Upon reaching the 6th standard, students are introduced to the NCERT Solutions for Class 6 Maths Chapter 3 syllabus, which is quite variant and includes different topics. The third chapter among the Maths syllabus, Playing with Numbers, particularly focuses on teaching students about multiples and divisors. As the chapter progresses, the students are introduced to topics such as common factors and multiples, divisibility rules, highest common factors, lowest common factors, etc. A proper comprehension of the chapter will enable students to understand prime and composite numbers, as well as their difference. You can also download NCERT Class 6 Maths and NCERT Class 6 Science to help you to revise complete syllabus and score more marks in your examinations. ch3-6Download NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers Exercise 3.1 1. Write all the factors of the following numbers. 2424 Ans: Given: number 2424 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 2424 as, 1×241×24 2×122×12 3×83×8 4×64×6 Therefore, the factors of 2424 will be 1,2,3,4,6,8,12,241,2,3,4,6,8,12,24. 1515 Ans: Given: number 1515 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 1515 as, 1×151×15 3×53×5 Therefore, the factors of 1515 will be 1,3,5,151,3,5,15. 2121 Ans: Given: number 2121 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 2121 as, 1×211×21 3×73×7 Therefore, the factors of 2121 will be 1,3,7,211,3,7,21. 2727 Ans: Given: number 2727 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 2727 as, 1×271×27 3×93×9 Therefore, the factors of 2727 will be 1,3,9,271,3,9,27. 1212 Ans: Given: number 1212 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 1212 as, 1×121×12 2×62×6 3×43×4 Therefore, the factors of 1212 will be 1,2,3,4,6,121,2,3,4,6,12. 2020 Ans: Given: number 2020 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 2020 as, 1×201×20 2×102×10 4×54×5 Therefore, the factors of 2020 will be 1,2,4,5,10,201,2,4,5,10,20. 1818 Ans: Given: number 1818 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 1818 as, 1×181×18 2×92×9 3×63×6 Therefore, the factors of 1818 will be 1,2,3,6,9,181,2,3,6,9,18. 2323 Ans: Given: number 2323 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 2323 as, 1×231×23 Therefore, the factors of 2323 will be 1,231,23. 3636 Ans: Given: number 3636 We need to write all the factors of the given number. A factor is a number that divides a given integer completely without leaving any reminder. We can write 3636 as, 1×361×36 2×182×18 3×123×12 4×94×9 6×66×6 Therefore, the factors of 3636 will be 1,2,3,4,6,9,12,18,361,2,3,4,6,9,12,18,36. 2. Write first five multiples of: 55 Ans: Given: number 55 We need to write the first five multiples of the given number. The product of a number and a counting number is a multiple of that number. Thus, 1×5=51×5=5 2×5=102×5=10 3×5=153×5=15 4×5=204×5=20 5×5=255×5=25 Therefore, the first five multiples of 55 will be 5,10,15,20,255,10,15,20,25. 88 Ans: Given: number 88 We need to write the first five multiples of the given number. The product of a number and a counting number is a multiple of that number. Thus, 1×8=81×8=8 2×8=162×8=16 3×8=243×8=24 4×8=324×8=32 5×8=405×8=40 Therefore, the first five multiples of 88 will be 8,16,24,32,408,16,24,32,40. 99 Ans: Given: number 99 We need to write the first five multiples of the given number. The product of a number and a counting number is a multiple of that number. Thus, 1×9=91×9=9 2×9=182×9=18 3×9=273×9=27 4×9=364×9=36 5×9=455×9=45 Therefore, the first five multiples of 99 will be 9,18,27,36,459,18,27,36,45. 3. Match the items in column 11 with the items in column 22: Column 11 Column 22i.3535a)Multiple of 88ii. 1515b)Multiple of 77iii.1616c)Multiple of 7070iv. 2020d)Factor of 3030v.2525e)Factor of 5050f) Factor of 2020 Ans: Given: two columns having different options We need to match the correct items in column 11 with the items in column 22. (i) 3535 We can write 3535 as, 7×57×5 So, 3535 is a multiple of 77(b). (ii) 1515 We know that a factor of 3030 is 1515. So, the correct option is (d). (iii) 1616 We can write 1616 as, 8×28×2 So, 1616 is a multiple of 88(a). (iv) 2020 We can write 2020 as, 20×120×1 So, 2020 is a factor of 2020 . So, the correct option is (f). (v) 2525 We know 25×2=5025×2=50 So, 2525 is a factor of 5050(e). Therefore, Column 11 Column 22 i.3535b) Multiple of 77ii.1515d) Factor of 3030iii.1616a) Multiple of 88iv.2020f) Factor of 2020v. 2525e) Factor of 5050 4. Find all the multiples of 99 up to 100100. Ans: Given: number 99 We need