the sum of 3 consecutive even numbers is 42

Remainder = 4. Hence, they are co-prime. (c) odd numbers? Hence, the HCF = 3. 12 x 1 = 36; (e) False 9 x 1 = 9; which is divisible by 11. (a) 23 is a prime number [ 23 = 1 x 23] Then, sum = 24 + 0 = 24 Hence, the required multiples of 8 are: 8, 16, 24, 32 and 40. =4 + 3 + 8 + 7 + 5 + 0 = 27 which is divisible by 3. Sum of the digits at odd places = 5 + 4 = 9 Example 3: And also, solving the question on the sum of even numbers without the formula takes so much time, which is also not good for the exam. (iv) 17 + 3 = 20 Hence, the required digits are 2 and 8. (b) 15 and 25 Factors of 5 are 1, 5 (i) All even numbers are composite numbers. Here, the number formed by the last three digits of the given number = 072. Here, the number formed by the last three digits of the given number = 159 (e) 4 Factors of 12 are 1, 2, 3, 4, 6, 12 Ex 3.7 Class 6 MathsQuestion 6. 8 x 5 = 40 Find the maximum value of weight which can measure the weight of the fertiliser exact number of times. 100 = (100 4) + 24 = 96 + 24 = 120 Sum of all the digits at even places = 0 + 9 + 1 + 7 = 17 Hence, 726352 is divisible by 4. Three boys.step off together from the same spot. By which other will that number be always divisible? From the above examples, we observe that the LCM of the two numbers, where one number is a factor of the other, is the greater number. Remainder = 0. (f) 14560 (a) The sum of any two odd numbers is even. Given number = 1729 The n th pronic number is the sum of the first n even integers, and as such is twice the n th triangular number and n more than the n th square number, as given by the alternative formula n 2 + n for pronic numbers. Here, 48 is divisible by 3 and 50 is divisible by 2. Then, the sum = 19 + 8 = 27 which is divisible by 3 (f) False Watch game, team & player highlights, Fantasy football videos, NFL event coverage & more (g) Given number = 21084 (i) Divisibility by 4 Sum = 105 + 107 = 212 Solution: Ex 3.4 Class 6 MathsQuestion 2. Hence, the given number is not divisible by 6. So, the missing digit = 8 Ex 3.3 Class 6 MathsQuestion 4. The digit at ones place of the given number is not even. Because only having the formula, or even the explanation of the topic is not enough. (e) False [ 2 is a prime number but it is even] And also, that is the best way to remember the formula for the sum of even numbers as well. Here, common factors are 5 and 7. (h) 91, 112, 49 Ex 3.7 Class 6 MathsQuestion 3. Find the common factors of: (i) False. Remainder = 6. Ex 3.4 Class 6 MathsQuestion 6. (c) 5500 Sol. (a) Factors of 24 are: Factors of 50 are: 1, 2, 5, 10, 50 The digit at ones place of the given number is not even. (ii) Divisibility by 8 Here, the number formed by last two digits is 96 which is divisible by 4. Ex 3.7 Class 6 MathsQuestion 2. Test of divisibility by 9: Hence, the given number is divisible by 11. Solution: Hence, 18 and 35 are co-prime. 21 = 3 x 7 (a) 24 = 2 x 3 x 4 (b) 6, 18 = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 which is divisible by 3. Hence, the common factors are 1 and 5. Sum of the digits = 4 + 2 + 7 + 6 = 19 Hence, the prime factors of 9999 = 3 x 3 x 11 x 101. 5 x 1 = 5; Sum of digits = 2 + 5 + 6 + 7 + 4 = 24 Find all the multiples of 9 upto 100. LCM = 2 x 3 x 5 = 30. For the given number to be divisible by 11 Hence, the HCF = 2 x 3 = 6. Required pairs of prime numbers less than 20 are: The sum of 2 consecutive even numbers is always an even number. Ex 3.1 Class 6 MathsQuestion 2. (b) True (j) Given numbers are 12, 45 and 75. Solution: Here, the common factor is 3. Ex 3.3 Class 6 MathsQuestion 1. 17 11 = 6 LCM of 3 and 4, and How to Find Least Common Multiple, What is Simple Interest? Solution: Reason: 0 is not the prime factor of any number. Hence, the required multiples of 5 are: 5, 10, 15, 20 and 25. (e) 216 and 215 (d) Given numbers are 17 and 68 Here, the common factors are 2 and 17. (a) Given number = 572 Hence, the given number is divisible by 6. Verify this statement with the help of some examples. Ex 3.4 Class 6 MathsQuestion 12. (j) 2150 Well, the sum of two consecutive integers, the sum of two consecutive even numbers, and the sum of two consecutive odd numbers were discussed separately above, the corresponding formulas were obtained. (i) Divisibility by 4 m3= 2 or 5 and m4= 5 or 2 Solution: Hence, they are not co-prime. 