Video transcriptLet's take the number 7 and divide it by 3. And I'm going to conceptualize dividing by 3 as let me see how many groups of 3 I can make out of the 7. So let me draw 7 things-- 1, 2, 3, 4, 5, 6, 7. So let me try to create groups of 3. So I can definitely create one group of 3 right over here. I can definitely create another group of 3. So I'm able to create two groups of 3. And then I can't create any more full groups of 3. I have essentially this thing right over here left over. So this right over here, I have this thing remaining. This right over here is my remainder after creating as many groups of 3 as I can. And so when you see something like this, people will often say 7 divided by 3. Well, I can create two groups of 3. But it doesn't divide evenly, or 3 doesn't divide evenly into 7. I end up with something left over. I have a leftover. I have a remainder of 1. So this is literally saying 7 divided by 3 is 2 remainder 1. And that makes sense. 2 times 3 is 6. So it doesn't get you all the way to 7. But then if you have your extra remainder, 6 plus that 1 remainder gets you all the way to 7. Let's do another one. Let's imagine 15 divided by 4. Let me draw 15 objects-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Now, let me try to divide it into groups of 4. So let's see, that's one group of 4. That's another group of 4. And then that's another group of 4. So I'm able to create three groups of 4. But then I can't create a fourth full group of 4. I am then left with this remainder right over here. I have a remainder right over here of 3. I have 3 left over. So we could say that 15 divided by 4 is 3 remainder 3. 4 goes into 15 three times. But that only gets us to 12. 4 times 3 is 12. To get all the way to 15, we need to use our remainder. We have to get 3 more. So 15 divided by 4, I have 3 left over. Now, let's try to think about this doing a little bit of our long division techniques. So let's say that I have 4. Let's say I want to divide 75 by 4. Well, traditional long division techniques. 4 goes into 7 one time. And If you're looking at place value, we're really saying the 4 is going into 70 ten times, because we're putting this in the tens place. And then we say, 1 times 4 is 4. But really, once again, since it's in the tens place, this is really representing a 40. But either way, we subtract 4 from the 7. We get a 3. And then we bring down this 5. And we say 4 goes into 35. Well, let's see. 4 times 8 is 32. 4 times 9 is 36. That's too big. So it goes 8 times. 8 times 4 is 32. You subtract 35 minus 32 is 3. And 4 doesn't go into 3 anymore. So here I have this 3 left over. I have a remainder of 3. So you could say that 75 divided by 4 is equal to 18 remainder 3. NCERT Solutions for Class 6 Math Chapter 2 Playing With Numbers are provided here with simple step-by-step explanations. These solutions for Playing With Numbers are extremely popular among Class 6 students for Math Playing With Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 6 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 6 Math are prepared by experts and are 100% accurate. Page No 2.10:Question 1:What are prime numbers? List all primes between 1 and 30. Answer:Those numbers with only two factors, i.e., 1 and the number itself, are known as prime numbers. Examples: 2, 3, 5, 7, 11 and 13 The prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Page No 2.10:Question 2:Write all prime numbers between: (i) 10 and 50 Answer:(i) The prime numbers between 10 and 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. Page No 2.10:Question 3:What is the smallest prime number? Is it an even number? Answer:The number 2 is the smallest prime number. It is an even prime number. Except 2, all the other even numbers are composite numbers. Page No 2.10:Question 4:What is the smallest odd prime? Is every odd number a prime number? If not, give an example of an odd number which is not prime. Answer:The smallest odd prime number is 3. No, every odd number is not a prime number. For example, 9 is an odd number but it is not a prime number because its three factors are 1, 3 and 9. Page No 2.10:Question 5:What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number. Answer:A number which has more than two factors is called a composite number. Yes, a composite number can be an odd number. The smallest odd composite number is 9. Page No 2.11:Question 6:What are the twin-primes? Write all pairs of twin-primes between 50 and 100. Answer:Twin primes: Two prime numbers are said to be twin primes if there is only one composite number between them. Twin primes between 50 and 100 are (59, 61) and (71, 73). Page No 2.11:Question 7:What are co-primes? Give examples of five of co-primes. Are co-primes always prime? If no, illustrate your answer by an examples. Answer:
Two numbers are said to be co-primes if they do not have any common factor other than 1. Two co-prime numbers need not be both prime numbers. Page No 2.11:Question 8:Which of the following pairs are always co-primes? (i) two prime numbers Answer:(i) Two prime numbers are always co-primes to each other. (ii) One prime and one composite number are not always co-prime. (iii) Two composite numbers are not always co-primes to each other. Page No 2.11:Question 9:Express each of the following as a sum of two or more primes? (i) 13 Answer:We can write the given numbers as the sums of two or more primes as follows: (i) 13 = 11 + 2 (ii) 130 = 59 + 71 (iii) 180 = 139 + 17 + 11 + 13 or 79 + 101 Page No 2.11:Question 10:Express each of the following numbers as the sum of two odd primes: (i) 36 Answer:We can express the given numbers as the sums of two odd primes as follows: (i) 36 = 7 + 29 or 17 + 19 (ii) 42 = 5 + 37 or 13 + 29 (iii) 84 = 17 + 67 or 23 + 61 Page No 2.11:Question 11:Express each of the following numbers as the sum of three odd prime numbers: (i) 31 Answer:We can express the given numbers as the sums of three odd prime numbers as follows: (i) 31 = 5 + 7 + 19 or 31 = 11 + 13 + 7 (ii) 35 = 5 + 7 + 23 or 35 = 17 + 13 + 5 (iii) 49 = 3 + 5 + 41 or 49 = 7 + 11 + 31 Page No 2.11:Question 12:Express each of the following numbers as the sum of twin primes: (i) 36 Answer:We can express the given numbers as the sums of twin primes which are as follows: (i) 36 = 17 + 19 Page No 2.11:Question 13:Find the possible missing twins for the following numbers so that they become twin primes: (i) 29 Answer:(i) The possible missing twins for 29 are 27 and 31. Since 31 is a prime and 27 is not, 31 is the missing twin. (ii) The possible missing twins for for 89 are 87 and 91. Since 87 and 91 are not primes, 89 has no twin. (iii) The possible missing twins for 101 are 99 and 103. Since 103 is a prime and 99 is not, 103 is the missing twin. Page No 2.11:Question 14:A list consists of the following pairs of numbers: (i) 51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73 Categorize them as pairs of: (i) co-primes Answer:(i) Co-primes: Two natural numbers are said to be co-prime numbers if they have 1 as their only common factor. (ii) Primes: Natural numbers which have exactly two distinct factors, i.e., 1 and the number itself are called prime numbers. (iii) Composite numbers: Natural numbers which have more than two factors are called composite numbers. Page No 2.11:Question 15:For a number, greater than 10, to be prime what may be the possible digit in the unit's place? Answer:For a number (greater than 10) to be a prime number, the possible digit in the unit's place may be 1, 3, 7, or 9. Example: 11, 13, 17, and 19 are prime numbers greater than 10. Page No 2.11:Question 16:Write seven consecutive composite numbers less than 100 so that there is no prime number between them. Answer:The required seven consecutive composite numbers are 90, 91, 92, 93, 94, 95 and 96. Page No 2.11:Question 17:State true (T) of false (F): (i) The sum of primes cannot be a prime. Answer:(i)
False. (ii) True. (iii) False. (iv) False. (v) False. (vi) False. (vii) True. Page No 2.11:Question 18:Fill in the blanks in the following: (i) A number having only two factors is called a .......................... . Answer:(i) A number having only two factors is called a prime number. Page No 2.15:Question 1:In which of the following expressions, prime factorization has been done? (i) 24 = 2 × 3 × 4 Answer:(i) 24 = 2 × 3 × 4 is not a prime
factorisation as 4 is not a prime number. Page No 2.15:Question 2:Determine prime factorization of each of the following numbers: (i) 216 Answer:(i) 216 ∴ Prime factorisation of 216 = 2 × 2 × 2 × 3 × 3 × 3 (ii) 420 ∴ Prime factorisation of 420 = 2 × 2 × 3 × 5 × 7 (iii) 468 ∴ Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13 (iv) 945 ∴ Prime factorisation of 945 = 3 × 3 × 3 × 5 × 7 (v) 7325 ∴ Prime factorisation of 7325 = 5 × 5 × 293 (vi) 13915 ∴ Prime factorisation of 13915 = 5 × 11 × 11 × 23 Page No 2.15:Question 3:Write the smallest 4-digit number and express it as a product of primes. Answer:The smallest 4-digit number is 1000. 1000 = 2 × 500 ∴ 1000 = 2 ×2 × 2 × 5 × 5 × 5 Page No 2.15:Question 4:Write the largest 4-digit number and give its prime factorization. Answer:The largest 4-digit number is 9999. Hence, the largest 4-digit number 9999 can be expressed in the form of its prime factors as 3 × 3 × 11 × 101. Page No 2.15:Question 5:Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors. Answer:The given number is 1729. Thus, the number 1729 can be expressed in the form of its prime factors as 7 ×13 ×19. Relation between its two consecutive prime factors: The consecutive prime factors of the given number are 7,
13, and 19. Page No 2.15:Question 6:Which factors are not included in the prime factorization of a composite number? Answer:1 and the number itself are not included in the prime factorisation of a composite number. Example: 4 is a composite number. Page No 2.15:Question 7:Here are two different factor trees for 60. Write the missing numbers:
Answer:(i) Since 6 = 2 × 3and 10 = 5 × 2, we have:
(ii) Since 60 = 30 × 2, 30 = 10 × 3 and 10 = 5 × 2, we have: Page No 2.20:Question 1:Test the divisibility of the following numbers by 2: (i) 6520 Answer:Rule : A natural number is divisible by 2 if its unit digit is 0, 2, 4, 6, or 8. (i) Here, the unit's digit = 0 (ii) Here, the unit's digit = 5 (iii) Here, the unit's digit = 4 Page No 2.20:Question 2:Test the divisibility of the following numbers by 3: (i) 70335 Answer:Rule : A number is divisible by 3 if the sum of its digits is divisible by 3. (i) Here, the sum of the digits in the given number = 7 + 0 + 3 + 3 + 5 = 18 which is divisible by 3. (ii) Here, the sum of the digits in the given number = 6 + 0 + 7 + 4 + 3 + 9 = 29
which is not divisible by 3. (iii) Here, the sum of the digits in the given number = 9 + 0 + 8 + 2 + 7 + 4 + 6 = 36 which is divisible by 3. Page No 2.20:Question 3:Test the divisibility of the following numbers by 6: (i) 7020 Answer:Rule : A number is divisible by 6 if it is divisible by 2 as well as 3. (i) Here, the unit's digit = 0 (ii) Here, the unit's digit
= 3 (iii) Here, the unit's digit = 0 Page No 2.20:Question 4:Test the divisibility of the following numbers by 4: (i) 786532 Answer:Rule : A natural number is divisible by 4 if the number formed by its last two digits is divisible by 4. (i) Here, the number formed by the last two digits is 32 which is divisible by 4. (ii) Here, the number formed by the last two digits is 31 which is not divisible by 4. (iii) Here, the number formed by the last two digits is 23 which is not divisible by 4. Page No 2.21:Question 5:Test the divisibility of the following numbers 8: (i) 8364 Answer:Rule: A number is divisible by 8 if the number formed by its last three digits is divisible by 8. (i) The given number = 8364 (ii) The given number = 7314 (iii) The given number = 36712 Page No 2.21:Question 6:Test the divisibility of the following numbers by 9: (i) 187245 Answer:
Rule: A number is divisible by 9 if the sum of its digits is divisible by 9. (i) The given number = 187245 (ii) The given number = 3478 (iii) The given number = 547218 Page No 2.21:Question 7:Test the divisibility of the following numbers by 11: (i) 5335 Answer:(i) The given number is 5,335. (ii) The given number is 7,01,69,803. (iii) The given number is 1,00,00,001. Page No 2.21:Question 8:In each of the following numbers, replace * by the smallest number to make it divisible by 3: (i) 75 * 5 Answer:We can replace the * by the smallest number to make the given numbers divisible by 3 as follows: Page No 2.21:Question 9:In each of the following numbers, replace * by the smallest number to make it divisible by 9: (i) 67 * 19 Answer:(i) Sum of the given digits = 6 + 7 + 1 + 9 = 23 (ii) Sum of the given digits = 6 + 6 + 7 + 8 + 4 = 31 (iii) Sum of the given digits = 5 + 3 + 8 + 8 = 24 Page No 2.21:Question 10:In each of the following numbers. replace * by the smallest number to make it divisible by 11: (i) 86 * 72 Answer:Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11. (i) 86 × 72 Sum of the digits at the odd places = 8 + missing
number + 2 = missing number + 10 (ii) 467 × 91 Sum of the digits at the odd places = 4 + 7 + 9 = 20 (iii) 9 × 8071
Sum of the digits at the odd places = 9 + 8 + 7 = 24 Page No 2.21:Question 11:Given an example of a number which is divisible by (i) 2 but not by 4. Answer:(i) A number which is divisible by 2 but not by 4 is 6. Page No 2.21:Question 12:Which of the following statements are true? (i) If a number is divisible by 3, it must be divisible by 9. Answer:(i) False. 12 is divisible by 3 but not by 9. Page No 2.24:Question 1:Find the H.C.F of the following numbers using prime factorization method: (i) 144,198 Answer:(i) 144 and 198 Prime factorisation of
144 = 2 × 2 × 2 × 2 ×3 × 3 (ii) 81 and 117 Prime factorisation of 81 = 3 ×3 × 3 × 3 (iii) 84 and 98
Prime factorisation of 84 = 2 × 2 × 3 × 7 (iv) 225 and 450 Prime factorisation of 225 = 3 × 3 × 5 × 5 (v) 170 and 238 Prime factorisation of 170 = 2 × 5 × 17 (vi) 504 and 980 (vii) 150, 140 and 210 Prime factorisation of 150 = 2 × 3 × 5 ×5 (viii) 84, 120 and 138 Prime factorisation of 84 = 2 × 2 ×
3 × 7 (ix) 106, 159, and 265 Prime factorisation of 106 = 2 × 53 Page No 2.24:Question 2:What is the H.C.F of two consecutive (i) Numbers Answer:(i) The common
factor of two consecutive numbers is always 1. (ii) The common factors of two consecutive even numbers are 1 and 2. (iii) The common factor of two consecutive odd numbers is 1. Page No 2.24:Question 3:H.C.F of co-prime numbers
4 and 15 was found as follow: Answer:No, it is not correct. Page No 2.27:Question 1:Determine the H.C.F of the following numbers by using Euclid's algorithm (i-x): (i) 300,450 Answer:(i) 300 and 450 Clearly, the last divisor is 150. (ii) 399 and 437 Clearly, the last divisor is 19. (iii) 1045 and 1520 Clearly, the last divisor is 95. Page No 2.27:Question 2:Show that the following pairs are co-prime: (i) 59,97 Answer:We know that two numbers are co-primes if their HCF is 1. (i) 59 and 97 Clearly, the last divisor is 1. Hence, the given numbers are co-primes. (ii) 875 and 1859 Clearly, the last divisor is 1. (iii) 288
and 1375 Clearly, the last divisor is 1. Page No 2.27:Question 3:What is the H.C.F of two consecutive numbers? Answer:The HCF of two consecutive numbers is 1.