to write all the multiples of 99 up to 100100. We know that the product of a number and a counting number is a multiple of that number. Therefore, 9×1=99×1=9 9×2=189×2=18 9×3=279×3=27 9×4=369×4=36 9×5=459×5=45 9×6=549×6=54 9×7=639×7=63 9×8=729×8=72 9×9=819×9=81 9×10=909×10=90 9×11=999×11=99 Therefore, all the multiples of 99 up to 100100 are 9,18,27,36,45,54,63,72,81,90,999,18,27,36,45,54,63,72,81,90,99. Exercise 3.2 1. What is the sum of any two: Odd numbers Ans: Given: two odd numbers We need to find the sum of two odd numbers. We know that odd numbers are the numbers which cannot be divided by 22 completely. Consider, 1,31,3 ⇒1+3⇒1+3 =4=4 Now consider, 13,4513,45 ⇒13+45⇒13+45 5858 Therefore, we can say that the sum of two odd numbers is always an even number. Even Numbers Ans: Given: two even numbers We need to find the sum of two even numbers. We know that even numbers are the numbers which can be divided by 22 completely. Consider, two even numbers 2,42,4 ⇒2+4⇒2+4 =6=6 Now, consider two more even numbers 98,4098,40 ⇒98+40⇒98+40 =138=138 Therefore, we can say that the sum of two even numbers is always an even number. 5. State whether the following statements are true or false: The sum of three odd numbers is even. Ans: Given: the sum of three odd numbers is even We need to state whether the statement is true or false. Consider, 1,29,991,29,99 ⇒1+29+99⇒1+29+99 =129=129 This is odd. Therefore, the given statement is False. The sum of two odd numbers and one even number is even. Ans: Given: the sum of two odd numbers and one even number is even We need to state whether the statement is true or false. Consider, 1,95,561,95,56 ⇒1+95+56⇒1+95+56 =152=152 This is even. Therefore, the given statement is True. The product of three odd numbers is odd. Ans: Given: the product of three odd numbers is odd We need to state whether the statement is true or false. Consider, 1,99,1051,99,105 ⇒1×99×105⇒1×99×105 =10395=10395 This is odd. Therefore, the given statement is True. If an even number is divided by 22, the quotient is always odd. Ans: Given: If an even number is divided by 22, the quotient is always odd. We need to state whether the statement is true or false. Consider, number 4444 Divide by 22, we will get 44÷244÷2 =22=22Thus, the quotient is not odd. Therefore, the given statement is False. All prime numbers are odd. Ans: Given: All prime numbers are odd We need to state whether the statement is true or false. We know that 22 is a prime number which is not odd. Therefore, the given statement is False. Prime numbers do not have any factors. Ans: Given: Prime numbers do not have any factors. We need to state whether the statement is true or false. We know that a factor is a number that divides a given integer completely without leaving any reminder. Consider, 2323 The factors of 2323 are 1,231,23. Therefore, all prime numbers have factors itself and 1.1. Therefore, the statement is False. Sum of two prime numbers is always even. Ans: Given: Sum of two prime numbers is always even. We need to state whether the statement is true or false. We know that all prime numbers are odd except 2.2. So, if we add two add numbers then the sum of those numbers is even. Consider, 2,52,5 Adding the numbers, we get 2+52+5 =7=7 It is not even. Therefore, the given statement is False. 2 is the only even prime number. Ans: Given: 2 is the only even prime number. We need to state whether the statement is true or false. We know that 22 is the only prime number. All other prime numbers are odd. Therefore, the given statement is True. All even numbers are composite numbers. Ans: Given: All even numbers are composite numbers We need to state whether the statement is true or false. We know that the composite numbers are the numbers having factors other than 11 and the number itself. Consider, 22 This is an even number. But it does not have any factors other than 11 and itself. Therefore, 22 is not a composite number. Therefore, the given statement is False. The product of two even numbers is always even. Ans: Given: The product of two even numbers is always even. We need to state whether the statement is true or false. Consider, two even numbers 2,42,4 Product of these numbers will be 2×42×4 =8=8 It is even. Now consider, 48,9848,98 Product of these numbers will be 48×9848×98 =4704=4704 It is also even. Therefore, the statement is True. 6. The numbers 1313 and 3131 are prime numbers. Both these numbers have the same digits 11 and 33. Find such pairs of prime numbers up to 100100. Ans: Given: The numbers 1313 and 3131 are prime numbers. Both these numbers have same digits 11 and 33 We need to find such pairs of prime numbers up to 100100. The prime numbers up to 100100 are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,972,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97 Therefore, the pairs of prime number will be 17and 71.17and 71. 