5 x 2 = 10; (e) Given numbers are 216 and 215 (h) Given numbers are 91, 112 and 49. Solution: Ex 3.6 Class 6 MathsQuestion 1. Sum of the digits at even places = 4 + 5 = 9 Since there is no common prime factors, so HCF of 4 and 15 is 0. Hence, the LCM of 9 and 4 = Product of 9 and 4. (b) 56 = 7 x 2 x 2 x 2 Hence, 56 = 7 x 2 x 2 x 2 is a prime factorisation. Solution: which is divisible by 3. Hence, the sum of numbers 97 and 99 i.e. (a) First five multiples of 5 are: The sum of all digits of 1258 = l + 2 + 5 + 8 = 16 which is not divisible by 3. Hence, the number 25110 is divisible by 45. Prime factorisations of 36 and 84 are: (d) 56 and 120 (d) The smallest prime number is . 27 = 1 x 27; Factors of 68 are 1, 2, 4, 17, 34, 68 The Tower of Hanoi (also called The problem of Benares Temple or Tower of Brahma or Lucas' Tower and sometimes pluralized as Towers, or simply pyramid puzzle) is a mathematical game or puzzle consisting of three rods and a number of disks of various diameters, which can slide onto any rod.The puzzle begins with the disks stacked on one rod in order of decreasing size, the Hence, the given number is divisible by 6. (b) 10824 Hence, the given number is divisible by 6. 9 x 7 = 63; 8 x 3 = 24; (c) To find the LCM of 12 and 48, we have 90, 91, 92, 93, 94, 95 and 96, Ex 3.2 Class 6 MathsQuestion 10. Sum of all the digits at odd places = 1 + 0 + 0 + 0 = 1 Hence, the LCM of 12 and 48 = 48. The greatest 4-digit number = 9999 Hence, all the factors of prime number 23 are: 1 and 23. (a) 18, 48 The sum of all the digits of the given number 1790184 5 x 3 = 15; 12 = 2 x 6; (b) The common factors of two consecutive even numbers are 1 and 2. Decimal numbers such as 0.1, 0.2, and 0.3 are not represented exactly in binary encoded floating point types. Prime factorisations of 18, 54 and 81 are: (v) (e) [ 25 x 2 = 50]. (c) Given numbers are 30 and 415 Required seven consecutive composite numbers are: Similarly, 10 = 5 x missing number Hence, 5500 is not divisible by 8. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3. Solution: Maximum value of weight which can measure the given weight exact number of time = HCF of 75 g and 69 kg Hence, the correct HCF of 4 and 15 is 1. (b) True [ 3 + 5 + 6 = 14 (even)] The product 9 and 4 = 9 x 4 = 36. To find the LCM of 8, 10 and 12, we have (c) 18, 60 NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. (i) False [ 2 is even but not composite number] Solution: Prime factorisations of 825, 675 and 450 are Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120 A number is divisible by 12. First three multiples of 18 are Hence, the required number is 928389. Since, the given number is not divisible by both 2 and 3, it is not divisible by 6. The prime number theorem then states that x / log x is a good approximation to (x) (where log here means the natural logarithm), in the sense that the limit (b) Given numbers are 15 and 37 Filed Under: CBSE Tagged With: CBSE Class 6 Playing With Numbers, CBSE Maths Playing With Numbers, learncbse Playing With Numbers, NCERT Class 6 Maths Solutions, NCERT Class 6 Playing With Numbers, NCERT Class 6 Solutions Playing With Numbers, NCERT Maths Playing With Numbers, NCERT Solutions for Class 6 Maths, NCERT Solutions For Class 6 Maths Chapter 1, NCERT Solutions For Class 6 Maths Playing With Numbers Exercise 3.1, Playing With Numbers CBSE Class 6 Solutions, RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, Playing with Numbers Class 6 Extra Questions, NCERT Class 6 Solutions Playing With Numbers, NCERT Solutions For Class 6 Maths Chapter 1, NCERT Solutions For Class 6 Maths Playing With Numbers Exercise 3.1, Playing With Numbers CBSE Class 6 Solutions, NCERT Solutions for Class 10 ScienceChapter 1, NCERT Solutions for Class 10 ScienceChapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 ScienceChapter 7, NCERT Solutions for Class 10 ScienceChapter 8, NCERT Solutions for Class 10 ScienceChapter 9, NCERT Solutions for Class 10 ScienceChapter 10, NCERT Solutions for Class 10 ScienceChapter 11, NCERT Solutions for Class 10 ScienceChapter 12, NCERT Solutions for Class 10 ScienceChapter 13, NCERT Solutions for Class 10 ScienceChapter 14, NCERT Solutions for Class 10 ScienceChapter 15, NCERT Solutions for Class 10 ScienceChapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. Solution: To find the LCM of 18, 24 and 32, we have Example 3: Solution: Get breaking Finance news and the latest business articles from AOL. I am the smallest number, having four different prime factors. Hence, 2150 is not divisible by 8. Factors of 15 are 1, 3, 5, 15 The sum of all the digits of 4335 = 4 + 3 + 3 + 5 = 15 which is divisible by 3. Since only 1 is common to 81 and 16 (b) 56 = 7 x 2 x 2 x 2 (c) 37 Solution: (d) To find the LCM of 9 and 45, we have Ex 3.2 Class 6 MathsQuestion 2. Hence, the required number is 1152. To find LCM