(Real Numbers) & (Polynomials) _ Study Module.pdf

1 CHAPTER Real Numbers School Level Number System You must have studied different types of numbers in your earlier classes such as natural n...

1 CHAPTER Real Numbers School Level Number System You must have studied different types of numbers in your earlier classes such as natural numbers, whole numbers, integers, rational, irrational and real numbers Let’s revise them. Real Natural 1, 2, 3... Whole Rational 0, 1, 2....333..., 3 4.77, –1.32 Integers...–3, –2, –1, 0, 1, 2, 3... Irrational 1.010010001... 5, 2, 10, π ‰ Natural numbers: Natural numbers are those which are used for counting and ordering i.e., N = {1, 2, 3, 4,...}. ‰ Whole numbers: Whole numbers are collection of all natural numbers including zero i.e., W = {0, 1, 2, 3, 4,...}. ‰ Integers: An integer is a number that can be written without fractional components. Integers consist of natural numbers, their negatives and zero i.e., Z = {....–4, –3, –2, –1, 0, 1, 2, 3, 4,....}. ‰ Prime numbers: A number which can only be divided by either 1 or itself is called a prime number. E.g: 2, 3, 5, 7, etc. NOTE: (i) Each prime number have only two factors. E.g: (a) 2 = 1 × 2 (b) 3 = 1 × 3 (ii) 2 is the only even prime number. ‰ Composite numbers: A number which has more than two factors is known as a composite number. E.g: 4, 6, 8, 10 etc. NOTE: 1 is neither a prime nor a composite number.  p ‰ Rational numbers: Rational numbers are those which can be expressed as , where p and q are integers and q ≠ q 0 & co-prime. Collection of all rational numbers are denoted by Q. 3 5 E.g: 1, 0, ,. etc. 2 3 ‰ Irrational numbers: All real numbers which are not rational numbers are termed as irrational numbers. A non-terminating and non-repeating decimal is an example of irrational number. Collection of all irrational number are denoted by P. E.g: 2, π etc. ‰ Real numbers : The collection of all the rational and irrational numbers is called real numbers and it is denoted by R. Fundamental Theorem of Arithmetic The fundamental theorem of arithmetic (FTA) which is also known as unique prime factorisation theorem, gives relationship between prime numbers and composite numbers. It states that: Theorem 1: Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occurs. E.g: 12600 = 23⋅32⋅52⋅7 Thus, the composite number 12600 is expressed as product of powers of primes in ascending order and this decomposition is unique. Corallary : The prime factorisation of a natural number is unique, except for the order of its factors. NOTE: In the prime factorisation of a number we write the primes in ascending order. i.e. If x = pa1 pb2 pc3.... pan, where pi are prime then p1 < p2 <... < pn, and a, b, c, d, & a are natural number’s. Knowledge Hub ™ If p a prime number. If p divides a2, then p divides a, where a is a positive integer. ™ Prime factor of a prime number is the number itself. Number of Zeros in a Number The number of zeros in a number is the minimum power of 2 and 5 in the prime factorisation of that number. i.e. Number of zeros in a number = Minimum (power of 2, power of 5). H.C.F. and L.C.M. of Numbers H.C.F (Highest common factor) : The H.C.F. of two or more numbers is the largest positive integer that divides each of the integers. L.C.M. (Least common multiple): The L.C.M. of two or more numbers is the smallest positive integer which is exactly divisible by each of the given numbers. To find HCF and LCM of given numbers using Prime Factorisation Method or Fundamental Theorem of Arithmetic, first express each number in the form of product of prime factors, then HCF = Product of the smallest powers of common factors. LCM = Product of the greatest power of each prime factor involved in the numbers. 2 Class-X MATHEMATICS P W Important relations ‰ L.C.M × H.C.F = Product of two numbers. ‰ L.C.M of two or more prime numbers is equal to their product. ‰ H.C.F of two or more prime numbers is always 1. ‰ For three positive integers (a, b and c), HCF (a, b, c) × LCM (a, b, c) ≠ a × b × c, where a, b, c are positive integers. ‰ The largest number which will divide a, b and c leaving remainders x, y and z respectively is, H.C.F. of (a – x), (b – y) and (c – z). ‰ The smallest number which when divided by a, b and c leaves the same remainder r in each case is (L.C.M of a, b and c) + r. EXAMPLE 1. Find the number of zeros in 38 × 2 × 25 × 39 × 50. As the given number have more than two factors, Sol. The number can be rewritten in prime factorisation \ (7 × 11 × 15 + 15) is a composite number. form as 4. By using prime factorization method, find the 2 × 2 × 2 × 3 × 5 × 5 × 5 × 5 × 13 × 19 L.C.M and H.C.F. of 1296 and 2520. = 23 × 3 × 54 × 13 × 19 Sol. 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 24 × 34 In prime factorisation, power of 2 = 3 & 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 23 × 32 × 5 × 7 power of 5 = 4. L.C.M. (1296, 2520) = 24 × 34 × 5 × 7 = 45360 So, number of zeros in the number H.C.F. (1296, 2520) = 23 × 32 = 72 = minimum (power of 2, power of 5) 5. Find the greatest number which can divide 1251, 9377 and 15628 leaving 1, 2 and 3, as = minimum (3, 4) = 3 remainders respectively. \ No. of zeros = 3 Sol. Numbers are: (1251 – 1), (9377 – 2) and 2. On an evening walk, three people step off (15628 – 3) or 1250, 9375 and 15625. together and their steps measure 40 cm, 42 cm Required number is the HCF of 1250, 9375 and and 45 cm, respectively. How much minimum 15625. distance each should cover so that each can Now, 1250 = 2 × 54 cover the same distance in complete steps? 9375 = 3 × 55 Sol. 40 = 23 × 51, 42 = 21 × 31 × 71 15625 = 56 45 = 32 × 51 HCF = 54 = 625 As three people step off together, minimum \ 625 is the largest number that divides 1251, distance covered by each one of them 9377 and 1568 leaving remainders 1, 2 and 3 = LCM of 40, 42, 45 respectively. 6. Find the smallest number which when divided by = 23 × 32 × 51 × 71 = 2520 cm. 16, 20 and 24 will leave a remainder 5 in each 3. 