7. Write down separately the prime and composite numbers less than 2020. Ans: Given: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,201,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 We need to write separately the prime and composite numbers less than 2020. We know that the composite numbers are the numbers having factors other than 11 and the number itself. Prime numbers are the numbers having factors 11 and itself. Therefore, prime numbers will be 2,3,5,7,11,13,17,192,3,5,7,11,13,17,19 The composite numbers will be 4,6,8,9,10,12,14,15,16,184,6,8,9,10,12,14,15,16,18 8. What is the greatest prime number between 11 and 1010? Ans: Given: prime numbers 2,3,5,72,3,5,7 We need to find the greatest prime number between 11 and 1010. Therefore, we can see that the greatest prime number among the given numbers is 7.7. 9. Express the following as the sum of two odd numbers: 4444 Ans: Given: 4444 We need to express the given numbers as the sum of two odd numbers. We can write the given number 4444 as 5+395+39 =44=44 3636 Ans: Given: 3636 We need to express the given numbers as the sum of two odd numbers. We can write the given number 3636 as 9+279+27 =36=36 2424 Ans: Given: 2424 We need to express the given numbers as the sum of two odd numbers. We can write the given number 2424 as 3+213+21 =24=24 1818 Ans: Given: 1818 We need to express the given numbers as the sum of two odd numbers. We can write the given number 1818 as 5+135+13 =18=18 10. Give three pairs of prime numbers whose difference is 22. Remark: Two prime numbers whose difference is 22 are called twin primes. Ans: Given: prime numbers We need to find three pairs of prime numbers whose difference is 22. We know that prime numbers are the numbers having factors 11 and itself. Therefore, the three pairs of prime numbers whose difference is 22 will be 5and 75and 7 11 and 1311 and 13 17and 1917and 19 11. Which of the following numbers are prime? 2323 Ans: Given: 2323 We need to find if the given number is prime. We know that prime numbers are the numbers having factors 11 and itself. Therefore, 2323 can be written as 1×231×23. Thus, it is a prime number. 5151 Ans: Given: 5151 We need to find if the given number is prime. We know that prime numbers are the numbers having factors 11 and itself. Therefore, 5151 can be written as 3×173×17. Thus, 5151 is not a prime number. 3737 Ans: Given: 3737 We need to find if the given number is prime. We know that prime numbers are the numbers having factors 11 and itself. Therefore, 3737 can be written as 1×371×37. Thus, it is a prime number. 2626 Ans: Given: 2626 We need to find if the given number is prime. We know that prime numbers are the numbers having factors 11 and itself. Therefore, 2626 can be written as 2×132×13. Thus, 2626 is not a prime number. 12. Write seven consecutive composite numbers less than 100100 so that there is no prime number between them. Ans: Given: Numbers less than 100100 We need to write seven consecutive composite numbers less than 100100 so that there is no prime number between them. We know that prime numbers less than 100100 are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,972,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97. Therefore, seven consecutive composite numbers less than 100100 so that there is no prime number between them will be 90,91,92,93,94,95,9690,91,92,93,94,95,96. 13. Express each of the following numbers as the sum of three odd primes: 2121 Ans: Given: 2121 We need to express the given number as the sum of three odd primes. We know that prime numbers are the numbers having factors 11 and itself. Therefore, we can write 2121 as 21=5+7+1121=5+7+11 3131 Ans: Given: 3131 We need to express the given number as the sum of three odd primes. We know that prime numbers are the numbers having factors 11 and itself. Therefore, we can write 3131 as 31=5+7+1931=5+7+19 5353 Ans: Given: 5353 We need to express the given number as the sum of three odd primes. We know that prime numbers are the numbers having factors 11 and itself. Therefore, we can write 5353 as 53=11+13+2953=11+13+29 6161 Ans: Given: 6161 We need to express the given number