of 63, 70 and 77, we use division method. Their steps measure 63 cm, 70 cm and 77 cm respectively. Verify this statement with the help of some examples. HCF of 825, 675 and 450 = 3 x 5 x 5 = 75 Hence, the required time = 432 seconds = 7 minutes 12 seconds i.e., 7 minutes 12 seconds past 7 a.m. Ex 3.7 Class 6 MathsQuestion 7. Hence, the HCF = 7. The numbers 13 and 31 are prime numbers. (a) 5, 20 Hence, they are co-prime. (d) False [ 6 + 2 = 3 (odd)] Test of divisibility by 5: unit place of the given number 25110 is 0. (c) 53 Hence, the required digits are 0 and 9. Write seven consecutive composite numbers less than 100 so that there is no prime number between them. Thus, S= n(n+1) Hence, 14560 is divisible by 8. Write the greatest 4-digit number and express it in terms of its prime factors. Also, evaluate the, Benefits of Having the Formula for the Sum of Even Numbers, Advantages of Having the Examples for the sum of Even Numbers, Mastering the Topic of Sum of Even numbers, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. (c) 12, 48 Factors of 215 are 1, 5, 43 (b) 51 Hence, the HCF = 1. Using divisibility tests, determine which of the following numbers are divisible by 2, by 3, by 4, by 5, by 6, by 8, by 9, by 10, by 11 (Say, Yes or No) Therefore, the product 48 x 49 x 50 = 117600 which is divisible by 6. (a) 5 (b) 8 (c) 9 Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 54, 72, 108, 216 Remainder = 0. Example: Determine the sum of even numbers from 1 to 200? (h) True The n th pronic number is also the difference between the odd square (2n + 1) 2 and the (n+1) st centered hexagonal number.. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. The sum of all the digits of the given number 438750 (b) First five multiples of 8 are: For example, 20, 34, 56, 46 12 are all even numbers. Solution: (a) numbers? LCM = 2 x 2 x 3 x 5 = 60. (a) 92 ___ 389 (c) 37 is a prime number [ 37 = 1 x 37] How to Calculate the Percentage of Marks? (i) Divisibility by 4 Hence, the missing numbers are 3 and 2. (b) 51 is not a prime number [ 51 = 1 x 3 x 17] (b) If a number is divisible by 9, it must be divisible by 3. Find first three multiples of: Find the longest tape which can measure the three dimensions of the room exactly. Give three pairs of prime numbers whose difference is 2. (a) 5445 (j) Given number = 17852 Also, from the perspective of the exam it is again very useful because the questions in which the students have performed all the steps in a correct manner, then and only then, the student gets the complete marks for that particular question. 9 x 8 = 72; (d) 61 can be expressed as 11 + 13 + 37, Ex 3.2 Class 6 MathsQuestion 11. So, it is not divisible by 2. (i) Given number = 639210 Sum of the digits at even places = 6 + 8 + 1 = 15 (b) 31 can be expressed as 5 + 7 + 19 20 = 4 x 5 Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Ex 3.6 Class 6 MathsQuestion 3. 24 = 2 x 12; (d) 70169308 Here, the number formed by the last two digits of the given number = 84. 1.2. Ex 3.4 Class 6 MathsQuestion 2. (b) 30, 42 Any integer which is not able to be divided exactly by 2 is an odd number. Factors of 17 are 1, 17 9 x 5 = 45 Solution: Hence, the common factors are 1 and 5. 10 = m3x m4 Hence, the required maximum value of weight = 3 kg. 450 = 2 x 3 x 3 x 5 x 5 9 x 4 = 36; Solution: Here, the last two digits of the given number is 0. Here, the common factors are 2 and 3. So, after 432 seconds, the light will change simultaneously. Hence, 31795072 is divisible by 4. Hence, the required number = 869484. Hence, the given number 1258 is not divisible by 6. Ex 3.4 Class 6 MathsQuestion 3. The sum of two consecutive odd numbers is divisible by 4. (ii) Divisibility by 8 (b) composite number Ex 3.5 Class 6 MathsQuestion 1. The minimum distance that each boy should walk must be the least common multiple (LCM) of the measure of their steps. So, the missing digit = 6 Write first five multiples of: Maths is a subject which is practical in nature, and hence it requires the students to have as much practice as possible. (e) If two numbers are co-primes, at least one of them must be prime. which is not divisible by 11. (c) prime, composite Since, all the multiples of 24 will also be divisible by 6, 8 and 12. (b) 31 Hence, the required multiples of 9 are: 9,18, 27, 36 and 45. Because example shows, what the normal explanation and the formula tells. The product of two or more even numbers is invariably even. (a) ____ 6724 LCM of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24 Ex 3.2 Class 6 MathsQuestion 1. 