Clearly, the last divisor is 1. Page No 2.27:Question 4:Write true (T) of false (F) for each of the following statements: (i) The H.C.F of two distinct prime numbers is 1. Answer:(i) True. Page No 2.31:Question 1:Find the largest number which 615 and 963 leaving remainder 6 in each case. Answer:We have to find the
largest number which divides (615 − 6) and (963 − 6) exactly. Resolving 609 and 957 into prime factors, we have: Page No 2.31:Question 2:Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively. Answer:We have to find the greatest number which divides (285 − 9) and (1,249 − 7) exactly. Resolving 276 and 1242 into prime factors, we have: ∴ HCF of 276 and 1242 is 2 × 3 × 23 = 138. Page No 2.31:Question 3:What is the largest number that divides 626,3127 and 15628 and leaves remainders of 1,2 and 3 respectively? Answer:We have to find the largest number which divides (626 − 1), (3,127 − 2), and (15,628 − 3) exactly. Resolving 625,
3125, and 15625 into prime factors, we have: Therefore, HCF of 625, 3125 and 15625 = 5 × 5 × 5 × 5 = 625 Page No 2.31:Question 4:The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions if the room exactly. Answer:Given: Length of the room = 8 m 25 cm = 825 cm The longest rod will be given by the HCF of 825, 675 and 450. Thus, the required length of the longest rod is 75 cm. Page No 2.31:Question 5:A rectangular courtyard is 20 m 16 cm long and 15 m 60 cm broad. It is to be paved with square stones of the same size. Find the least possible number of such stones. Answer:Length of the rectangular courtyard = 20 m 16 cm = 2,016 cm Least possible side of square stones used to pave the rectangular courtyard
is 24 cm. =Area of rectangular courtyardArea of square stone =2016 cm ×1560
cm(24 cm)2=5460 Page No 2.31:Question 6:Determine the longest tape which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm. Answer:Given: The length of the longest tape will be the HCF of 700, 385, and 1,295. ∴ Required length of the longest tape = 35 cm Page No 2.31:Question 7:105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip. Answer:We have to find the largest possible number of animals. Thus, we will have to find the HCF of 105, 140, and 175. ∴ Required HCF = 5 × 7 = 35 Page No 2.31:Question 8:Two brands of chocolates are available in packs of 24 and 15 respectively. If i need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind i would need to buy? Answer:Let the brand ‘A’ contain 24 chocolates in one packet and brand ‘B’ contain 15 chocolates in one packet. ∴ LCM of 15 and 24 is: Required LCM = 2 × 2 × 2 × 3 × 5 = 120 Therefore, minimum 120 chocolates of each kind should be purchased. Number of boxes of brand ‘A’ which needs to be purchased = 120 ÷ 24 = 5 Number of boxes of brand ‘B’ which needs to be purchased = 120 ÷ 15 = 8 Page No 2.31:
Question 9:During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many each would you need to buy? Answer:To find the required number of pencils and crayons, we need to find the LCM of 24 and 32. ∴ Required LCM of 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 = 96 Page No 2.31:Question 10:Reduce each of the following fractions to the lowest terms: (i) 161207 (ii) 296481 Answer:(i) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF. Now, 161÷23207÷23=79
(ii) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF. Now, 296÷37481÷37=813 Page No 2.31:Question 11:A merchant has 120 litres of oil of one kind, 180 liters of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin? Answer:
The maximum capacity of the required tin is the HCF of the three quantities of oil. Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5 Page No 2.34:Question 1:Determine the L.C.M of the numbers given below: (i) 48,60 Answer:(i) Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3 Page No 2.37:Question 1:What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time? Answer:We have to find prime factorisation of 24, 36, and 54. Prime factorisation of 24 = 2 × 2 × 2 × 3 Thus, 216 is the smallest number exactly divisible by 24, 36, and 54. Page No 2.37:Question 2:What is the smallest number that both 33 and 39 divide leaving remainders of 5? Answer:We have to find prime factorisation of 33 and 39. Prime factorisation of 33 = 3
× 11 Thus, 429 is the smallest number exactly divisible by 33 and 39. Page No 2.37:Question 3:Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive). Answer:To find the
required least number, we have to find the LCM of the numbers from 1 to 10. Prime factorisation of 4 = 2 × 2 ∴ Required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520 Page No 2.37:Question 4:What is the smallest number that, when divide by 35, 56 and 91 leaves remainders of 7 in each case? Answer:We have to find the prime factorisation of 35, 56, and 91. Prime factorisation of 35 = 5 × 7 Thus, 3,640 is the smallest number exactly
divisible by 35, 56, and 91. Page No 2.37:Question 5:In a school there are two sections - section A and section B of Class VI. There are 32 students in section A and 36 in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B. Answer:We have to find the LCM of 32 and 36. Required LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288 Page No 2.37:Question 6:In a morning walk three persons step off together. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? Answer:We have to find the LCM of 80 cm, 85 cm, and 90 cm. ∴ Required LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 ×