7 × 11 × 15 + 15 is a composite number. Explain. case. Sol. 7 × 11 × 15 + 15 = (7 × 11 + 1) × 15 Sol. The required smallest number = (77 + 1) × 15 = 78 × 15 = 2 × 3 × 13 × 15 = (L.C.M of 16, 20 and 24) + 5 = 240 + 5 = 2 × 3 × 13 × 15 = 245. Real Numbers 3 Revisiting Irrational and Rational Numbers Theorem 2: Let p be a prime number. If p divides m2, then p divides m, where m is a positive integer. EXAMPLE 1. Prove that 2 is an irrational number. Again on squaring both sides, we get a2 = 4c2 Sol. Proof: Let's assume, to the contrary, that 2 is rational. ⇒ 2b2 = 4c2[ a2 = 2b2] a ⇒ b2 = 2c2 Then 2 = where a and b are co-prime, integers b This means that 2 divides b2 and therefore 2 & b ≠ 0. divides b (again using Theorem 2). So, 2b = a. Therefore, a and b have at least 2 as a common factor. On squaring both sides, we get But this contradicts the fact that a and b do not a2 = 2b2 have a common factors other than 1. \ 2 divides a2. This contradiction has arisen because of our wrong Now, using Theorem 2, it follows that 2 divides a. assumption that 2 is rational. So, we can write a = 2c for some integer c. So, we conclude that 2 is irrational. Knowledge Hub ™ If pq is irrational, then p + q will definitely be irrational. ™ If p is a positive prime number, then p is an irrational number. ™ Decimal expansions of irrational numbers are non-terminating and non-reccuring. ™ Decimal expansions of rational numbers are either terminating or non-terminating and recurring. ™ Sum of a rational number and an irrational number always gives irrational number. ™ Sum of two irrational numbers does not always give an irrational number. ™ Product of a non-zero rational number with an irrational number always gives an irrational number. ™ Product of two irrational numbers does not always give an irrational number. EXAMPLE 1. Prove that 2 + 3 is irrational. x2 − 5 As x, 5 and 2 are rational numbers, is a Sol. Let us assume, to the contrary, that 2 + 3 is a rational number. 2 rational number equals to 'x' & x is a rational number. x2 − 5 Hence, 6= is a rational number, which is ⇒x= 2+ 3 2 On squaring both sides, we get in contradiction to the fact that 6 is an irrational number. x2 = 2 + 3 + 2 3 ⋅ 2 =+ 5 2 6 Hence our assumption is wrong. So, we conclude x2 − 5 that 2 + 3 is an irrational number. ⇒ x2 = 5+2 6 ⇒ 6 = 2 4 Class-X MATHEMATICS P W  Summary ™ Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes factors and this factorisation is unique, apart from the order in which the prime factors occur. ™ The number of zeros in a number is the minimum power of 2 and 5 in the prime factorisation of that number. i.e. Number of zeros in a number = Minimum (power of 2, power of 5). ™ Theorem: Let p be a prime number. If p divides m2, then p divides m, where m is a positive integer. NCERT Exercise Product of LCM and HCF = 182 × 13 = 2366 Exercise-I Clearly, LCM (26, 91) × HCF (26, 91) 1. Express each number as a product of its prime = Product of 26 and 91. factors: (b) By prime factorisation, we get: (a) 140 (b) 156 510 = 2 × 3 × 5 × 17 (c) 3825 (d) 5005 92 = 2 × 2 × 23 (e) 7429 \ HCF of 510 and 92 = 2 Sol. (a) 22 × 5 × 7 (b) 22 × 3 × 13 and LCM of 510 and 92 (c) 32 × 52 × 17 (d) 5 × 7 × 11 × 13 = 22 × 3 × 5 × 17 × 23 = 23460 (e) 17 × 19 × 23 Now, HCF × LCM = 2 × 23460 = 46920 …(i) 2. Find the LCM and HCF of the following pairs of Product of numbers = 510 × 92 = 46920 (ii) integers and verify that LCM × HCF = product From (i) and (ii), we get: of the two numbers. (a) 26 and 91 LCM × HCF = Product of numbers (b) 510 and 92 Hence, verified. (c) 336 and 54 (c) By prime factorisation, we get: Sol. (a) 26 and 91 336 = 2 × 2 × 2 × 2 × 3 × 7 On expressing 26 and 91 as product of its prime 54 = 2 × 3 × 3 × 3 factors, we have \ HCF of 336 and 54 = 2 × 3 = 6 26 = 2 × 13 and LCM of 336 and 54 = 24 × 33 × 7 = 3024 91 = 7 × 13 Now, LCM × HCF = 3024 × 6 = 18144 …(i) Hence, LCM (26, 91) = 2 × 7 × 13 = 182 Product of numbers = 336 × 54 = 18144…(ii) and HCF (26, 91) = 13 From (i) and (ii), we get: Verification: LCM × HCF = Product of number Product of 26 and 91 = 26 × 91 = 2366 Hence, verified. Real Numbers 5  3. Find the LCM and HCF of the following integers Therefore, it is clear from above that, 6n is not by applying the prime factorisation method. divisible by 5 for any natural number n and hence, (a) 12, 15 and 21 [CBSE 2020] it proves that 6n can never end with the digit 0 for (b) 17, 23 and 29 any natural number n. (c) 8, 9 and 25 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Sol. (a) By prime factorisation, we get: 12 = 2 × 2 × 3 Sol. For 7 × 11 × 13 + 13 15 = 3 × 5 = 13(7 × 11 + 1) = 13 × 78 21 = 3 × 7 = 13 × 13 × 3 × 2 HCF of 12, 15 and 21 = 3 Hence it is composite number and LCM = 2 × 2 × 3 × 5 × 7 = 420. For 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 (b) By prime factorisation, we get: Taking 5 as a common factor, we get, 17 = 17 × 1 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) 23 = 23 × 1 = 5(1008 + 1) = 5 × 1009 29 = 29 × 1 Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number. \ HCF of 17, 23 and 29 = 1 7. There is a circular path around a sports field. and LCM = 17 × 23 × 29 = 11339. Monika takes 18 minutes to drive one round (c) By prime factorisation, we get: of the field, while Rohit takes 12 minutes for 8 = 2 × 2 × 2 × 1 the same. Suppose they both start at the same 9 = 3 × 3 × 1 point and at the same time, and go in the same 25 = 5 × 5 × 1 direction. After how many minutes will they meet \ HCF of 8, 9 and 25 = 1 again at the starting point? and LCM of 8, 9 and 25 = 23 × 32 × 52 = 1800. Sol. We need to find the number of minutes after which they will meet again at the starting point. For this, 4. Given that HCF (306, 657) = 9, find there will be a smallest number that is divisible by LCM (306, 657). both 18 and 12 and that will be the time when both Sol. HCF (a, b) × LCM (a, b) = ab meet again at the starting point. To find this we have ab to take LCM of both numbers. ⇒ LCM (a, b) = HCF(a, b) Therefore, LCM (18, 12) = 2 × 3 × 3 × 2 = 36 \ LCM (306, 657) Hence, Monika and Rohit will meet again at the 306 × 657 starting point after 36 minutes. = =34 × 657 =22338 9 5. Check whether 6n can end with the digit 0 for Exercise-2 any natural number n. 1. Prove that 5 is irrational. Sol. For a number 6n to end with the digit zero (0), it Sol. Let us assume to the contrary that 5 is a rational must be divisible by 5, as we already know that any number. number having unit place as 0 or 5 is divisible by 5. a Prime factorization of 6n = (2 × 3)n i.e. 5= (where, a and b are co-prime, integers b As we can see that, the prime factorization of 6n and b ≠ 0) doesn’t contain prime number 5. ⇒ b 5=a 6 Class-X MATHEMATICS P W  On squaring both sides, we have 1 a So, = , where a and b are co-prime, 2 b (b 5 ) 2 = a 2 ⇒ 5b2 = a2...(i) integers and b ≠ 0. As, a2 is divisible by 5, so a is also divisible by 5 b (by theorem 2). ⇒ a 2 =⇒ b 2= a So, we can write a = 5c, for some integer c and on Q b and a are integers. substituting the value of a in equation (i), we get b 5b2 = (5c)2 ⇒ b2 = 5c2 So, is a rational number. a As, b2 is divisible by 5, it means b is also divisible So, 2 is a rational number which contradicts by 5 (by theorem 2). the fact that 2 is irrational. Therefore, a and b have at least 5 as a common This contradiction occured due to we assumed factor. But this contradicts the fact that a and b 1 that is rational. are co-prime. This contradiction has arisen because 2 1 of our incorrect assumption that 5 is a rational Hence, we conclude that is irrational. 2 number. (b) Let us assume that 7 5 is rational. Hence, 5 is an irrational number. \ There exists co-prime, integers a and b (b ≠ 2. Prove that 3 + 2 5 is irrational. 