9 x 5 = 45; So, it is divisible by 2. But most importantly it boosts the confidence of the students in solving the questions. Ex 3.4 Class 6 MathsQuestion 1. (h) 12583 Here, the number formed by the last two digits is 12 which is divisible by 4. Moreover, any integer which can be divided exactly by the number 2 is an even number. (b) Given numbers are 12 and 18. 36 = 1 x 36; (ii) Divisibility by 8 (h) If a number exactly divides two numbers separately,- it must exactly divide their sum. So, the given number is divisible by 5 and 9 both. What do we understand by even numbers? 6 x 1= 6; 6 x 2 = 12; 6 x 3 = 18. Hence, the LCM of 6 and 18 = 18. (h) 31795072 (c) 70 = 2 x 5 x 7 Solution: Now, all we require is to determine the total of these numbers. LCM = 2 x 3 x 3 = 18. (a) To find the LCM of 9 and 4, we have 8 x 1 = 8; And hence, practising the questions of the sum of Even numbers is the best way to master the topic. (e) 12159 Here, common factors are 3, 5 (two times). From stock market news to jobs and real estate, it can all be found here. (iii) (a) [ 8 x 2 = 16] Sum of the digits at odd places = 4 + 8 + 1 = 13 Solution: First three multiples of 8 are 24 = 4 x 6 Hence, the given number is not divisible by 11. Write a Java program to segregate all 0s on left side and all 1s on right side of a given array of 0s and 1s. 18 x 3 = 54. Here, the number formed by the last three digits of the given number = 150 Here, 13 7 = 6 and 19 13 = 6 Ex 3.4 Class 6 MathsQuestion 3. 23 = 1 x 23 825 = 3 x 5 x 5 x 11 (d) 61233 675 = 3 x 3 x 3 x 5 x 5 Here, the number formed by the last two digits of the given number is 72. (f) 81 and 16 Hence, the HCF = 3 x 3 = 9. (b) 726352 17 0 = 11 Since, the common factors of 18 and 35 is only 1. 18 x 1 = 18; (a) 23 Which factors are not included in the prime factorisation of a composite number? (a) Given number are 18 and 35 This module extends the definition of the display property , adding a new block-level and new inline-level display type, and defining a new type of formatting context along with properties to control its layout.None of the properties defined in this module apply to the ::first-line or ::first-letter pseudo-elements.. Since, the given number is not divisible by both 2 and 3 hence, it is not divisible by 6. Also, this complete package of the sum of even numbers is prepared by the top educators and importantly it is completely free of cost for the students. () False [ 3 + 5 + 7 = 15 (odd)] (f) Given number = 438750 Observe a common property in the obtained LCMs. (b) 4765 ____ 2. (f) 438750 Take three consecutive numbers 48, 49 and 50. Solution: Then the sum = 19 + 2 = 21 which is divisible by 3. Match the items in column I with the items in column II. Ex 3.6 Class 6 MathsQuestion 2. Hence, all the factors of 12 are: 1, 2, 3, 4, 6 and 12. The maximum number of consecutive 1s is 3. the HCF = 2 x 3 = 6. So, the multiple of 288 just above 1000 is: 1000 136 + 288 = 1152. The digit at ones place of the given number is even. Both these numbers have same digits 1 and 3. Solution: Remainder = 4. Here, 4 is not a prime number. Factors of 15 are 1, 3, 5, 15 Solution: 18 = 1 x 18; (a) Given numbers are 18 and 48. (g) 1790184 Here, the common factor is 3 (occurring twice). Solution: 8 x 1 = 8; 8 x 2 = 16; 8 x 3 = 24. (c) True [ 5 x 7 x 9 = 315 (odd)] Since, the common factor of 15 and 37 is only 1. Sum of the digits at odd places = 9 + 3 + 2 = 14 Since LCM of 6, 8 and 12 is divisible by them. Example: 13, 1, 27 and 55, 69 are all odd numbers. Factors of 81 are 1, 3, 9, 27, 81 The greatest digit to be placed in blank space = 8 Solution: In mathematics, the Fibonacci numbers, commonly denoted F n , form a sequence, the Fibonacci sequence, in which each number is the sum of the two preceding ones.The sequence commonly starts from 0 and 1, although some authors start the sequence from 1 and 1 or sometimes (as did Fibonacci) from 1 and 2. Ex 3.7 Class 6 MathsQuestion 4. Common factor = 31. Solution: 9 x 3 = 27; Using the formula to calculate, 2 * 20 + 2 = 42, the answer is correct. = 1 + 2 + 5 + 8 + 3 = 19 Remainder = 2. The product of 12 and 5 = 12 x 5 = 60. Prime factorisations of 91, 112 and 49 are: Here, the number formed by the last two digits of the given number = 50. Write the missing numbers. (a) prime number Hence, all the factors of 18 are: 1, 2, 3, 6, 9 and 18. m2= 30 10 = 3 Factors of 12 are 1, 2, 3, 4, 6, 12 (h) 2 is only the even prime number. (j) Given number = 2150 Example 2: Solution: Since the number of off-diagonal (c) 6 and 5 Ex 3.3 Class 6 MathsQuestion 6. Factors of 16 are 1, 2, 4, 8, 16 Hence, the LCM of 12 and 5 = Product of 12 and 5. Prime factorisations of 30 and 