17 = 12,240 Page No 2.37:Question 7:Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21. Answer:First, we have to find the LCM of 8, 15, and 21. Prime factorisation of 8 = 2 × 2 × 2 The number nearest to 1,00,000 and exactly divisible by each of 8, 15, and 21 should also be exactly divisible by their LCM (i.e. 840). We have to divide 1,00,000 by 840. Remainder = 40 ∴ Number just greater than 1,00,000 and exactly divisible by 840 = 1,00,000 + (840 − 40) = 1,00,000 + 800 = 1,00,800 ∴ Required number = 1,00,800 Page No 2.37:Question 8:A school bus picking up children in a colony of flats stops at every sixth block of flats. Another school bus starting from the same place stops at every eighth blocks of flats. Which is the first bus stop at which both of them will stop? Answer:First bus stop at which both the buses will stop together = LCM of 6th block and 8th block Prime
factorisation of 6 = 2 × 3 Page No 2.37:Question 9:Telegraph poles occur at equal distances of 220 m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole? Answer:We have to find the LCM of 220 m and 300 m. Prime factorisation of 220 = 2 × 2 × 5 × 11 Hence, 3,300 m far is the next heap that lies at the foot of a pole. Page No 2.37:Question 10:Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. Answer:First, we have to find the LCM of 28 and 32. It is given that when we divide the number by 28, the remainder is 8 and
when we divide the number by 32, the remainder is 12. Page No 2.40:Question 1:For each of the following pairs of numbers, verify the property: (i) 25,65 Answer:(i) Given numbers are 25 and 65. Prime factorisation of 25 = 5 × 5 HCF of 25 and 65 = 5 ∴ Product of the number = Product of their HCF and LCM (Verified) (ii) Given numbers are 117 and 221. Prime factorisation of 117 = 3 × 3 × 13 HCF of 117 and 221 = 13 ∴ Product of the number = Product of their HCF and LCM (Verified) (iii) Given numbers are 35 and 40. Prime factorisation of 35 = 5 × 7 HCF of 35 and 40 = 5 ∴ Product of the number = Product of their HCF and LCM (Verified) (iv) Given numbers are 87 and 145. Prime factorisation of 87 = 3 × 29 HCF of 87 and 145 = 29 ∴ Product of the number = Product of their HCF and LCM (Verified) (v) Given numbers are 490 and 1,155. Prime factorisation of 490 = 2 × 5 × 7 × 7 HCF of
490 and 1,155 = 5 × 7 = 35 ∴ Product of the number = Product of their HCF and LCM (Verified) Page No 2.40:Question 2:Find the H.C.F and L.C.F of the following pairs of numbers: (i) 117,221 Answer:(i) 117 and 221 Prime factorisation of 117 = 3 × 3 × 13 ∴ Required HCF of 117 and 221 = 13 (ii) 234 and 572 Prime
factorisation of 234 = 2 × 3 × 3 × 13 Required HCF of 234 and 572 = 2 × 13 = 26 (iii) 145 and 232 Prime factorisation of 145 = 5 × 29 Required HCF of 145 and 232 = 29 (iv) 861 and 1,353 Prime factorisation of 861 = 3 × 7 × 41 Required HCF of 861 and 1,353 = 3 × 41 = 123 Page No 2.40:Question 3:The L.C.M and H.C.F of two numbers are 180 and 6 respectively. If ones of the numbers is 30, find the other number. Answer:Given: Product of the two numbers = Product of their HCF and LCM ∴ 30 × other number = 6 × 180 Page No 2.40:Question 4:The H.C.F of two numbers is 16 and their product is 3072. Find their L.C.M. Answer:Given: Product of the two numbers = Product of their HCF and LCM ∴ 3,072 = 16 × LCM LCM = =307216=192 Page No 2.40:Question 5:The H.C.F of two numbers is 145, their L.C.M is 2175. If one number is 725, find the other. Answer:HCF of two numbers = 145 Product of the given two numbers = Product of their LCM and HCF Other number = 145 x 2175725 =435 Thus, the required number is 435. Page No 2.40:Question 6:Can two numbers have 16 as their H.C.F and 380 as their L.C.M.? Give reason. Answer:No. Page No 2.40:Question 1:Which of the following numbers is a perfect number? (a) 4 Answer:(d) 6 Sum of the factors of 6 = 1 + 2 + 3 + 6 = 12 = 2 × 6 Page No 2.40:Question 2:Which of the following are not twin-primes? (a) 3,5 Answer:(d) 17, 23 Page No 2.40:Question 3:Which of the following are co-primes? (a) 8,10 Answer:(b) 9, 10 Page No 2.40:Question 4:Which of the following is a prime number? (a) 263 Answer:(a) 263 Page No 2.40:Question 5:The number of primes between 90 and 100 is (a) 0 Answer:(b) 1 Page No 2.40:
Question 6:Which of the following numbers is a perfect number? (a) 16 Answer:(d) 28 Sum of factors of 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28 Page No 2.40:Question 7:Which of the following is a prime number? (a) 203 Answer:(b) 139 139 = 1 × 139 Page No 2.40:Question 8:The total number of even prime numbers is (a) 0 Answer:(b)
1 Page No 2.40:Question 9:Which one of the following is a prime number? (a) 161 Answer:(c) 373 373 = 1 × 373 Page No 2.40:Question 10:The least prime is (a)
1 Answer:(b) 2 Page No 2.40:Question 11:Which one of the following numbers is divisible by 3? (a) 27326 Answer:(c) 73,545 Page No 2.40:Question 12:Which of the following numbers is divisible by 4? (a) 8675231 Answer:(b) 98,43,212 Page No 2.40:Question 13:Which of the following numbers is divisible by 6? (a) 7908432 Answer:(a) 79,08,432 and (c) 4,59,82,014 A number divisible by 6 must also be divisible by 3 and 2 as 6 is a multiple of 3 and 2. In 79,08,432, the sum of the digits = 7 + 9 + 0 + 8 + 4 + 3 + 2 = 33 In number 4,59,82,014, the sum of the digits = 4 + 5 + 9 + 8 + 2 + 0 + 1 + 4 = 33. Page No 2.40:Question 14:
Which of the following numbers is divisible by 8? (a) 87653234 Answer:(d) 9,87,41,032 Page No 2.40:Question 15:Which of the following numbers is divisible by 9? (a) 9076185 Answer:(a) 90,76,185 Page No 2.41:Question 16:Which of the following numbers is divisible by 11? (a) 1111111 Answer:(b) 2,22,22,222 Page No 2.41:Question 17:If 1*548 is divisible by 3, then * can take the value (a) 0 Answer:(a) 0 Page No 2.41:Question 18:5*2 is a three digit number with * as a missing digit. If the number is divisible by 6, the missing digit is.