as the sum of three odd primes. We know that prime numbers are the numbers having factors 11 and itself. Therefore, we can write 6161 as 61=3+5+5361=3+5+53 14. Write five pairs of prime numbers less than 2020 whose sum is divisible by 55. Ans: Given: 2,3,5,7,11,13,17,192,3,5,7,11,13,17,19 We need to write five pairs of prime numbers less than 2020 whose sum is divisible by 55. We know that prime numbers are the numbers having factors 11 and itself. A number is divisible by 55 if the unit place digit is either 00 or 5.5. Therefore, the pairs will be 3+7=103+7=10 2+3=52+3=5 13+2=1513+2=15 13+7=2013+7=20 3+17=203+17=20 5+5=105+5=10 15. Fill in the blanks: A number which has only two factors is called a __________. Ans: We need to fill in the blanks with appropriate numbers or words. A number which has only two factors is called a prime number. A number which has more than two factors is called a _________. Ans: We need to fill in the blanks with appropriate numbers or words. A number which has more than two factors is called a composite number. 11 neither ________ nor _________. Ans: We need to fill in the blanks with appropriate numbers or words. 11 neither prime nor composite number. The smallest prime number is ________. Ans: We need to fill in the blanks with appropriate numbers or words. The smallest prime number is 2.2. The smallest composite number is ________. Ans: We need to fill in the blanks with appropriate numbers or words. The smallest composite number is 44. The smallest even number is __________. Ans: We need to fill in the blanks with appropriate numbers or words. The smallest even number is 22. Exercise-3.3 1. Using divisibility test, determine which of the following numbers are divisible by 2;2;by 3;3; by 3;3; by 5;5; by 6;6; by 8;8; by 9;9; by 10;10; by 11.11.(say yes or no) NumberDivisible by223344556688991010111112812899099015861586275275668666866392106392104297144297142856285630603060406839406839 Ans: We need to determine that which of the given numbers are divisible by 2;2; by 3;3; by 4;4; by 5;5; by 6;6; by 8;8; by 9;9; by 10;10; by 11.11. NumberDivisible by2233445566889910101111128128YesNoYesNoNoYesNoNoNo990990YesYesNoYesNoNoYesYesYes15861586YesNoNoNoNoNoNoNoNo275275NoNoNoYesNoNoNoNoYes66866686YesNoNoNoYesNoNoNoNo639210639210YesNoNoYesYesNoNoYesYes429714429714YesYesNoNoYesNoYesNoNo28562856YesYesNoNoYesYesNoNoNo30603060YesYesYesYesYesNoYesYesNo406839406839NoYesNoNoNoNoNoNoNo 2. Using divisibility test, determine which of the following numbers are divisible by 4; by 8:8: 572572 Ans: Given: 572572 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 726352 Ans: Given: 726352 726352 We need to determine whether the given number is divisible by 4;4; by 8.8. As the last two digits of the given number are divisible by 4,4, hence the given number is divisible by 4.4. The given number is divisible by8,8, as the last three digits are divisible by 8.8. 5500 Ans: Given:55005500 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 6000 Ans: Given: 60006000 We need to determine whether the given number is divisible by 4;4; by 8.8. As the last two digits of the given number are divisible by 4,4, hence the given number is divisible by 4.4. The given number is divisible by 8,8, as the last three digits are divisible by 8.8. 12159 Ans: Given: 1215912159 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are not divisible by 4;4; hence the number is not divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 14560 Ans: Given: 1215912159 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is divisible by 8,8, as the last three digits are divisible by 8.8. 21084 Ans:Given: 2108421084 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 31795072 Ans: Given: 3179507231795072 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is divisible by 8,8, as the last three digits are divisible by 8.8. 1700 Ans: Given: 17001700 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are divisible by 4;4; hence the number is divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 2150 Ans: Given: 21502150 We need to determine whether the given number is divisible by 4;4; by 8.8. Here the last two digits of the number are not divisible by 4;4; hence the number is not divisible by 4.4. The given number is not divisible by 8,8, as the last three digits are not divisible by 8.8. 3. Using divisibility test, determine which of the following numbers are divisible by 6:6: 297144 Ans: Given: 297144297144 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is divisible by 33 as the sum of digits is divisible by 3.3. As the number is divisible by both 22 and 3,3,hence the given number is divisible by 6.6. 