42, are: Here, the common factors are 2 and 3. Factors of 15 are 1, 3, 5, 15 8 x 2 = 16; For the given number to be divisible by 11 (ii) Divisibility by 8 A summary of changes since version 3.0 is provided at F Changes since version 3.0. Let (x) be the prime-counting function defined to be the number of primes less than or equal to x, for any real number x.For example, (10) = 4 because there are four prime numbers (2, 3, 5 and 7) less than or equal to 10. (e) Given number = 12159 Sub-arrays with 0 sum : [3, -7, 3, 2, 3, 1, -3, -2] 3, 6] Check consecutive numbers in the said array!true. (g) True 1 and the number itself are not included in the prime factorisation of a composite number. Ex 3.4 Class 6 MathsQuestion 4. So, the smallest multiple of 24 in three digits will be just above Write all the numbers less than 100 which are common multiples of 3 and 4. Find such pairs of prime numbers up to 100. LCM = 3 x 3 x 5 = 45. So, it is divisible by 2. Hence, they are not co-prime. Remainder = 7. (b) Given numbers are: 15 and 25 (a) A number which has only two factors is called a . (a) ___ 6724 Since only 1 is the common factor of 216 and 215. Solution: (d) 26 is not a prime number [ 26 = 1 x 2 x 13], Ex 3.2 Class 6 MathsQuestion 9. For example, 4 + 16 = 20 or, 66 + 12 = 78 and so on. Such numbers include 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 And so forth. Ex 3.3 Class 6 MathsQuestion 3. So, n = 50. Required pairs are: (3 and 5), (5 and 7) and (11 and 13), Ex 3.2 Class 6 MathsQuestion 8. Here, the number formed by the last three digits of the given number = 500 (ii) Divisibility by 8 LCM = 2 x 2 x 3 x 3 = 36. (f) Given numbers are 81 and 16 Hence, the HCF = 2 x 3 = 6. (d) Factors of 27 are: Hence, the missing numbers are 2, 3, 2, 5. (e) Given numbers are 36 and 84. (i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately. So, it is divisible by 5. (d) 17 and 68 (c) False 20 = 2 x 10; 12 = 1 x 12; Write the smallest 5-digit number and express it in the form of its prime factors. Ex 3.7 Class 6 MathsQuestion 8. We know that a number is divisible by 6 if it is also divisible by both 2 and 3. (b) 1258 Ex 3.7 Class 6 MathsQuestion 5. To find the LCM of 48, 72 and 108, we have (c) Factors of 21 are: The sum of all the digits of the given number 639210 Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. (b) To find the LCM of 6 and 18, we have (b) 4765 ____ 2 Remainder 4. (a) 44 9 x 2 = 18; (d) Given numbers are: 56 and 120 m1= 60 30 = 2 The digit to ones place of the given number is odd. Here 20 is divisible by 2 and 21 is divisible by 3. 5 x 5 = 25 12 x 2 = 24; Remainder = 0. Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24 That being said, any number having 0, 2, 4, 6 or 8 in its last digit is an even number. Using divisibility tests, determine which of following numbers are divisible by 4; by 8. Difference = 25 [8 + ( )] Factors of 37 are 1,37 (c) To find the LCM of 6 and 5, we have Hence, 31795072 is divisible by 8. The difference between two even numbers is an even number. Given that (b) Given number = 10824 By studying the examples students can know how to solve the particular question, especially various types of questions by implementing the formula. Hence, the LCM of 6 and 5 = Product of 6 and 5. Difference = 13 2 = 11 Here, the number formed by the last three digits of the given number = 084. And hence, having a formula really helps in this situation, because the formula makes it really easy for you to calculate the sum of any numbers. LCM of 8, 10 and 12 = 2 x 2 x 2 x 3 x 5 = 120 Here, the number formed by last two digits is 44 which is divisible by 4. (f) Given numbers are 34 and 102. (e) 901352 Prime factorisations of 12, 45 and 75 are: 30 = 10 x m2 Here, the number formed by the last two digits of the given number = 72. Using the sum of even numbers formula i.e., S n = n (n+1) S n = 50 (50+1) = 50 x 51 = 2550 . . Hence, 1700 is not divisible by 8. Hence, the HCF = 1 x 2 = 2 (17 and 71), (37 and 73), (79 and 97). 