(a) 2 Answer:(a) 2 Page No 2.41:Question 19:What least value should be given to * so that the number 6342*1 is divisible by 3? (a) 0 Answer:(c) 2 Page No 2.41:Question 20:What least value should be given to * so that the number 915*26 is divisible by 9? (a) 1 Answer:(b) 4 Page No 2.41:Question 21:What least number be assigned to * so that number 653*47 is divisible by 11? (a) 1 Answer:(a) 1 Sum of the digits at odd places = 6 + 3 + 4 = 13 Page No 2.41:Question 22:What least number be assigned to * so that the number 63576*2 is divisible by 8? (a) 1 Answer:(c) 3 Page No 2.41:Question 23:Which one of the following numbers is exactly divisible by 11? (a) 235641 Answer:(d) 4,15,624 Page No 2.41:Question 24:If 1*548 is divisible by 3, which of the following digits can replace *? (a) 0 Answer:(a) 0 Page No 2.41:Question 25:The sum of the prime numbers between 60 and 75 is (a) 199 Answer:(d) 272 Page No 2.41:
Question 26:The HCF of two consecutive natural numbers is (a) 0 Answer:(b) 1 Page No 2.41:Question 27:The HCF of two consecutive even numbers is (a) 1
Answer:(b) 2 Page No 2.41:Question 28:The HCF of two consecutive odd numbers is (a) 1 Answer:(a) 1 Page No 2.41:Question 29:The HCF of an even number and an odd number is (a) 1 Answer:(d) non-existent Page No 2.41:Question 30:The LCM of 24,36 and 40 is (a) 4 Answer:(c) 360 Page No 2.41:Question 31:If x and y are two co-primes, then their LCM is (a) xy Answer:(a) xy Page No 2.41:Question 32:If the HCF of two number is 16 and their product is 3072, then their LCM is (a) 182 Answer:(b) 192 Page No 2.41:Question 33:The least number divisible by 15,20,24,32 and 36 is (a) 1440 Answer:(a) 1,440 ∴ LCM of 15, 20, 24, 32 and 36 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1,440 Hence, 1,440 is the least number that is divisible by 15, 20, 24, 32 and 36. Page No 2.41:Question 34:The smallest number which when diminished by 3 is divisible by 11,28,36 and 45 is (a) 1257 Answer:(d) None of these Page No 2.41:Question 35:Three numbers are in the ratio 1:2:3 and their HCF is 6, the numbers are (a) 4,8,12 Answer:(c) 6, 12, 18 Page No 2.41:Question 36:The ratio of two numbers is 3:4 and their HCF is 4. Their LCM is (a) 12 Answer:(d) 48 Page No 2.42:Question 1:Which of the following numbers is prime? Answer:(a) 23 = 1 × 23, Page No 2.42:Question 2:Which of the following numbers are twin primes? Answer:Twin primes are pairs of primes which differ by two. Page No 2.42:Question 3:What smallest digit be written in the blank space of the mumber ......6724 so that the number formed is divisible by 3? Answer:A number is divisible by 3 if the sum of its digits is divisible by 3. Page No 2.42:Question 4:Which of the following numbers is divisible by 6? Answer:A number divisible by 2 and 3 is also divisble by 6. Page No 2.42:Question 5:Which of the following numbers is divisible by 11? Answer:A number is divisible by 11 if the difference of the sums of alternating digits is divisible by
11. Page No 2.42:Question 6:Which of the following numbers is a perfect number? Answer:A perfect number is a
positive number that equals the sum of its divisors, excluding itself. Page No 2.42:Question 7:Which of the folowing numbers is not divisible by 4? Answer:A number is divisible by 4 if the number's last two digits are divisible by 4. Page No 2.42:Question 8:The
smallest prime just greater than the HCF of 84 and 144 is Answer:84 = 1 × 2 × 2 × 3 × 7 = 22 × 31 × 71 Page No 2.42:
Question 9:Every counting number has an infinite number of Answer:Multiples are what we get after multiplying the number by any number. Page No 2.42:Question 10:What least number should be replaced by * so that the number
37610*2 is exactly divisible by 9 ? Answer:A number is divisible by 9 if the sum of its digits is divisible by 9. Page No 2.42:Question 11:Define a perfect number. Write two perfect numbers Answer:A perfect number is a positive number that equals the sum of its divisors, excluding itself. Page No 2.42:Question 12:Make a list of seven consecutive numbers, none of which is prime. Answer:The seven consecutive numbers, none of which is prime are: Page No 2.42:Question 13:The HCF of two numbers is 23 and their product is 55545. Find their LCM Answer:Product of two numbers = HCF of two numbers × LCM of two numbers Page No 2.42:Question 14:Find the smallest 5-digit number which is exactly divisible by 20, 25, 30. Answer:20 = 1 × 2 × 2 × 5 = 22 × 51 Page No 2.42:Question 15:Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case. Answer:First we subtract the required remainder from 615 and 963. Page No 2.42:Question 16:The length, breadth and height of a room are 1050 cm, 750 cm and 425 cm respectively. Find the length of the longest tape which can measure the three dimensions of the room exactly Answer:1050 = 1 × 2 × 3 × 5 × 5 × 7 = 21 × 31 × 52 × 71 Page No 2.42:Question 17:Find the greatest number of four digits which is exactly divisible by each of 8, 12, 18 and 30. Answer:8 = 1 × 2 × 2 × 2 = 23 Page No 2.42:Question 18:Find the least number of five digits which is exactly divisible by each of 8, 12, 18, 40 and 45 Answer:8 = 1 × 2 × 2 × 2 = 23 Page No 2.42:Question 19:Reduce 289 391 to the lowest terms Answer:289391=17×1717×23=1723 Page No 2.42:Question 20:Three tankers contain 403 lltres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times. Answer:The maximum capacity of three containers is equal to the HCF of 403, 434 and 465. Page No 2.42:Question 21:A number which has only two factors is called a .............. Answer:A number which has only two factors is called a prime number Page No 2.43:Question 22:The smallest composite number is ...... Answer:A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself). Page No 2.43:Question 23:Two perfect numbers are ......... and ........... Answer:A perfect number is a positive number that