0) such that: Sol. Let us assume to the contrary, that 3 + 2 5 is rational. a a 7 5 = ⇒ 5= b 7b i.e., where a, b are integers, co-prime & b ≠ 0 such a Since a and b are integers, we get is rational that 7b and so 5 is rational. a a 3+ 2 5 = ⇒ 2 5= −3 But this contradicts the fact that 5 is irrational. b b 1a  This contradiction has arisen because of our ⇒= 5  − 3 2 b  incorrect assumption that 7 5 is rational. 1a  Hence, we conclude that 7 5 is irrational. Since, a and b are integers, thus,  − 3  is a 2 b  (c) Let us assume that 6 + 2 is rational. rational number and therefore, 5 should also be \ There exists co-prime, integers a and b (b ≠ 0) a rational number. such that But this contradicts the fact that 5 is irrational. a a 6 + 2 = ⇒ 2 = –6 Hence, we conclude that 3 + 2 5 is irrational. b b a 3. Prove that the following are irrationals Since a and b are integers, we get – 6 is b 1 rational and so 2 is rational. (a) (b) 7 5 2 But this contradicts the fact that 2 is (c) 6 + 2 irrational. This contradiction has arisen because of our 1 Sol. (a) Let us assume to the contrary that is a incorrect assumption that 6 + 2 is rational. 2 rational number. Hence, we conclude that 6 + 2 is irrational. Real Numbers 7  2. Match the following columns: Fill in the Blanks Column-I Column-II 1. 144 as a product of its prime factors is _______ P. Remainder when 667 is (i) 448 2. 5 − 3 is an _______ Number. divided by 22 3. The greatest possible number with which when we Q. Remainder when 448 is (ii) Yes divide 37 and 58, leaving the respective remainder divided by 449 of 2 and 3, is _______ R. Remainder can be zero? (iii) No 4. The product of two numbers is equal to the _______ S. Divisor can be zero? (iv) 7 of their HCF and LCM. 5. The two numbers are 396 and 576 and their LCM is (a) P-(iv), Q-(iii), R-(ii), S-(i) 6336. The HCF of the numbers is _______ (b) P-(i), Q-(ii), R-(iii), S-(iv) (c) P-(ii), Q-(iv), R-(i), S-(iii) True and False Statements (d) P-(iv), Q-(i), R-(ii), S-(iii) 1. LCM of 6, 72 and 120 is 720. 2. 8n may end with the digit 0 for any natural number n. Assertion & Reason Type Questions 3. If 19 divides a3 (where a is a positive integer), then Direction (Q. 1-3): In the following questions, a statement 19 divides a. of Assertion (A) is followed by a statement of Reason (R). 4. The quotient of two integers is always a rational Mark the correct choice as: number. (a) Both Assertion (A) and Reason (R) are true 5. Product of 2 irrational numbers may or may not be and Reason (R) is the correct explanation of an irrational number. Assertion (A). (b) Both Assertion (A) and Reason (R) are true Match the Following but Reason (R) is not the correct explanation of Assertion (A). 1. Match the following columns : (c) Assertion (A) is true but Reason (R) is false. Column-I Column-II (d) Assertion (A) is false but Reason (R) is true P. H.C.F. of 306 and 657 (i) 2 1. Assertion (A): If H.C.F. of two numbers is 16 and Q. H.C.F. of the smallest (ii) 5 their product is 3072, then their L.C.M. is 162. composite number and the smallest prime Reason (R): If p, q are two positive integers, then number H.C.F. (p, q) × L.C.M. (p, q) = p × q. R. H.C.F. of 475 and 495 (iii) 9 2. Assertion (A): 2 is an example of a rational number. Reason (R): The square roots of all positive integers S. Power of 2 in 144 (iv) 4 are irrational numbers. (a) P-(i), Q-(iv), R-(iii), S-(ii) 3. Assertion (A): For any two positive integers p and (b) P-(iii), Q-(ii), R-(i), S-(iv) q, HCF (p, q) × LCM (p, q) = p × q (c) P-(iii), Q-(i), R-(ii), S-(iv) Reason (R): If the HCF of two numbers is 5 and (d) P-(iv), Q-(i), R-(iii), S-(iv) their product is 150, then their LCM is 40. 8 Class-X MATHEMATICS P W  1. The LCM of two numbers is 2400. Which of (a) Only Statement-I is true the following can not be their HCF? (b) Both Statement-I & Statement-II is true  [CBSE 2022] (c) Only Statement-II is true (a) 300 (b) 400 (d) Neither Statement-I or Statement-II is true (c) 500 (d) 600 7. HCF of two numbers is 27 and their LCM is 162. 2. The number of zeroes in number n, if n = 23 × 32 If one of the numbers is 54, then other number is × 52 × 7, is equal to  [CBSE 2020] (a) 2 (b) 1 (a) 36 (b) 35 (c) 5 (d) 6 (c) 9 (d) 81 3. The least number that is divisible by all the natural 8. If a = 2 3 × 3, b = 2 × 3 × 5, c = 3 p × 5 and numbers from 1 to 10 (both inclusive) is [Exemplar] LCM (a, b, c) = 23 × 32 × 5, then p = (a) 10 (b) 100 (a) 6 (b) 4 (c) 504 (d) 2520 (c) 9 (d) 2 4. If two positive integers a and b can be expressed as 9. What is the greatest possible speed at which a girl p = ab2 and q = a3b; a, b being prime numbers, then can walk 95 m and 171 m in an exact number of LCM (p, q) is [Exemplar] minutes? [CBSE 2022] (a) ab 2 (b) a b 2 (a) 17 m/min. (b) 19 m/min. 3 (c) a b 2 (d) a3b3 (c) 23 m/min. (d) 13 m/min. 5. Find the greatest number which when divides 70 10. Find the smallest number which is a perfect square and 125, leaves remainder 5 and 8 respectively. and is divisible by each of 16, 20 and 24. (a) 875 (b) 10 (a) 600 (b) 2800 (c) 13 (d) 1680 (c) 3600 (d) 2400 6. Statement-I: It two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime 11. The L.C.M. of a and 18 is 36. The H.C.F. of a and numbers, then HCF of a & b is x3y2. 18 is 2. Find the value of number a. Statement-II: If n is a natural number then 12n can (a) 4 (b) 5 not end with digit 5. (c) 6 (d) 1 2 pa Very Short Answer Type Questions 2 qb 1. If HCF (336, 54) = 6, find LCM (336, 54) 2 br  [CBSE 2019] 2. Find the prime factors of 1080. 17 3. Numbers p, q and r are missing in the following 4. Find the total number of factors of prime numbers. factorisation. Find them.  [CBSE 2020] Real Numbers 9  5. There is a number 6n, where n is a natural number. Check if there is any value of n ∈ N exists, for Long Answer Type Questions which 6n is divisible by 7. 1. If a, b, c, d be positive rationals such that 6. Find a number which is greater than 3 and when it gets divided by 4, 5 and 6 always leaves 3 as remainder. a + b =+ c d , then prove that either a = c and 7. Express 429 as a product of its prime factors. b = d or b and d are squares of rationals. [CBSE 2019] 2. Prove that 7 + 11 is an irrational number. 8. For two numbers 150 and 100, it is given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100). Case Study Type Questions 9. Find a rational number between 2 and 3.  [CBSE 2019] Case Study-I The department of Mathematics is conducting an Short Answer Type Questions International Seminar. In the seminar, the number of participants in Mathematics, Physics and Computer 1. Find the number which is nearest to 100000 and Science are 60, 84 and 108 respectively. The arrangement greater than 100000 which is exactly divisible by was made such that, the same number of participants, all each of 8, 15 and 21. of them being in the same subject, are to be seated in each 2. In a school 437 girls and 342 boys have been room. A separate room was alloted for all the official other divided into groups, so that each group has the same than participants. number of students and no group has boys and girls mixed. What is the least number of groups needed? 