1258 Ans: Given: 12581258 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is not divisible by 33 as the sum of digits (=16)(=16) is not divisible by 3.3. As the number is divisible by 22 but not by 3,3, hence the given number is not divisible by 6.6. 4335 Ans: Given: 43354335 We need to find whether the given number is divisible by 6.6. The units place digit is an odd number; hence the given number is not divisible by 2.2. The given number is divisible by 33 as the sum of digits (=15)(=15) is divisible by 3.3. As the number is not divisible by both 22 and 3,3, hence the given number is not divisible by 6.6. 61233 Ans: Given: 6123361233 We need to find whether the given number is divisible by 6.6. The units place digit is an odd number; hence the given number is not divisible by 2.2. The given number is divisible by 33 as the sum of digits(=15)(=15) is divisible by 3.3. As the number is not divisible by both 22 and 3,3, hence the given number is not divisible by 6.6. 901352 Ans: Given: 901352901352 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is not divisible by 33 as the sum of digits (=20)(=20) is not divisible by 3.3. As the number is not divisible by both 22 and 3,3, hence the given number is not divisible by 6.6. 438750 Ans: Given: 438750438750 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is divisible by 33 as the sum of digits (=27)(=27) is divisible by 3.3. As the number is divisible by both 22 and 3,3, hence the given number is divisible by 6.6. 1790184 Ans: Given: 17901841790184 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is divisible by 33 as the sum of digits (=30)(=30) is divisible by 3.3. As the number is divisible by both 22 and 3,3, hence the given number is divisible by 6.6. 12583 Ans: Given: 1258312583 We need to find whether the given number is divisible by 6.6. The units place digit is an odd number; hence the given number is not divisible by 2.2. The given number is not divisible by 33 as the sum of digits (=19)(=19) is not divisible by 3.3. As the number is not divisible by both 22 and 3,3, hence the given number is not divisible by 6.6. 639210 Ans: Given: 438750438750 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is divisible by 33 as the sum of digits (=21)(=21) is divisible by 3.3. As the number is divisible by both 22 and 3,3, hence the given number is divisible by 6.6. 17852 Ans: Given: 1785217852 We need to find whether the given number is divisible by 6.6. The units place digit is an even number; hence the given number is divisible by 2.2. The given number is not divisible by 33 as the sum of digits (=23)(=23) is not divisible by 3.3. As the number is divisible by 22 but not divisible by 3,3, hence the given number is not divisible by 6.6. 4. Using divisibility test, determine which of the following numbers are divisible by 11:11: 5445 Ans: Given: 54455445 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 4+5=94+5=9 and sum of digits at even places is 4+5=94+5=9 Difference of both sums is 9−9=0.9−9=0. As the difference is 0,0, therefore, the number is divisible by 11.11. 10824 Ans: Given: 1082410824 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 4+8+1=134+8+1=13 and sum of digits at even places is 2+0=22+0=2 Difference of both sums is 13−2=1113−2=11 As the difference is 11,11, therefore, the number is divisible by 11.11. 7138965 Ans: Given: 71389657138965 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 5+9+3+7=245+9+3+7=24 and sum of digits at even places is 6+8+1=156+8+1=15 Difference of both sums is 24−15=924−15=9 As the difference is 9,9, therefore, the number is not divisible by 11.11. 70169308 Ans: Given: 7016930870169308 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 8+3+6+0=178+3+6+0=17 and sum of digits at even places is 0+9+1+7=170+9+1+7=17 Difference of both sums is 17−17=0.17−17=0. As the difference is 0,0, therefore, the number is divisible by 11.11. 10000001 Ans: Given: 1000000110000001 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 1+0+0+0+=11+0+0+0+=1 and sum of digits at even places is 0+0+0+1=1.0+0+0+1=1. Difference of both sums is 1−1=0.1−1=0. As the difference is 0,0, therefore, the number is divisible by 11.11. 901153 Ans: Given: 54455445 We need to find whether the given number is divisible by 1111 or not. The sum of digits at odd places is 3+1+0=43+1+0=4 and sum of digits at even places is 5+1+9=15.5+1+9=15. Difference of both sums is 15−4=11.15−4=11. As the difference is 11,11, therefore, the number is divisible by 11.11. 5. Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisible by 3:3: _6724 Ans: Given: _6724_6724 We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by 3.3. A number is divisible by 33 if the sum of digits is divisible by 3.3. Hence, Smallest digit is 2 → 26724 = 2 + 6 + 7 + 2 + 4 = 212 → 26724 = 2 + 6 + 7 + 2 + 4 = 21 And Largest digit is 8 → 86724 = 8 + 6 + 7 + 2 + 4 = 27 8 → 86724 = 8 + 6 + 7 + 2 + 4 = 27 4765_2 Ans: Given: 4765_24765_2 We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by 3.3. A number is divisible by 33 if the sum of digits is divisible by 3.3. Hence, Smallest digit is 0 → 476502 = 4 + 7 + 6 + 5 + 0 + 2 = 24 0 → 476502 = 4 + 7 + 6 + 5 + 0 + 2 = 24 And Largest digit is 9 → 476592 = 4 + 7 + 6 + 5 + 0 + 2 = 339 → 476592 = 4 + 7 + 6 + 5 + 0 + 2 = 33 6. Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisible by 11:11: 92____38992____389 Ans: Given: 92____38992____389 We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by 11.11. Let this number is x� A number is divisible by 1111 if the difference of the sum of digits at odd and even places is either or 11.11. Hence, 92x389 →92�389 → Sum of digits at even places = 9 + x + 8 = 17 + x = 9 + � + 8 = 17 + � Sum of digits at odd places = 2 + 3 + 9 = 14= 2 + 3 + 9 = 14 Number will be divisible by 11 if 17 + x −14 = 11 17 + � −14 = 11 ⇒3+ x = 11⇒3+ � = 11 ⇒ x = 11−3 =8⇒ � = 11−3 =8 Also 17 + x −14 = 0 17 + � −14 = 0 ⇒ 3+x = 0⇒ 3+� = 0 ⇒x = 0−3 =−3⇒� = 0−3 =−3 Which is not possible So the only digit that can be placed in this blank space is 88 So number is 928389928389 8____94848____9484 Ans: Given: 8____94848____9484 We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by 11.11. Let this number is x� A number is divisible by 1111 if the difference of the sum of digits at odd and even places is either or 11.11. Hence, 8x9484 →8�9484 → Sum of digits at even places = 8 + 9 + 8 = 25= 8 + 9 + 8 = 25 Sum of digits at odd places = x + 4 + 4 = 8 + x = � + 4 + 4 = 8 + � Number will be divisible by 1111 if 25− (8+x) = 1125− (8+�) = 11 25−8 − x = 1125−8 − � = 11 ⇒ 17 − x = 11⇒ 17 − � = 11 ⇒−x = 11−17⇒−� = 11−17 ⇒−x = −6⇒−� = −6 ⇒x = 6⇒� = 6 So the only digit that can be placed in this blank space is 66 So number is 869484869484 Exercise 3.6 1.Find the H.C.F. of the following numbers: 18, 4818, 48 Ans: Given: 18, 4818, 48 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 1818 will be = 2 x 3 x 3 = 2 x 3 x 3 Factors of 4848 will be = 2 x 2 x 2 x 2 x 3 = 2 x 2 x 2 x 2 x 3 Take common factors of 1818 and 4848 , we get H.C.F. of 18, 4818, 48 2×3=62×3=6 30, 4230, 42 Ans: Given: 30,4230,42 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 3030 will be = 2 x 3 x 5 = 2 x 3 x 5 Factors of 4242 will be = 2 x 3 x 7 = 2 x 3 x 7 Take common factors of 3030 and 4242 , we get H.C.F of 3030 and 4242 2×3=62×3=6 18, 6018, 60 Ans: Given: 18,6018,60 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 1818 will be = 2 x 3 x 3 = 2 x 3 x 3 Factors of 6060 will be = 2 x 2 x 3 x 5 = 2 x 2 x 3 x 5 Take common factors of 1818 and 6060 , we get H.C.F. of 1818 , 6060 = 2 x 3 = 2 x 3 = 6 = 6 27, 6327, 63 Ans: Given: 27,6327,63 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 2727 will be = 3 x 3 x 3 = 3 x 3 x 3 Factors of 6363 will be = 3 x 3 x 7 = 3 x 3 x 7 Take common factors of 2727 and 6363 ,we get H.C.F. of 2727 , 6363 = 3 x 3 = 3 x 3 = 9 = 9 36,8436,84 Ans: Given: 36,8436,84 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 3636 will be = 2 x 2 x 3 x 3 = 2 x 2 x 3 x 3 Factors of 8484 will be = 2 x 2 x 3 x 7 = 2 x 2 x 3 x 7 Take common factors of 3636 and 8484 , we get H.C.F. of 3636 , 8484 = 2 x 2 x 3 = 2 x 2 x 3 = 12 = 12 34,10234,102 Ans: Given: 34,10234,102 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 3434 will be = 2 x 17 = 2 x 17 Factors of 102102 will be = 2 x 3 x 17 = 2 x 3 x 17 Take common factors of 3434 and 102102 ,we get H.C.F. of 34,10234,102 = 2 x 17 = 2 x 17 = 34 = 34 70, 105, 17570, 105, 175 Ans: Given: 70, 105, 17570, 105, 175 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 7070 will be = 2 x 5 x 7 Factors of 105105 will be = 3 x 5 x 7 Factors of 175175 will be = 5 x 5 x 7 Take common factors of 7070 , 105105 and 175175 , we get H.C.F. of 70, 105, 17570, 105, 175 = 7 x 5 = 7 x 5 = 35 = 35 91, 112, 4991, 112, 49 Ans: Given: 91, 112, 4991, 112, 49 We need to find H.C.F of the given numbers . We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 