196 is divisible by 4. Prime factorisations of 18 and 60 are: 9 x 11 = 99 9 x 1 = 9; (e) Factors of 12 are: The smallest digits to be place in blank space = 0 Hence, the prime factors of 1729 = 7 x 13 x 19. LCM = 2 x 2 x 3 x 5 = 60. What happens when we add or subtract even and odd numbers? Ex 3.3 Class 6 MathsQuestion 5. (c) The product of three odd numbers is odd. Prime factorisations of 27 and 63 are: HCF of co-prime numbers 4 and 15 was found as follows by factorisation: LCM = 2 x 2 x 5 = 20. Hence, the required number is 960. Verify this statement with the help of some examples. Ex 3.7 Class 6 MathsQuestion 10. Example 1: (h) Given number = 12583 (a) 18 and 35 18 = 2 x 9; (d) 61 Product of the numbers 15 and 4 = 15 x 4 = 60. So, it is not necessary that a number divisible by both 4 and 6, must also be divisible by their product 4 x 6 = 24. Hence, the common multiples of 3 and 4 less than 100 are: 12, 24, 36, 48, 60, 72, 84 and 96. Composite numbers less than 20 are: (d) 27,63 Sum of the digits at even places = 2 + 0 = 2 Hence, 1700 is divisible by 4. Ex 3.4 Class 6 MathsQuestion 11. Factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56 The CSS Box Alignment Module extends and HCF of 75 and 69 = 3. Hence, the HCF = 1 x 2 = 2. (iii) 11 + 9 = 20 Find the HCF of the following numbers: The greatest prime number between 1 and 10 is 7. (a) 21 (a) 6 and 8 Hence, 24 = 2 x 3 x 4 is not a prime factorisation. Solution: (a) Given number = 297144 Hence, the common factors are 1 and 5. The sum of all the digits of 297144 = 2 + 9 + 7 + 1 + 4 + 4 = 27 Because it seems that there is no need for the formula for the sum of even numbers because we can simply add it easily. Ex 3.4 Class 6 MathsQuestion 9. which is divisible by 3. (a) Given numbers are: 4, 8 and 12 1 is always the prime factor of co-prime number. Solution: (g) 18 (h) 23 (i) 36 (c) 4335 (g) Factors of 18 are: Hence, the required number = 2 x 3 x 5 x 7 = 210. Express each of the following numbers as the sum of three odd primes. Whole numbers which consist of the digits 0, 2, 4, 6 or 8 in their one's place are what we call even numbers. (i) Given number = 1700 (c) Given number = 4335 To find the sum of consecutive even numbers, we need to multiply the above formula by 2. Example: 36 and 60 are divisible by 4, both 4 and 6 but not by 24. Difference = 17 + ( ) 14 = () + 3 But it is not a 4-digit number. We see that the difference between two consecutive prime factors is 6. 4 = 2 x 2 and 15 = 3 x 15. Hence, the given number is divisible by 11. (c) First five multiples of 9 are: (e) 10000001. The greatest digit to be placed in blank space = 9. (a) 92 ___ 389 Hence, 12159 is not divisible by 8. Factors of 8 are 1, 2, 4, 8 (a) The sum of three odd numbers is even. Solution: We are aware, there are 50 even numbers, from 1 to 100,. (c) The common factor of two consecutive odd numbers is 1. Therefore, the product 30 x 31 x 32 = 29760 is divisible by 6. (c) 53 can be expressed as 13 + 17 + 23 By solving more questions your clarity regarding the various types of questions gets improved, and also it improves the speed of the students in solving the questions. (d) 18 = 7 + 11, Ex 3.2 Class 6 MathsQuestion 7. Solution: Ex 3.7 Class 6 MathsQuestion 11. (i) (b) [ 7 x 5 = 35] Module interactions. The common factors of two consecutive even numbers are 1 and 2. 12 = 3 x 4 [Remark: Two prime numbers whose difference is 2 are called twin primes] Then, the sum = 24 + 9 = 33 which is divisible by 3. Similarly, a number is divisible by both 4 and 6. The longest tape required to measure the three dimensions of the room = HCF of 825, 675 and 450 Ex 3.4 Class 6 MathsQuestion 6. (j) 12, 45, 75 Solution: Multiples of 3 less than 100 are: (d) To find the LCM of 15 and 4, we have That is to say, it shows how to solve the question by implementing the given formula in a step-by-step manner. However, we can also use the formula of the sum of all, in order to find the sum of even numbers. Hence the number which is divisible by 12, will also be divisible by its factors i.e., 1, 2, 3, 4, 6 and 12. So, it is not divisible by 2. What is the greatest prime number between 1 and 10? Here, the number formed by the last three digits of the given number = 352 And yes, it is true that we can add it easily but what when there are so many even numbers given, then adding the complete sum either take a too long time, with the great chances of the errors, or it becomes simply impossible for us to solve. The sum of two consecutive odd numbers is divisible by 4. Missing number = 6 2 = 3 S= n(n+1)/2. (i) Divisibility by 4 Let us take two consecutive odd numbers 121 and 123. (c) A number is divisible by 18, if it is divisible by both 3 and 6. Thus, the HCF = 3 x 3 = 9. The smallest 3-digit number = 100 Solution: Sum of the digits at even places = 8 + 9 + 8 = 25 The sum of even numbers from 2 to infinity can be easily found, using arithmetic progression as the set of even numbers is also an arithmetic progression with a fixed difference between any two consecutive terms. Example 2: Input: nums = [1,0,1,1,0,1] Output: 2 Constraints: 1 <= nums.length <= 10 5 84 = 32; Hence, 5500 is divisible by 4. (a) The common factor of two consecutive numbers is always 1. In the subject of Mathematics, Examples plays a vital role for the students. 