1. Total number of participants will be (a) 160 (b) 84 3. Pooja multiplied a number 484 with a certain number to obtain the result 3823a. Find the value of a. (c) 220 (d) none of these 4. If first 100 multiples of 10 are multiplied with each 2. Find the LCM of 60, 84 and 108. other, then find the number of zeroes at the end of (a) 3780 (b) 840 the product. (c) 544320 (d) 12 5. Three people running around a rectangular track, can 3. If in each room, the same number of participants complete one turn in 2 hours, 4 hours and 5.5 hours are to be seated and all of them being in the same respectively. When will they meet at the starting point? subject, then find the least number of rooms 6. 4 bells ring together at 9.00 a.m. They ring after required for them. 7 seconds, 8 seconds, 11 seconds and 12 seconds (a) 16 (b) 20 respectively. How many times will they ring together (c) 14 (d) 21 again in the next 3 hours? 7. Find HCF and LCM of 404 and 96 and verify that Case Study-II HCF × LCM = Product of the two given numbers. Birbal, Akbar and Mohit are playing a game. Birbal climbs  [CBSE 2018] 5 stairs and gets down 2 stairs in one turn. Akbar goes up 2+ 3 by 7 stairs and comes down by 2 stairs every time. Mohit 8. Prove that is an irrational number, given 5 5 goes 10 stairs up and 3 stairs down each time. that 3 is an irrational number. [CBSE 2019] Doing this they have to reach the nearest point of the 100th 9. Prove that (3 − 5) is an irrational number. stairs and they will stop once they find it impossible to go 10. Prove that 3 is an irrational number. forward. They can not cross the 100th stair in any way. [CBSE 2020] 1. Who reaches the nearest point? 11. Prove that p + q is irrational, where p, q are (a) Birbal (b) Akbar primes. [Exemplar] (c) Mohit (d) All together 10 Class-X MATHEMATICS P W  2. How many times can they meet in between on the (b) Akbar and Mohit will meet for the first time on same stair? the 35th stair. (a) 3 (b) 4 (c) Birbal and Mohit will meet for the first time on (c) 5 (d) 0 the 21st stair. 3. Who takes the least number of steps to reach near a (d) Birbal and Akbar will meet for the first time on hundred? 21st stair. (a) Birbal (b) Akbar 5. What is the second stair where any two out of three (c) Mohit (d) All will meet together? 4. What is the first stair where any two out of three (a) Birbal and Akbar will meet on 21st stair. will meet together? (b) Akbar and Mohit will meet on the 35th stair. (a) Birbal and Akbar will meet for the first time on (c) Birbal and Mohit will meet on the 21st stair. the 15th stair. (d) Birbal and Mohit will meet on the 35th stair. ANSWER KEY Quick Recall Fill in the Blanks 1. 24 × 32 2. Irrational Number 3. The number will be HCF(35,55) = 5 4. Product Product of the numbers (396 × 576) 5. HCF = = = 36 L.C.M. 6336 True and False Statements 1. False 2. False 3. True 4. False 5. True Match the Following 1. (c) 2. (d) Assertion & Reason Type Questions 1. (d) 2. (c) 3. (c) Multiple Choice Questions 1. (c) 2. (a) 3. (d) 4. (c) 5. (c) 6. (c) 7. (d) 8. (d) 9. (b) 10. (c) 11. (a) Subjective Questions Case Study Type Questions Case Study-I 1. (d) 2. (a) 3. (d) Case Study-II 1. (a) 2. (d) 3. (c) 4. (a) 5. (c) Real Numbers 11  Competitive Level Euclid’s Division Lemma* Theorem 3: Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b. Here, when a is divided by b, i.e a is dividend, b is the divisor, q and r are obtained as quotient and remainder respectively. Let’s have a look at some examples. (i) Consider number a = 27 and b = 5, 27 = 5 × 5 + 2 Comparing this with a = bq + r, we get q = 5, r = 2 and 0 ≤ r < b (as 0 ≤ 2 < 5). (ii) Consider positive integers a = 4 and b = 14. 4 = 14 × 0 + 4 Comparing this with a = bq + r, we get q = 0, r = 4 and 0 ≤ r < b (as 0 ≤ 4 < 14). (iii) Consider positive integers a = 18 and b = 6 18 = 6 × 3 + 0 Comparing this with a = bq + r, we get q = 3, r = 0 and 0 ≤ r < b as 0 ≤ 0 < 6 NOTE: q and r can be zero. The relation a = bq + r, 0 ≤ r < b is just a restatement of long division process of number 'a' by a number 'b' in which 'q' is obtained as quotient and 'r' is obtained as remainder. Thus, dividend = divisor × quotient + remainder ⇒ a = bq + r. Knowledge Hub ™ Lemma is a proven statement used for proving an another statement. ™ Euclid's Division Lemma hold not only for positive integers but for all integers a & b, b ≠ 0. EXAMPLE 1. Prove that the square of any positive integer = 5(5m2) cannot be of the form 5q + 2 or 5q + 3 for any = 5q  [Let q = 5m2, q is an integer] integer q. [Exemplar, CBSE 2015, 2020] Case-II: (5m + 1)2 = 25m2 + 10m + 1 Sol. We know that any positive integer can be of the = 5(5m2 + 2m) + 1 form 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4, for some integer m. = 5q + 1 [Let q = 5m2 + 2m, q is an integer] Here 0 ≤ r < 5, ⸪ 0 ≤ r < b Case-III: (5m + 2)­2 = 25m2 + 20m + 4 Therefore, we have = 5(5m2 + 4m) + 4 Case-I: (5m)2 = 25m2 = 5q + 4 [Let q = 5m2 + 4m, q is an integer] 12 Class-X MATHEMATICS P W  Case-IV: (5m + 3)2 = 25m2 + 30m + 9 = 4m2 + 1 + 4m + 4n2 + 1 + 4n = 25m2 + 30m + 5 + 4 On rearranging, we get = 5(5m2 + 6m + 1) + 4 = 5q + 4 [Let q = 5m2 + 6m + 1, q is an integer] = 4m2 + 4n2 + 4m + 4n + 1 + 1 Case-V: (5m + 4)2 = 25m2 + 40m + 16 = 4(m2 + n2) + 4(m + n) + 2 = 25m2 + 40m + 15 + 1 = 2[2(m2 + n2) + 2(m + n) + 1] = 5(5m2 + 8m + 3) + 1 = 5q + 1 [Let q = 5m2 + 8m + 3, q is an integer] = 2q [Let q = 2(m2 + n2) + 2(m + n) + 1] This proves that for any integer q, the square As x2 + y2 is of the form 2q, this means that it is of any positive integer cannot be of the form 5q + 2 or 5q + 3. divisible by 2 and hence it is even. 2. If x and y are both odd positive integers, then Again x2 + y2 = 4[(m2 + n2) + (m + n)} + 2 prove that x2 + y2 is even but not divisible by 4. [Exemplar] = 4q + 2 [Let q = (m2 + n2) + (m + n)] Sol. For some integers m and n, ⇒ x2 + y2 is even and leaves remainder 2 on Let x = 2m + 1 and y = 2n + 1 dividing by 4. We have x2 + y2 \ x2 + y2 = (2m + 1)2 + (2n + 1)2 Therefore, x2 + y2 is not divisible by 4. Euclid's Division Algorithm Euclid’s Algorithm is a method to compute highest common factor (HCF) or the greatest common divisor (GCD) of two given positive integers a & b. So, let's understand Euclid's division algorithm more clearly. For obtaining HCF of two positive integers, say a and b, with a > b, follow the steps given below: Step-1: Apply Euclid's division lemma to a and b, so that we can find whole numbers q and r, such that a = bq + r (0 ≤ r < b). Step-2: If r = 0, b is the HCF of a and b. If r ≠ 0, apply the division lemma to b and r. Step-3: The process is continued till we get the remainder zero. The divisor we get at this stage will be the required HCF. NOTE: Euclid’s division lemma and algorithm are so closely interlinked that often lemma is called as the division algorithm. Knowledge Hub ™ An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. ™ The basic principle behind this algorithm is the fact that, HCF (a, b) = HCF (b, r) where, a, b, & r are dividend, divisor & remainder respectively. Real Numbers 13  EXAMPLE 1. Using Euclid’s algorithm find the HCF of 25152 = HCF (148, 50) = HCF (346, 148) = HCF (840, and 2026. 