9191 will be = 7 x 13 = 7 x 13 Factors of 112112 will be = 2 x 2 x 2 x 2 x 7 = 2 x 2 x 2 x 2 x 7 Factors of 49 49 will be = 7 x 7 = 7 x 7 Take common factors of 91, 11291, 112 and 49 49 , we get H.C.F. of 91, 112, 4991, 112, 49 = 1×7 = 1×7 = 7 = 7 18, 54, 8118, 54, 81 Ans: Given: 18, 54, 8118, 54, 81 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 1818 will be = 2 x 3 x 3 = 2 x 3 x 3 Factors of 5454 will be =2×3×3×3=2×3×3×3 Factors of 81 81 will be = 3 x 3 x 3 x 3 = 3 x 3 x 3 x 3 Take common factors of 18, 5418, 54 and 81 81 , we get H.C.F. of 18, 54, 8118, 54, 81 = 3 x 3 = 3 x 3 = 9 = 9 12, 45, 7512, 45, 75 Ans: Given: 12, 45, 7512, 45, 75 We need to find H.C.F of the given numbers. We know that the bigger factor that splits two or more numbers is the highest common factor. Thus, Factors of 1212 will be = 2 x 2 x 3 = 2 x 2 x 3 Factors of 4545 will be = 3 x 3 x 5 = 3 x 3 x 5 Factors of 75 75 will be = 3 x 5 x 5 = 3 x 5 x 5 Take common factors of 12, 4512, 45 and 7575 ,we get H.C.F. of 12, 45, 7512, 45, 75 = 1 x 3 = 1 x 3 = 3 = 3 2. What is the H.C.F. of two consecutive: Numbers? Ans: We have to find H.C.F. of two consecutive numbers: We know that the bigger factor that splits two or more numbers is the highest common factor. Let us take consecutive numbers 22 and 33 , Thus, Factors of 22 = 2 x 1 = 2 x 1 Factors of 33 = 3 x 1 = 3 x 1 Hence, the H.C.F. of two consecutive numbers is 11 . Even Numbers? Ans: We have to find H.C.F. of two even consecutive numbers: We know that the bigger factor that splits two or more numbers is the highest common factor. Let us take even consecutive numbers 22 and 44 . Factors of 22 will be =1×2=1×2 Factors of 44 will be =2×2=2×2 Take common factors of 22 and 44 , we get H.C.F. of 22 and 44 =2=2 Hence, H.C.F. of two even consecutive numbers is 22 . Odd Numbers? Ans: We have to find H.C.F. of two odd consecutive numbers: We know that the bigger factor that splits two or more numbers is the highest common factor. Let us take odd consecutive numbers 33 and 55 , Factors of 33 will be =1×3=1×3 Factors of 55 will be =1×5=1×5 Take common factors of 33 and 55 , we get H.C.F. of 33 and 55 =1=1 Hence, H.C.F. of two consecutive odd numbers is 11 . 3. H.C.F. of co-prime numbers 44 and 1515 was found as follows by factorization: 4=2 x 24=2 � 2 and 15=3 x 515=3 � 5 since there is no common prime factor, so H.C.F. of 44 and 1515 is 00 . Is the answer correct? If not, what is the correct H.C.F.? Ans: Given: Factors of 44 =2×2=2×2 Factors of 1515 =3×5=3×5 H.C.F. of co-prime numbers 44 and 1515 is 00 . But this statement is wrong. ∵∵ 11 is the common factor of each and every number. Factors of 44 will be =1×2×2=1×2×2 Factors of 1515 will be =1×3×5=1×3×5 Hence, H.C.F of co-prime numbers 44 and 1515 is 11 NCERT Class 6 Maths Chapter 3 – Free PDF With the help of NCERT solutions for Class 6 Maths Chapter 3, the student is able to prepare for their examinations with ease. Various solved examples provided here, help students understand how they are to proceed with the calculation for these sums. Also, they allow students to grasp this basic knowledge that will prove to be useful in the later years. You can download a free PDF of Class 6 Maths Chapter 3 available on this page below. It helps you to understand the prime role that different numbers and their multiples play in solving an equation. Formulas used and applied in this chapter are primary equations that will later be useful when the student moves forward to solve more complex equations. NCERT Class 6 Maths Chapter 3 Exercises Exercise 3.1 – Introduction Purpose of this introduction is to provide students with the basic knowledge of what can be classified as a divisor and what is a factor. This part is what the entire chapter is mostly based on. Exercise 3.2 – Factors and Multiples NCERT solutions for Class 6th Maths Chapter 3 – Playing with Numbers, then moves on to the first set of exercises that asks students to spot multiples and factors. It also states that every number is a multiple and factor of itself. Exercise 3.3 – Prime and Composite Numbers This section of the chapter elaborates on which numbers can be defined as prime numbers and which as composite numbers. Segment also specifies how 2 is the smallest of all the prime numbers, and 1 can neither be a composite nor a prime number. Exercise 3.4 – Tests for Divisibility of Numbers Main focus of this part of Chapter 3 in NCERT Maths book Class 6 is to make students understand which numbers are divisible by 10, 5, 2, 3 6, 4, 