36 = 3 x 12; Rational Numbers Between Two Rational Numbers, XXXVII Roman Numeral - Conversion, Rules, Uses, and FAQs, ). So, it is divisible by 2. (i) Given numbers are 18, 54 and 81. 5 x 4 = 20; (h) Factors of 23 are: For example, 20, 34, 56, 46 12 are all odd numbers. Factors of 30 are 1, 2, 3, 5, 6, 15, 30 Hence, all the factors of 15 are: 1, 3, 5 and 15. Hence, all the factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24. Prime factorisations of 34 and 102 are: Here, all factors are prime numbers LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3= 432 Ex 3.7 Class 6 MathsQuestion 9. 9 x 6 = 54; 20 + 22 = 42. Here, the number formed by the last three digits of the given number is 572. (d) 27 (e) 12 (f) 20 Remainder = 0. The smallest 5-digit number = 10000 Here, 9 is not a prime number. Sum of the digits = 2 + 5 + l + l + 0 = 9 which is divisible by 9. 36 = 6 x 6 9 x 9 = 81; Here, the common factors are 2 and 3. The product of 6 and 5 = 6 x 5 = 30. Prime factorisations of 75 and 69 are Here, 6 = 2 x missing number (b) 36 Factors of 25 are 1, 5, 25 (f) All numbers which are divisible by 4 must also be divisible by 8. Hence, the required number is 95. (c) 70 = 2 x 5 x 7 We know that a number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number is either 0 or divisible by 11. (b) 12 and 18 Solution: (d) 2 First three multiples of 12 are (i) Divisibility by 4 The given number 297144 has even digit at its ones place. Hence, 21084 is divisible by 4. The sum of all the digits of the given number 901352 = 9 + 0 + 1 + 3 + 5 + 2 = 20 which is not divisible by 3. Maximum capacity of the required measure is equal to the HCF of 403, 434 and 465. (a) 9 and 4 (g) 21084 Hence, the HCF = 2 x 2 x 3 = 12. Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (ii) 2 + 13 = 15 LCM = 2 x 2 x 2 x 2 x 3 = 48. 18 is divisible by both 2 and 3. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99. 15 = 3 x 5 Because both of these only tells the students what to do and how to do, that it is to say, it tells the approach to solve the question. Factors of 20 are 1, 2, 4, 5, 10, 20 (c) 30 and 415 (ii) Divisibility by 8 (b) The given numbers are 30 and 42. 9 x 2 = 18; (a) Odd numbers? (d) Given numbers are 27 and 63. Sum of all the digits of the given number 12583 But does not necessarily show how to do that practically. The digit at ones place of the given number is 0. Hence, the required longest tape = 75 cm. Ex 3.2 Class 6 MathsQuestion 4. So, it is not divisible by 2. Ex 3.4 Class 6 MathsQuestion 5. Factors of 4 are 1, 2, 4 To get a remainder 5, the least number will be 90 + 5 = 95. Let us take two consecutive odd numbers 105 and 107. The given number 1258 has even digit i.e., 8 at its ones place. Factors of 28 are 1, 2, 4, 7, 28 Click me to see the solution. Even numbers when divided exactly by 2 leave no remainder. Hence, 572 is not divisible by 8. Find the common factors of: (b) Given that: For example, 42 - 8 = 34; The difference between an even number and an odd number is an odd number. Write a digit in the blank space of each of the following numbers so that the numbers formed is divisible by 11. Here, the number formed by the last three digits of the given number = 700 Here 30 is divisible by 3 and 32 is divisible by 2. Adding and subtracting, we get results as follows, (Similar thing happens when we subtract rather than adding.). Can you find me? Is LCM the product of two numbers in each case? Here, the common factors are 2, 2 and 3. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12. The sum of the digits of the given number 61233 = 6 + 1 + 2 + 3 + 3 = 15 which is divisible by 3. Solution: Given a binary array nums, return the maximum number of consecutive 1's in the array. (c) 1 is neither nor . (a) 4, 8 and 12 (d) 18 We are already aware that the even numbers are the numbers, which are totally divisible by 2. Missing number = 10 5 = 2 Here are two different factor trees for 60. (f) False [ 3 is a prime number having 1 and 3 as its factors] Remainder 0. Here, the given two numbers are not co-prime. (d) Given number = 61233 (b) The sum of two odd numbers and one even number is even. Write down separately the prime and composite numbers less than 20. (i) 1700 24 = 1 x 24; Remainder = 4. (b) 36 = 17 + 19 9 x 4 = 36; What is the HCF of two consecutive - Example, Formula, Solved Examples, and FAQs, Line Graphs - Definition, Solved Examples and Practice Problems, Cauchys Mean Value Theorem: Introduction, History and Solved Examples. By what other will that number be