346) = HCF (2026, 840) = HCF (25152, 2026). Sol. Two given numbers are 25152 and 2026. Now, 2. A trader has 1232 Dettol soaps and 64 Pears we start dividing larger number by the smaller soaps. He packs them in such a way that each number. box has only one type of soap. If each box has Step-1: Since, 25152 > 2026, thus we apply the the same quantity of the soaps, calculate the division algorithm to 25152 and 2026. number of soaps in each box so that the total number of boxes is the lowest. 25152 = 2026 × 12 + 840 Sol. To find the number of soaps in each box such Step-2: Since, r = 840 ≠ 0, thus we apply division that the number of boxes is the minimum, we find algorithm to 2026 and 840. HCF (1232, 64). 2026 = 840 × 2 + 346 Let's find their HCF using Euclid's algorithm. Step-3: Since, r = 346 ≠ 0, thus we apply division We have, 1232 = 64 × 19 + 16 algorithm to 840 and 346. Again, 64 = 16 × 4 + 0 840 = 346 × 2 + 148 So, the HCF of 1232 and 64 is 16. Step-4: Since, the r = 148 ≠ 0, thus we apply division algorithm to 346 and 148. Therefore, the number of soaps in each box such that the number of boxes is the least is 16. 346 = 148 × 2 + 50 3. Find the HCF of 420, 130 and 600, by using Step-5: Since, r = 50 ≠ 0, thus we apply division Euclid's division algorithm. algorithm to 148 and 50. Sol. To find the HCF of 420, 130 & 600, we will first 148 = 50 × 2 + 48 find the HCF of 420 & 130 by Euclid’s division Step-6: Since, r = 48 ≠ 0, thus we apply division algorithm. algorithm to 50 and 48. By using division algorithm, we get 50 = 48 × 1 + 2 420 = 130 × 3 + 30 (r = 30 ≠ 0) Step-7: Since, r = 2 ≠ 0, thus we apply division 130 = 30 × 4 + 10 (r = 10 ≠ 0) algorithm to 48 and 2. 30 = 10 × 3 + 0 (r = 0) 48 = 2 × 24 + 0 So, HCF (420, 130) = 10 The remainder, r has now become zero, so here we stop our procedure. Since the divisor in this Now, we will find the HCF of 10 and 600 step is 2, thus the HCF of 25152 and 2026 is 2. 600 = 10 × 60 + 0 (r = 0) Observing the above solution, we can say that \ HCF (10, 600) = 10 2 = HCF (48, 2) = HCF (50, 48) Hence, HCF (420, 130, 600) = 10. p Theorem 4: Let x is a rational number whose decimal expansion terminates, then x can be expressed as q , where p and q are co-primes, and the prime factorisation of q is of the form 2m5n, where n, m are non-negative integers. 875 7 × 53 7 E.g: 0.0875 = = 4 = 4 4 4 10 2 ×5 2 ×5 Here, q = 24 × 5 which is of the form 2m × 5n, m, n ∈ W 14 Class-X MATHEMATICS P W  a Theorem 5: Let x = be a rational number in which the prime factorization of b is of the form 2m 5n, where m, n b are non-negative integers, then x has a decimal expansion which terminates. 64 26 29 512 9 9 × 22 36 36 E.g: (i) = = = = 0.512 (ii) = =2 2 = 2 2 = 0.36 125 5 3 (2 × 5) 3 10 3 25 5 × 2 (2 × 5) (10) a So, a rational number of the form , where b is of the form 2m5n can be converted to an equivalent rational number of b the form p where q is a power of 10. Therefore, the decimal expansion of such rational numbers terminate. q a Theorem 6: Let x = be a rational number in which the prime factorization of b is not of the form 2m5n, where m, n b are non-negative integers, then x has a decimal expansion which is recurring i.e. non-terminating repeating. 1 E.g: = 0.142857142857... 7 As, denominator 7 is not of the form 2m5n, therefore this rational number will not terminate. It can be concluded from the above discussion that the decimal expansion of every rational number is either terminating or non-terminating repeating. Relation Between HCF & LCM of Three Numbers a ⋅ b ⋅ c ⋅ HCF (a, b, c) (i) LCM (a, b, c) = HCF (a, b) ⋅ HCF (b, c) ⋅ HCF (a, c) a ⋅ b ⋅ c ⋅ LCM (a, b, c) (ii) HCF (a, b, c) = LCM (a, b) ⋅ LCM (b, c) ⋅ LCM (a, c) HCF & LCM of Fractions p a For fractions and q b L.C.M of numerators L.C.M.(p, a ) (i) L.C.M of fraction = = H.C.F of denominators H.C.F. (q, b) H.C.F of numerators H.C.F. (p, a ) (ii) H.C.F of fraction = = L.C.M of denominators L.C.M.(q, b) 20 52 E.g: Find the LCM and HCF of numbers &. 27 87 Sol. 20 = 22 × 5; 27 = 33; 52 = 22 × 13; 87 = 3 × 29 20 52 LCM(20,52) 22 × 5 × 13 260 \ LCM of = and = = 27 87 HCF(27,87) 3 3 20 52 HCF(20,52) 22 4 HCF of and = = = 3 27 87 LCF(27,87) 3 × 29 783 Real Numbers 15 H.C.F and L.C.M. of Decimals Follow the following procedure to find LCM & HCF Step-1: Convert each of the decimal numbers to numbers with same decimal places. Step-2: Remove the decimal point and find the HCF and LCM. Step-3: In the resultant HCF or LCM put the decimal point as there was in numbers after step-1. E.g. Find the LCM and HCF of 0.34 and 2.7. Sol. 0.34 and 2.7 can be rewritten as 0.34 and 2.70. Now, removing the decimal point we get- 34 and 270 HCF (34, 270) = 2 LCM (34, 270) = 270 × 17 = 4590 Putting the decimal point after two places as was earlier. HCF (0.34, 2.7) = 0.02 LCM (0.34, 2.7) = 45.9 Important Concepts Related to H.C.F & L.C.M of Numbers U Total number of numbers (upto a certain number a) which are divisible by a certain integer b is the quotient q, when a is divided by b. E.g. Find how many numbers up to 594 are divisible by 15? Sol. We divide 594 by 15. 594 = 39 × 15 + 9 The quotient obtained is the required number. Therefore, there are 39 such numbers upto 594 which are divisible by 15. E.g. Find how many numbers up to 600 are divisible by 7 and 5 together? Sol. As, L.C.M. of 7 and 5 = 35 \ We divide 600 by 35 600 = 17 × 35 + 5 Therefore, there are 17 such numbers upto 600 which are divisible by 7 and 5 together. Two numbers when get divided by a certain divisor leave remainders r1 and r2 and when their sum is divided by the same divisor, gives remainder r3. Then, the divisor is given by (r1 + r2 – r3). E.g. Two numbers when divided by a certain divisor leave remainders 473 and 298, respectively. When their sum is divided by the same divisor, the remainder is 236. Find the divisor. Sol. The required divisor = 437 + 298 – 236 = 499 Hence, the required divisor is 499. U The smallest number which when divided by a, b and c leave the remainders x, y and z respectively. Such that (a – x) = (b – y) = (c – z) = k(say) is (L.C.M. of a, b and c) – k. 16 Class-X MATHEMATICS P W E.g. Find the smallest number which when divided by 12, 15 and 21 leave the remainders 7, 10 and 16 respectively. Sol. Since (12 – 7) = (15 – 10) = (21 – 16) = 5, therefore the required smallest number, = (L.C.M of 12, 15 and 21) – 5 = 420 – 5 = 415. U The largest number which will divide a, b and c leaving the same remainder in each case. (a) When the value of remainder is not given: Required number = H.C.F. of |(a – b)|, |(b – c)| and |(c – a)| (b) When the value of remainder r is given: Required number = H.C.F. of (a – r), (b – r) and (c – r). E.g. Find