8 9, and 11. Learn more about how this divisibility test is helpful. Exercise 3.5 – Common Factors and Common Multiples Understanding common multiples and common factors are mandatory for gaining a grasp over the subject-matter of numbers. Focus of this exercise is to teach how certain numbers are the multiple and factor of more than one number. Exercise 3.6 – Some More Divisibility Rules There are certain divisibility rules that the student will need to know to gain a proper grasp over the subject of factors and multiples. Refer to Chapter 3 solutions from NCERT to gain more detailed knowledge on the topic. Exercise 3.7 – Prime Factorization This section of the NCERT Class 6 Maths Chapter 3 talks about prime factors and how which number when multiplied forms the original number. Learn about prime factor number play in this segment. Exercise 3.8 – Highest Common Factor HCF or highest common factors is an important part that students are required to concentrate on. The reason for that being that this section is where most examination questions are derived from. Exercise 3.9 – Lowest Common Factor LCF or Lowest Common Factor also tends to be preferred by teachers to be included in examinations. Calculation process of the LCF is quite simple once the students understand the equations and how they are framed. Exercise 3.10 – Problems of HCF and LCM This section comprises theory and practical problems for the student to solve. At the end of the chapter, students are provided with a brief revision of the entire topic. Key Learnings from the Chapter 3 of Class 6 Maths NCERT Solutions Chapter 3 of Class 6 Maths NCERT Solutions is mainly based on the topics such as multiples, divisors, and factors, identification of multiples and factors, how to determine whether a number is a prime or a composite number using the factors, and how to find the HCF and LCM of numbers using the factors method. NCERT Solutions for Class 6 Maths Chapter 3 consists of sums related to these concepts Below are the key learnings from Chapter 3 of Class 6 Maths NCERT Solutions. A prime number can be defined as a number (except 1) which has only two factors 1 and the number itself. Composite numbers can be defined as those numbers that have more than two factors. 1 is a factor of every number and it is neither composite nor prime. If the sum of all the factors of a number is equal to that number, then it is called a perfect number. The number 6 is a perfect number as 6 = 1 + 2 + 3 and the number 28 is also a perfect number as 1 + 2 + 4 + 7 + 14 = 28. Every number is a factor of itself and a multiple of itself. Every factor of a number is an exact divisor of that number. Every factor is less than or equal to that number but every multiple of a number is greater than or equal to that number. The number of factors of a given number is finite but the number of multiples of a given number is infinite. If two numbers are divisible by a number, their sum and difference are also divisible by the same number. The HCF of two or more numbers is the highest common factor of the numbers. The Least Common Multiple (LCM) of two or more numbers is the smallest of their common multiple of the numbers. Benefits of NCERT Class 6 Maths Chapter 3 Benefits of availing Class 6 Maths Chapter 3 solutions from Study Studio have been listed below - The solutions have been constructed as per examination format to assist students in scoring well. All questions and solutions are written in simple language. Solutions to every question are precise and to the point. All solutions are informative and effective in helping students score well. With the help of the Class 6th Maths, Chapter 3 offered by Study Studio, students can now aim for excellence in their examinations. Upon reaching the 6th standard, students are introduced to the NCERT Solutions for Class 6 Maths Chapter 3 syllabus, which is quite variant and includes different topics. The third chapter among the Maths syllabus, Playing with Numbers, particularly focuses on teaching students about multiples and divisors. As the chapter progresses, the students are introduced to topics such as common factors and multiples, divisibility rules, highest common factors, lowest common factors, etc. A proper comprehension of the chapter will enable students to understand prime and composite numbers, as well as their difference. You can also download NCERT Class 6 Maths and NCERT Class 6 Science to help you to revise complete syllabus and score more marks in your examinations.