17} + 7^{34} = 7^1 + 7^2 = 56 \), Hence the unit digit is 6.

Question. (d) What will be the last digit of \( (73)^{75^{64^{76}}} \)
Answer: Let \( (73)^{75^{64^{76}}} = (73)^x \) where \( x = 75^{64^{76}} = (75)^{\text{even power}} \).
Cyclicity of 3 is 4. To find the last digit we have to find the remainder when x is divided by 4.
\( x = (75)^{\text{even power}} = (76-1)^{\text{even power}} \), where n is divided by 4 so remainder will be 1. Therefore, the last digit of \( (73)^{75^{64^{76}}} \) will be \( 3^1 = 3 \).

Euclid’s division lemma

Euclid's division lemma states that "For any two positive integers a and b, there exist integers q and r such that \( a = bq + r, 0 \le r < b \)."
e.g.(i) Consider number 23 and 5, then : \( 23 = 5 \times 4 + 3 \).
Comparing with \( a = bq + r \), we get, \( a = 23, b = 5, q = 4, r = 3 \) and \( 0 \le r < b \) (as \( 0 < 3 < 5 \)).
(ii) Consider positive integers 18 and 4. \( 18 = 4 \times 4 + 2 \).
For 18 (= a) and 4 (= b) we have q = 4, r = 2 and \( 0 \le r < b \).
In the relation \( a = bq + r \), where \( 0 \le r < b \) is nothing but a statement of the long division of number a by b in which q is the quotient obtained and r is the remainder.

Euclid’s divison algorithm

In mathematics, the Euclid's Algorithm is an efficient method for computing the greatest common divisor (GCD) or highest common factor (HCF). So, let us state Euclid's division algorithm clearly.
To obtain the HCF of two positive integers, say c and d, with c > d follow the steps below :

• Step-1 : Apply Euclid's division lemma, to c and d. So, we find whole numbers, q and r such that \( c = dq + r, 0 \le r < d \).
• Step-2 : If r = 0, d is the HCF of c and d. If \( r \neq 0 \), apply the division lemma to d and r.
• Step-3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

This algorithm works because \( \text{HCF}(c,d) = \text{HCF}(d,r) \) where the symbol \( \text{HCF}(c,d) \) denotes the HCF of c and d.

Question. Use Euclid's division algorithm to find the HCF of 441, 567, 693.
Answer: In order to find the HCF of 441, 567 & 693, we first find the HCF of 441 & 567 by Euclid's division algorithm.
Using division algorithm, we get
\( 567 = 441 \times 1 + 126 \)
\( 441 = 126 \times 3 + 63 \)
\( 126 = 63 \times 2 + 0 \)
So, \( \text{HCF}(567, 441) = 63 \).
Now, we find the HCF of 63 and 693.
\( 693 = 63 \times 11 + 0 \)
\( \text{HCF}(63, 693) = 63 \).
Hence \( \text{HCF}(441, 567, 693) = 63 \).

Fundamental theorem of arithmetic

The fundamental theorem of arithmetic (FTA) tells us something important about the relationship between composite numbers and prime numbers. It is usually stated as follows : "Every composite number can be expressed as a product of primes, and their decomposition is unique, apart from the order in which the prime factors occur."
e.g. \( 12600 = 2^3 \cdot 3^2 \cdot 5^2 \cdot 7 \). Thus we have expressed the composite number 12600 as product of powers of primes in ascending order and this decomposition is unique.

Points to remember

• For any three positive integers p, q, r, \( \text{HCF}(p,q,r) \times \text{LCM}(p,q,r) \neq p \times q \times r \). However, the following results hold good. \( \text{LCM}(p,q,r) = \frac{p \cdot q \cdot r \cdot \text{HCF}(p,q,r)}{\text{HCF}(p,q) \cdot \text{HCF}(q,r) \cdot \text{HCF}(p,r)} \).
• All odd numbers can be obtained by the formula, \( f(n) = 2n + 1, n = 1,2,3,\dots \). while even numbers can be obtained by the formula, \( g(n) = 2n, n = 1,2,3,\dots \).
• \( \text{L.C.M. of pair of coprimes} = \text{Product of co-primes} \).
• The denominator of the rational number must be in the form \( 2^m \cdot 5^n \) (where m & n are non-negative integers) so as to have the decimal expansion of that rational number as terminating.
• Any positive odd integer is of the form \( 6q + 1 \), or \( 6q + 3 \) or \( 6q + 5 \), where q is some integer.
• The square of any positive integer is either of the form \( 3m \) or \( 3m + 1 \) for some integer m.
• The cube of any positive integer is of the form \( 9m, 9m + 1 \) or \( 9m + 8 \).
• The square of any positive odd integer is of the form \( 8m + 1 \) for some integer m.