the largest number which on dividing 34, 90 and 104 leaves the same remainder in each case. Sol. The required largest number. = H.C.F. of |(a – b)|, |(b – c) and (c – a)| = H.C.F. of |(34 – 90)|, |(90 – 104)| and |(104 – 34)| = H.C.F. of 56, 14 and 70 = 14 E.g. Find the largest number which will divide 155 and 192 and leaves the remainder 7 in each case. Sol. The required largest number = H.C.F. of (155 – 7) and (192 – 7) = H.C.F. of 148 and 185 = 37 Number of Factors of a Given Number Let N be a composite number, and its prime factors be a, b, c, d,....... etc. and k, l, m, n.... etc. be the powers of a, b, c, d.... etc. respectively i.e., N can be expressed as N = ak ⋅ bl ⋅ cm ⋅ d n.... Then, the number of total factors or divisors of N is (k + 1) (l + 1) (m + 1) (n + 1).... E.g: Find the total number of factors of 1080 Sol. 1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5 540 = 23 × 33 × 51 Therefore total number of factors of 1080 is (3 + 1) (3 + 1) (1 + 1) = 32 Number of Odd Factors To find the number of odd factors of a number N, firstly express the number N as N = (a1p × a2q × a3r ×....) × (ex) (where, a1, a2, a3,... are the odd prime factors and e is the even prime factor) Then the total number of odd factors = (p + 1)(q + 1)(r + 1) E.g: The number of odd factors of 72 is.... Sol. 72 = 23 × 32 \ Number of odd factors = (2 + 1) = 3 Number of Even Factors Number of even factors of a number = (Total number of factors of the given number) – (Total number of odd factors of the given number). Real Numbers 17 E.g: Find the number of even factors (or divisors) of 72 Sol. Total number of factors = (3 + 1) (2 + 1) = 12 Number of odd factors = (2 + 1) = 3 Number of even factors = Total number of factors – No. of odd factors = 12 – 3 = 9 Sum of Factors of a Given Number Let N be a composite number and its prime factors be a, b, c, d.... etc. and k, l, m, n.... etc. be the powers of a, b, c, d.... etc. respectively i.e., N can be expressed as N = ak. bl. cm. dn.... (a k +1 − 1)(bl +1 − 1)(c m +1 − 1)(d n +1 − 1)... then the sum of all the factors or divisors of N = (a − 1)(b − 1)(c − 1)(d − 1)... E.g: Find the sum of the factors of 180. Sol. 180 = 22 × 32 × 5 (22+1 − 1)(32+1 − 1)(51+1 − 1) 7 × 26 × 24 \ Sum of the factors of 180 = = = 546 (2 − 1)(3 − 1)(5 − 1) 1× 2 × 4 Product of Factors n The product of factors of a composite number N is given by N 2 , where n is the total number of factors of the composite number N. E.g: N = 24 = 23 × 3 No. of factor is (3 + 1) (1 + 1) = 8 8 Product of factors of = N 24 = 2 244 Irrational Numbers in Ascending and Descending Orders Irrational numbers with same index can be compared. To compare irrational numbers with different indexes, we have to convert them into same index and then compare. Writing irrational numbers in ascending and descending order: (i) To write irrational numbers (with same index) from smaller to greater is known as ‘ascending order’. (ii) To write irrational numbers (with same index) from greater to smaller is known as ‘descending order’. Converting irrational numbers with different indexes to irrational numbers with same index (i) Write the orders of irrational numbers given. (ii) Find the least common multiple of indexes of irrational numbers. (iii) Make the indexes same with the help of least common multiple found. (iv) Compare the radicands. E.g: Write 3 24 , 3 56 , 3 29 in ascending order. Sol. Q Radicals of all the given numbers are same and ascending order of radicands 24, 56 & 29 is 24, 29 & 56. So, 3 24 < 3 29 < 3 56 18 Class-X MATHEMATICS P W E.g: Write 2 3 , 3 4 , 4 2 in desending order. Sol. LCM of the indexes 2, 3 & 4 of the given numbers is 12. 1 6 1 Now, 2 3 3= = 2 312= (36 )12= 12 6 3= 12 729 1 4 1 12 3 = = (44 )12 4 4= 412 3 = 4 4= 12 256 1 3 1 3 2 12 3 4 2 2= 2 = (2 )= = 4 12 2= 12 8 Thus, all the number are in same index & 729 > 256 > 8. So, 12 729 > 12 256 > 12 8 i.e. 2 3>34>42 Unit Digit of a Number Raised any Power U Use the following table to find the unit digit of a number, ending with 0, 1, 5 or 6, raised to any power Any Power of Unit digit 0 0 1 1 5 5 6 6 E.g: (i) The unit digit of (30)35 = 0. (ii) The unit digit of (71)13 = 1. (iii) The unit digit of (35)12 = 5 (iv) The unit digit of (26)23 = 6. U The unit digit of a number, ending with 4 or 9 and raised to a power is given by- Number Unit digit when power is odd Unit digit when power is even 4 4 6 9 9 1 E.g: (i) The unit digit of (74)53 = 4 (ii) The unit digit of (74)24 = 6 (iii) The unit digit of (69)24 = 1 (iv) The unit digit of 6933 = 9 U The unit digit of the power of 2, 3, 7 and 8 repeat after every cycle of 4 steps therefore to find the unit digit of any power of 2, 3, 7 and 8, first divide the power by 4 and look at the remainder, (0, 1, 2 or 3) and use the following table: Remainder Unit digit of 0 1 2 3 2 6 2 4 8 3 1 3 9 7 7 1 7 9 3 8 6 8 4 2 E.g: (i) The unit digit of 8725 Q remainder when 25 is divided by 4 is 1. So, unit digit of 8725 = 7 (ii) The unit digit of 7232. Q remainder when 32 is divided by 4 is 0. So, unit digit of 7232 = 6. Real Numbers 19  Last Two Digits of a Product If X × Y is the product where X =….ab and Y =….cd then the last two digits of the product can be found with the following process: Step-1: Unit digit of X × Y is the unit digit in the product of b and d. If any, the excess digit will be carried forward to next step. Step-2: Add the product of a and d, and c and b i.e. find a × d + c × b. Step-3: The unit’s digit in above step becomes ten’s digit of the number. If there was any excess digit in Step 1 then it will be added to a × d + c × b and the unit’s digit in the resultant will be ten's digit of the product. E.g: X = 5647 and Y = 6384 Step-1: 7 × 4 = 28. Unit digit of the product XY = 8 and 2 is excess digit. Step-2: a × d + c × b = 4 × 4 + 8 × 7 = 72 Step-3: Add 2 (excess digit) to 72 = 74. Ten’s digit of XY = 4. So, last two digit of XY = 5647 × 6384 is 48 i.e. 4 and 8. Last Two Digit of a Number Raised to Some Power Let the number is XY. Case-1: If unit digit of X is 1 Unit digit of X Y will be 1. Ten’s digit is the unit digit in the product of ten’s digit of X and the unit’s digit of Y. E.g: Let the number is 7123 Unit digit = 1 Tens digit = unit digit in 7 × 3 = 21 i.e., 1. So, last two digit = 11. Case-2: If unit digit in X is 3, 7 or 9 In this case first we convert X to X1 such that the unit’s digit in X1 is 1, then we use Case 1. Y  Y  Y 4  4    In case of 3 and 7: X will be converted to X1 = X4 so that = X (= X ) ( X1 ) 4 . Y  Y  Y 2  2    In case of 9: X will be converted to X1 = X2 so that = X (= X ) ( X1 ) 2 . E.g: Last two digit of 1728 Q Unit digit of 17 is 7 28 = So 17 28 (17 = 4 4 ) (83521)7 Unit’s digit of 1718 = Unit digit of (83521)7 = 1 & Ten’s digit of 1718 = unit digit of (2 × 7) = Unit digit of (14) = 4 Case-3: If unit digit in X is 2, 4, 6 or 8 In this case we use the following identities. (i) 210 ends in 24. (ii) 24(odd +ve no.) ends in 24. (iii) 24(even +ve no.) ends in 76. (iv) 76n, where n is any natural number, ends in 76. 