equals the sum of its divisors, excluding itself. Page No 2.43:Question 24:The HCF of two consecutive odd numbers is .......... Answer:Since, the common factor in
two consecutive odd numbers is only 1. Page No 2.43:Question 25:The prime triplet is ................ Answer:A set of three prime numbers which form an arithmetic sequence with common difference two is called a prime triplet. Page No 2.43:Question 1:The greatest five digit number exactly divisible by 9 and 13 is Answer:LCM of 9 and 13 = 9 × 13 = 117 Page No 2.43:Question 2:From the numbers 2, 3, 4, 5, 6, 7, 8, 9 how many pairs of co-primes can be formed? Answer:We can form 19 pairs of co primes from the 2, 3, 4, 5, 6, 7, 8, 9 which are given below, Page No 2.43:Question 3:If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of a + b is Answer:A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maximum. Page No 2.43:Question 4:The HCF of 100 and 101 is . Answer:100 = 1 × 2 × 2 × 5 × 5 Page No 2.43:Question 5:The LCM of 100 and 101 is Answer:100 = 1 × 2 × 2 × 5 × 5 Page No 2.43:Question 6:The
greatest four digit number which when divided by 18 and 12 leaves a remainder of 4 in each case is Answer:18 = 1 × 2 × 3 × 3 = 21 × 32 Page No 2.43:Question 7:The GCD of two numbers is 17 and their LCM is 765. How many pairs of values can the numbers assume? Answer:GCD of two numbers is 17 Page No 2.43:Question 8:The number of factors of 1080 is Answer:1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5 =
23 × 33 × 51 Page No 2.43:Question 9:The HCF of first 100 natural numbers is Answer:The HCF of first 100 natural numbers is 1 because there are some prime numbers like 2, 3, 5 and so on which can't
have common factor other than 1. Page No 2.43:Question 10:The least number exactly divisible by 36 and 24 is Answer:36 = 2 × 2 × 3 × 3 = 22 × 32 Page No 2.43:Question 11:Find the HCF of all natural numbers from 200 to 478. Answer:The HCF of all natural numbers from 200 to 478 is 1 because there are some prime numbers like 211, 233 and so on which can't have common factor other than 1. Page No 2.43:Question 12:If x is prime, y is a composite number such that x + y = 240 and their LCM is 4199. Find x and y. Answer:We know that the LCM of a prime number and a composite number is equal to their product. Page No 2.43:Question 13:The LCM of two numbers is 1024 and one of them is a prime number. Find their HCF. Answer:LCM of two numbers is 1024 = 210 Page No 2.43:Question 14:The HCF of two numbers is 4 and their LCM is 400. How many pairs of values can the numbers assume? Answer:HCF of two numbers is 4 Page No 2.44:Question 15:Find the greatest number that can divide 101 and 115 leaving remainders 5 and 7 respectively. Answer:First we will subtract 5 and 7 from 101 and 115 recpectively. Page No 2.44:Question 16:Find the least three digit number which when divided by 20, 30, 40 and 50 leaves remainder 10 in each case. Answer:20 = 1 × 2 × 2 × 5 = 22 × 51 Page No 2.44:Question 17:Find the largest number that divides 59 and 54 leaving remainders 3 and 5 respectively Answer:First we will subtract 3 and 5 from 59 and 54 recpectively. Page No 2.44:Question 18:Can two numbers have 12 as their HCF and 512 as their LCM? Justify your answer Answer:HCF of two numbers is a factor of the LCM of those numbers Page No 2.44:Question 19:Write all prime numbers between 50 and 100. Answer:The prime numbers between 50 and 100 are given below: Page No 2.44:Question 20:Find the least 5-digit number which is exactly divisible by 20, 25 and 30. Answer:20 = 1 × 2 × 2 × 5 = 22 × 51 Page No 2.44:Question 21:The least number which when divided by 6, 9, 12 and 18 leaves no remainder is Answer:6 = 2 × 3 = 21 × 31 Page No 2.44:Question 22:If the product of two numbers is 360 and their HCF is 6, then their LCM is Answer:Product of two numbers = HCF of two numbers × LCM of two numbers Page No 2.44:Question 23:The least number which when divided by 5, 7 and 8 leaves 3 as remainder in each case is Answer:LCM of 5, 7 and 8 is 5 × 7 × 8 = 280. Page No 2.44:Question 24:The LCM of two numbers is 26. The possible values of HCF are ....... Answer:26 = 1 × 2 × 13 Page No 2.44:Question 25:Two perfect numbers are ......... and ........... Answer:A perfect number is a positive number that equals the sum of its
divisors, excluding itself. and 28.Page No 2.5:Question 1:Define (i) factor Give four examples of each. Answer:
(i) Factor: A factor of a number is an exact divisor of that number. Examples of factors are:
(ii) Multiple: When a number 'a' is multiplied by another number 'b', the product is the multiple of both the numbers 'a' and 'b'. Examples of multiples:
Page No 2.5:Question 2:Write all factors of each of the following numbers: (i) 60 Answer:(i) 60 = 1 × 60 (ii) 76 = 1 × 76 (iii) 125 = 1 × 125 (iv) 729 = 1 × 729 ∴ The factors of 729 are 1, 3, 9, 27, 81, 243 and 729. Page No 2.5:Question 3:
Write first five multiples of each of the following numbers: (i) 25 Answer:(i) The first five multiples of 25 are as follows: 25 × 1 = 25 25 × 2 = 50 25 × 3 = 75 25 × 4 = 100 25 × 5 = 125 (ii) The first five multiples of 35 are as follows: 35 × 1 = 35 35 × 2 = 70 35 × 3 = 105 35 × 4 = 140 35 × 5 = 175 (iii) The first five multiples of 45 are as follows: 45 × 1 = 45 45 × 2 = 90 45 × 3 = 135 45 × 4 = 180 45 × 5 = 225 (iv) The first five multiples of 40 are as follows: 40 × 1 = 40 40 × 2 = 80 40 × 3 = 120 40 × 4 = 160 40 × 5 = 200 Page No 2.5:Question 4:Which of the following numbers have 15 as their factor? (i) 15625 Answer:(i)15 is not a factor of 15,625 because it is not a divisor of 15,625. (ii) 15 is a factor of 1,23,015 because it is a divisor of
1,23,015. Page No 2.5:Question 5:Which of the following numbers are divisible by 21? (i) 21063 Answer:We know that a given number is divisible by 21 if it is divisible by each of its factors. (i) Sum of the digits of the
given number = 2 + 1 + 0 + 6 + 3 = 12 which is divisible by 3. Again, a number is divisible by 7 if the difference between twice the one's digit and the number formed by the other digits is either 0 or a multiple of 7. (ii) Sum
of the digits of the given number = 2 + 0 + 1 + 6 + 3 = 12 which is divisible by 3. Again, a number is divisible by 7 if the difference between twice the one's digit and the number formed by the other digits is either 0 or multiple of 7. Page No 2.5:Question 6:Without actual division show that 11 is a factor of each of the following numbers: (i) 1111 Answer:(i) 1,111 The sum of the digits at the odd places = 1 + 1 = 2 The sum of the digits at the even places = 1 + 1 = 2 The difference of the two sums = 2 − 2 = 0 ∴ 1,111 is divisible by 11 because the difference of the sums is zero. (ii) 11,011 The sum of the digits at the odd places = 1 + 0 + 1 = 2 The sum of the digits at the even places = 1 + 1 = 2 The difference of the two sums = 2 − 2 = 0 ∴ 11,011 is divisible by 11 because the difference of the sums is zero. (iii) 1,10,011 The sum of the digits at the odd places = 1 + 0 + 1 = 2 The sum of the digits at the even places = 1 + 0 + 1 = 2 The difference of the two sums = 2 − 2 = 0 ∴ 1,10,011 is divisible by 11 because the difference of the sums is zero. (iv) 11,00,011 The sum of the digits at the odd places = 1 + 0 + 0 + 1 = 2 The sum of the digits at the even places = 1 + 0 + 1 = 2 The difference of the two sums = 2 − 2 = 0 ∴ 11,00,011 is divisible by 11 because the difference of the sums is zero. Page No 2.5:Question 7:Without actual division show that each of the following numbers is divisible by 5: (i) 55 Answer:A number will be divisible by 5 if the unit's digit of that number is either 0 or 5. (i) In 55, the unit's digit is 5. Hence, it is divisible by 5. (ii) In 555, the unit's digit is 5. Hence, it is divisible by 5. (iii) In 5,555, the unit's digit is 5. Hence, it is divisible by 5. (iv) In 50,005, the unit's digit is 5. Hence, it is divisible by 5. Page No 2.5:Question 8:Is there any natural number having no factor at all? Answer:No, because each natural number is a factor of itself. Page No 2.5:Question 9:Find numbers between 1 and 100 having exactly three factors. Answer:The numbers between 1 and 100 having exactly three factors are 4, 9, 25, and 49. The factors of 4 are 1, 2 and 4. The factors of 9 are 1, 3 and 9. The factors of 25 are 1, 5 and 25. The factors of 49 are 1, 7 and 49. Page No 2.5:Question 10:Sort out even and odd numbers: (i) 42 Answer:A number which is exactly divisible by 2 is called an even number. A number
which is not exactly divisible by 2 is called an odd number. Page No 2.7:Question 1:Find the common factors of: (i) 15 and 25 Answer:(i) 15 and 25 Again, 25 = 1 × 25 Therefore, the common factors of the two numbers are 1 and 5. (ii) 35 and 50 Again, 50 = 1 × 50 Therefore, the common factors of the two numbers are 1 and 5. (iii) 20 and 28 Again, 28 = 1 × 28 Therefore, the common factors of the two numbers are 1, 2 and 4. Page No 2.7:Question 2:Find the common factors of: (i) 5, 15 and 25 Answer:(i) 5, 15 and 25 (ii) 2, 6 and 8 Therefore, the common factors of 2, 6 and 8 are 1 and 2. Page No 2.7:Question 3:Find first three common multiples of 6 and 8. Answer:Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, … Therefore, the first three common multiples of 6 and 8 are 24, 48 and 72. Page No 2.7:Question 4:Find first two common multiples of 12 and 18. Answer:Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, … Therefore, the first two common multiples of 12 and 18 are 36 and 72. Page No 2.7:Question 5:A number is divisible by both 7 and 16. By which other number will that number be always divisible?
Answer:Since the number is divisible by 7 and 16, they are the factors of that number. Page No 2.7:Question 6:A number is divisible by 24. By what other numbers will that number be divisible? Answer:Since the number is divisible by 24, it will be divisible by all the factors of 24. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. Hence, the number is also divisible by 1, 2, 3, 4, 6, 8 and 12. View NCERT Solutions for all chapters of Class 6 |