20 Class-X MATHEMATICS P W E.g: 21024 = 21020 × 24 = (210)102 × 24 Now, 210 end in 24 and any number, which ends in 24, raised to power 102 (even) ends in 76. Thus, (210)102 ends in 76. Also, 24 ends 16. Thus, Last two digit of 21024 = Last two digit of 76 × 16 = 16 Case-4: If unit digit in X is 5 In this case we use the below table: Tens digit of the number Unit’s digit of the number Last two digits Even number Odd number 25 Even number Even number 25 Odd number Odd number 75 Odd number Even number 25 E.g: Find the last two digit of (235)237 Sol. Q Digit at ten’s place is 3 which is odd & the power is also odd So, (235)237 ends in 75. Identities Related to an + bn or an – bn (i) an – bn is always divisible by a – b. (ii) an – bn is divisible by a + b if n is even. (iii) an + bn is divisible by a + b is n is odd. E.g: Find the remainder when 1516 – 616 is divided by 21. Sol. Q n is even, so 15 + 6 = 21 divides 1516 – 616. Hence remainder is 0. Highest Power of a Number in a Factorial U Factorial: Factorial of a non-negative integer n is denoted by n! and defined as, n! = n(n – 1)(n – 2)(n – 3)... 3.2.1 E.g: 5! = 5 × 4 × 3 × 2 × 1; 2! = 2 × 1 Note: 0! = 1 U Greatest integer function: Greatest integer function of a real number x is denoted by [x] and defined as the greatest integer less than or equal to x. E.g: [3.2] = 3; = 3; [–1.5] = – 2 U Highest power of a prime in a factorial: Let p is a prime number, then the highest power of p in n! n  n   n   n  =   +  2  +  3  +  4  +… , where [ ] is greatest integer function.  p  p   p   p  U Highest power of a composite number in a factorial: No. of highest power of a composite number a in n! is equal to the no. of highest power of the highest prime factor of ‘a’ in n!.  Highest power of p in n! NOTE: Highest power of pa, where p is prime, in n! is equal to   , were [ ] is greatest integer function.  a  Real Numbers 21 Examples:  40   40   40  (i) Maximum power of 3 in 40! =   +  2  +  3  = [13.33] + [4.44] + [1.48] = 13 + 4 + 1 = 18  3  3  3  (ii) Maximum power of 6 in 50!  6= 2 × 3 , here 3 is highest prime factor of 6.  50   50   50  ⇒ Maximum power of 6 in 50! = Maximum power of 3 in 50 ! =   +  2  +  3  = [16.66] + [5.55] + [1.85]  3  3  3  = 16 + 5 + 1 = 22  Maximum power of 3 in 40! 18  =  (iii) Maximum power of 9 in 40! 2 =    2= = 9  Excluded from NCERT 255 = 102 × 2 + 51 Exercise-I 102 = 51 × 2 + 0 1. Use Euclid’s division algorithm to find the HCF \ HCF (867, 255) = 51 of 2. Show that any positive odd integer is of the form (a) 135 and 225 6q + 1, or 6q + 3, or 6q + 5, where q is some (b) 196 and 38220 integer. (c) 867 and 255 Sol. According to Euclid’s Division Lemma if we have Sol. (a) 135 and 225 two positive integers a and b, then there exist As we can see, 225 is greater than 135. Hence, unique integers q and r which satisfies the condition on applying Euclid’s division algorithm, we a = bq + r where 0 ≤ r < b. have Let x be any positive odd integer which when 225 = 135 × 1 + 90 divided by 6 gives q as quotient and r as remainder. 135 = 90 × 1 + 45 According to the Euclid’s division lemma 90 = 45 × 2 + 0 x = 6q + r where 0 ≤ r < 6. We have obtained zero as remainder. The So r can be either 0, 1, 2, 3, 4 and 5. divisor of this step becomes our H.C.F. Therefore, x = 6q or 6q + 1 or 6q + 2 or 6q + 3 or Hence, the HCF of 135 and 225 is 45. 6q + 4 or 6q + 5. (b) By Euclid’s Division Algorithm, we have Case-1: 6q + 0: 6 is divisible by 2, so it is an even 38220 = 196 × 195 + 0 number. r = 0 so H.C.F. will be 196. Case-2: 6q + 1: 6 is divisible by 2, but 1 is not \ HCF (38220, 196) = 196. divisible by 2 so it is an odd number. (c) By Euclid’s Division Algorithm, we have Case-3: 6q + 2: 6 is divisible by 2, and 2 is also 867 = 255 × 3+ 102 divisible by 2 so it is an even number. 22 Class-X MATHEMATICS P W  Case-4: 6q + 3: 6 is divisible by 2, but 3 is not Case-1: a = 3q divisible by 2 so it is an odd number. ⇒ a2 = 9q2 Case-5: 6q + 4: 6 is divisible by 2, and 4 is also = 3 × 3q2 divisible by 2 so it is an even number. = 3m (Assuming m = q2) Case-6: 6q + 5: 6 is divisible by 2, but 5 is not Case-2: a = 3q + 1 divisible by 2 so it is an odd number. ⇒ a2 = (3q + 1)2 = 9q2 + 6q + 1 And therefore, any odd integer can be expressed in = 3(3q2 + 2q) + 1 the form 6q + 1 or 6q + 3 or 6q + 5. = 3m + 1 (Assuming m = 3q2 + 2q) 3. An army contingent of 616 members is to Case-3: a = 3q + 2 march behind an army band of 32 members in a parade. The two groups are to march in the ⇒ a2 = (3q + 2)2 = 9q2 + 12q + 4 same number of columns. What is the maximum = 9q2 + 12q + 3 + 1 number of columns in which they can march? = 3(3q2 + 4q + 1) + 1 Sol. Given, = 3m + 1. (Assuming m = 3q2 + 4q + l) Number of members in army contingent = 616 Therefore, the square of any positive integer (say, a2) is always of the form 3m or 3m + 1. Hence, Number of members in army band = 32 proved. If the two groups have to march in the same number of columns, we need to find out the highest common 5. Use Euclid’s division lemma to show that the factor (HCF) between the two groups. cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. We shall find the H.C.F using Euclid’s algorithm Sol. Let us suppose, x be any positive integer and As, 616 is greater than 32, therefore divisor = 3. 616 = 32 × 19 + 8 On using Euclid’s division algorithm, 32 = 8 × 4 + 0 x = 3q + r, where q ≥ 0 and 0 ≤ r < 3. Therefore Now, we have obtained zero as remainder. r = 0, 1, 2. Therefore, HCF (616, 32) = 8. Now, on putting the value of r, we get Hence, the maximum number of columns in which x = 3q (if r = 0) the two groups can march is 8. x = 3q + 1 (if r = 1) 4. Use Euclid’s division lemma to show that the x = 3q + 2 (if r = 2) square of any positive integer is either of the On cubing both sides of all the three above form 3m or 3m + 1 for some integer m. expressions, we get [Hint: Let x be any positive integer then it is of Case (i): When x = 3q the form 3q, 3q + 1 or 3q + 2. Now square each x3 = (3q)3 = 27q3 of these and show that it can be rewritten in the = 9(3q3) = 9m;  [where m = 3q3] form 3m or 3m + 1.] Case (ii): When x = 3q + 1 Sol. Let a be a positive integer, q be the quotient and r x3 = (3q + 1)3 be the remainder. = (3q)3 + 13 + 3 × 3q × 1(3q + 1) Dividing a by 3 using the Euclid’s Division Lemma, = 9(3q3 + 3q2 + q) +1 We have: = 9m + 1; [where m = 3q3 + 3q2 + q] a = 3q + r, where 0 ≤ r < 3 Case (iii): When x = 3q + 2 Putting r = 0, 1 and 2, we get: x3 = (3q + 2)3 a = 3q (if r = 0) = (3q)3 + 23 + 3 × 3q × 2(3q + 2) a = 3q + 1 (if r = 1) = 9(3q3 + 6q2 + 4q) + 8 a = 3q + 2 (if r = 2) = 9m + 8 [where m = 3q3 + 6q2 + 4q] Real Numbers 23  Hence, the cube of any positive integer is of the 29 29 = (e) form 9m, 9m + 1 or 9m + 8. 343 73 Here, denominator = 73, which is not of the Exercise-2 29 form 2n × 5m. So, the rational number has 1. Without actually performing the long division, 343 a non-terminating repeating decimal expansion. state whether the following rational numbers will have a terminating decimal expansion or a non- 23 (f ) terminating repeating decimal expansion: 2 × 52 3 17 Here, denominator = 23 × 52, which is of the form 13 (a) (b) 3125 8 2n × 5m (n = 3, m = 2). So, the rational number (c) 64 (d) 15 23 has a terminating decimal expansion. 455 1600 2 × 52 3 29 23 129 (e) (f) (g) 2 7 5 343 (23 52 ) 2 ×5 ×7 129 6 Here, denominator = 22 × 57 × 75, which is not (g) 2 7 5 (h) (2 5 7 ) 15 of the form 2n × 5m. So, the rational number 35 77 129 (i) (  j) has a non-terminating repeating 50 210 2 × 57 × 75 2 decimal expansion. 13 13 6 2 Sol. (a) = (h) = 3125 55 15 5 Here, denominator = 55, which is of the form Here, denominator = 51, which is of the form 2n × 5m (n = 0, m = 5). So, the rational number 2n × 5m (n = 0, m = 1). So, the rational number 13 6 has a terminating decimal expansion. has a terminating decimal expansion. 3125 15 17 17 35 35 7 (b) = (i)= = 8 23 50 2 × 5 2 2×5 Here, denominator = 23, which is of the form Here, denominator = 2 × 5, which is of the form 2n × 5m (n = 3, m = 0). So, the rational number 2n × 5m (n = 1, m = 1). So, the rational number 17 35 has a terminating decimal expansion. has a terminating decimal expansion. 8 50 64 64 77 11 11 (c) = (j) = = 455 5 × 7 × 13 210 30 2 × 3 × 5 Here, denominator = 5 × 7 × 13, which is not Here, denominator = 2 × 3 × 5, which is not of the of the form 2n × 5m. So, the rational number 77 form 2n × 5m. So, the rational number has a 64 210 has a non-terminating repeating decimal 455 non-terminating repeating decimal expansion. expansion. 2. Write down the decimal expansions of those 15 3× 5 3 rational numbers which have terminating (d) = = decimal expansions. 1600 26 × 52 26 × 5 Here, denominator = 26 × 5, which is of the form (a) 13 (b) 17 3125 8 2n × 5m (n = 6, m = 1). So, the rational number (c) 64 (d) 15 15 has a terminating decimal expansion. 455 1600 1600 24 Class-X MATHEMATICS P W  23 3. The following real numbers have decimal (e) 29 (f) expansions as given below. In each case, decide 343 (23 52 ) whether they are rational or not. If they are 129 (g) 2 7 5 (h) 6 rational and of the form p , what can you say (2 5 7 ) 15 q about the prime factors of q? 35 77 (a) 43.123456789 (i) (  j) 50 210 (b) 0.120120012000120000... 13 13 13 × 25 416 (c) 43.123456789 Sol. (a) = = = 3125 55 55 × 25 105 Sol. (a) R ational, prime factors of q will be either 2 13 416 or 5 or both only. ⇒= = 0.00416 3125 100000 (b) Not rational. (b) 2.125 (d) 0.009375 (f) 0.115 (h) 0.4 (i) 0.7 (c) Rational, prime factors of q will also have a factor other than 2 or 5. 1. The decimal expansion of the rational number 987 5. Statement-I: will have terminating decimal 14587 10500 , will terminate after: [Exemplar] 1250 expansion. (a) One decimal place Statement-II: For any natural number a and b there exist unique whole numbers q and r such that, a = bq (b) Two decimal places + r, (0 ≤ r ≤ b). (c) Three decimal places (a) Only I is true (b) Both I & II is true (d) Four decimal places (c) Only II is true (d) Neither I or II is true 2. If the HCF of 65 and 117 is in the form 6. Select the one having a terminating decimal 65m – 117, then the value of m is: expansion: [Exemplar CBSE 2019] 77 22 (a) (b) (a) 4 (b) 2 210 7 (c) 1 (d) 3 23 125 (c) (d) 3. (n2 – 1) is divisible by 8, if n is: [Exemplar] 8 441 (a) An integer (b) A natural number 7. For some integer q, every odd integer is of the form:  [Exemplar] (c) An odd integer (d) An even integer (a) q (b) q + 1 4. Decimal expansion of a rational number is 327.7081. (c) 2q (d) 2q + 1 a When the number is expressed in form, where 8. Among the following rational numbers, which one b has a non-terminating repeating decimal expansion? a & b are co-primes what would be in the prime 45 71 factors of b? (a) (b) 3125 512 (a) 2 and 7 (b) 3 and 8 24 (c) 2 and 5 (d) 2, 3 and 5 (c) (d) None of these 200 Real Numbers 25  9. If a is a rational number, then 52a – 22a is divisible by: 19. A certain number has exactly eight factors including (a) Both 3 and 7 (b) 9 1 and itself. Two of its factors are 21 and 35. The (c) 3 (d) None of these number is: (a) 105 (b) 210 10. The number 313 – 310 is divisible by: (c) 420 (d) 525 (a) 3 and 5 (b) 2, 3 and 13 20. The last digit in the finite decimal representation of (c) 2, 3 and 10 (d) 3 and 10 2004 1 11. Find the smallest number which on being divided the number   is: 5 by 15, leaves 5 as remainder, on being divided by 25, leaves a remainder of 15 and on being divided (a) 2 (b) 4 by 35 leaves a remainder of 25. (c) 6 (d) 8 (a) 1050 (b) 525 21. What is the greatest positive integer ‘n’ which (c) 540 (d) 515 makes n3 + 100 divisible by ‘n + 10’? (a) 890 (b) 810 12. The number 10A – 1 is divisible by 11 for: (c) 1000 (d) 900 (a) All value of A 22. Which of the following can never bе а common (b) Even values of A factor of 287 + x and 378 + x where x is any natural (c) Odd values of A number? (d) A must be multiple of 11 (a) 26 (b) 13 13. A six digit number which consists of only one type (c) 91 (d) 7 of digits, either 1, 2, 3, 4, 5, 6, 7, 8 or 9, eg 111111, 23. Number 1146600 can be written as the product of 222222... etc. This six digit number is always two factors, in how many ways? divisible by: (a) 100 (b) 273 (a) 11 (b) 13 (c) 216 (d) 108 (c) 7 (d) All of these 24. Find unit’s digit in a = 717 + 734. 14. The difference of the squares of two odd natural (a) 7 (b) 8 numbers is divisible by: (c) 10 (d) 6 (a) 8 (b) 6 25. Last two digits of 33288 will be: (c) 14 (d) 16 (a) 41 (b) 81 15. For any positive integer a, a3 – a is divisible by: (c) 56 (d) 93 (a) 2 (b) 6 26. Find the unit’s digit of (90 + 91 + 92 + 93 +... + 92009). (c) 4 (d) None of these (a) 9 (b) 5 16. Find the number of natural numbers lying between (c) 0 (d) 7 200 and 400 which are divisible by 4 and 5. (a) 10 (b) 11 27. Find the greatest power of 7 contained in 926!. (c) 9 (d) None of these (a) 148 (b) 1078 17. Find the number of perfect cubes in the sequence (c) 152 (d) None of these 11, 22, 33, 44,..., 100100. 28. How many zeroes will be there at the end of 36!36!. (a) 32 (b) 34 (a) 6! (b) 8 × 36! (c) 37 (d) 40 (c) 36! (d) 83 × 36! 18. Find the largest number which will divide 398, 436 and 542 and leave 7, 11 and 15 as remainders, 29. Find the value of n for which 2200 – 2192.31 + 2n is respectively. a perfect square. (a) 17 (b) 16 (a) 198 (b) 208 (c) 20 (d) 19 (c) 232 (d) 146 26 Class-X MATHEMATICS P W  Competitive Corner 1. When 31513 and 34369 are divided by a certain (a) 7 (b) 2 three digit number, the remainders are equal, then (c) 8 (d) 11 the remainder is.... [Andhra Pradesh 2017] 10. The expression 14m – 6m will always divisible by (a) 86 (b) 97 [Delhi 2018] (c) 374 (d) 113 (a) 8 (b) 20 2. The sum of squares of two consecutive even (c) 14 (d) 6 numbers added by 4 is always divisible by: [Chandigarh 2017]