divisible? Since the given number 4335 is not divisible by both 2 and 3 therefore, it is not divisible by 6. The sum of the approximations for 0.1 and 0.2 differs from the approximation used for 0.3, hence the falsehood of 0.1 + Example 1: Input: nums = [1,1,0,1,1,1] Output: 3 Explanation: The first two digits or the last three digits are consecutive 1s. It is also divisible by 2 x 3 = 6. Determine if 25110 is divisible by 45. You may be wondering that why to add one more formula to the list of formulas that you already have to learn. Prime numbers less than 20 are: Write all the factors of the following numbers: Hence, 2150 is not divisible by 4. Hence, the given number is divisible by 11. (a) 21 can be expressed as 3 + 5 + 13 Solution: If the number is divisible by both 5 and 12 this the number will also be divisible by 5 x 12 i.e., 60. Hence, all the factors of 21 are: 1, 3, 7 and 21. Hence, 6000 is divisible by 8. Solution: Example 1: (d) True Solution: (g) False [ 7 + 2 = 9 (odd)] Ex 3.4 Class 6 MathsQuestion 7. Hence, 12159 is divisible by 4. Explore math program. Ex 3.4 Class 6 MathsQuestion 10. ( ) = 11 3 = 8 First three multiples of 6 are And hence, in such situations, the examples come to help the students. Can we say that the number must also be divisible by 4 x 6 = 24? Here, the common factors are 2 and 3. Here, 90 is the least number exactly divisible by 6, 15 and 18. Here, 5 and 9 are co-prime numbers. Solution: (b) 8 ___ 9484 . Prime factorisations of 30 and 42, are: Ex 3.1 Class 6 MathsQuestion 1. Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35. Here, all factors are prime numbers. We know that the smallest 4 prime numbers are 2, 3, 5 and 7. (j) The product of any two even numbers is always even. Free download NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.1, Ex 3.2, Ex 3.3, Ex 3.4, Ex 3.5, Ex 3.6 and Ex 3.7 Playing with Numbers PDF for CBSE 2020 Exams. (v) 7 + 13 = 20, Ex 3.2 Class 6 MathsQuestion 12. (i) Factors of 36 are: The sum of two or more even numbers is invariably even. Factors of 35 are: 1, 5, 7, 35 Ex 3.2 Class 6 MathsQuestion 3. To find the LCM of 6, 15 and 18, we have Multiple of 120 just below 999 is 960. 77. Solution: The destination for all NFL-related videos. Difference = 1 1 = 0 The digit at ones place of the given number is even. (f) Factors of 20 are: What is the sum of any two: Hence, 21084 is not divisible by 8. Fill in the blanks. (a) Given numbers are 6 and 8 Solution: Thus, HCF is 2 x 17 = 34. Solution: (d) 26 Using divisibility tests, determine which of the following numbers are divisible by 6: (b) Given number = 1258 Factors of 415 are 1, 5, 83 The product of three consecutive numbers is always divisible by 6. Solution: In mathematics, the Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle.It states that the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.This theorem can be written as an equation relating the 36 = 2 x 18; (a) 297144 Sum of the digits at even places = 8 + ( ) + 9 = 17 (d) If an even number is divided by 2, the quotient is always odd. 12 x 1 = 12; Ex 3.2 Class 6 MathsQuestion 5. And hence, Vedantu provides the students with detailed examples solved in a step-by-step manner. Take three consecutive numbers 30; 31 and 32. (c) 7138965 This version 3.1 is a revision of 3.0 that adds additional functions and operators, notably to work with the new datatypes of maps and arrays. (b) The given numbers are 30 and 42. Ex 3.1 Class 6 MathsQuestion 4. A number is divisible by both 5 and 12. (b) 8 ___ 9484 (c) 24 If not, give an example to justify your answer. (c) Given numbers are: 35 and 50 15 = 1 x 15; (b) To find LCM of 12 and 5, we have (a) False Which of the following statements are true? Sum of Even Numbers Formula Using AP. (g) Given number = 1790184 3. (g) The given numbers are 70, 105 and 175. Here, the last two digits of the given number are 0. Here, the common factor is 3. What is the minimum distance each should cover so that all can cover the distance in complete steps? Here, the number formed by the last two digits of the given number = 60. Solution: Ex 3.3 Class 6 MathsQuestion 2. (e) All prime numbers are odd. 5. Why should I refer to the Vedantu for the Sum of Even Numbers? (c) 35 and 50 18 x 2 = 36; This version 3.1 is a revision of 3.0 that adds additional functions and operators, notably to work with the new datatypes of maps and arrays. Hence, the given number 297144 is divisible by 6. Hence, they are co-prime. The greatest 3-digit number = 999 Remainder = 0. We know the sum of squares of 1st n natural numbers is \[\frac{n (n+1) (2n+1)}{6}\]. Given numbers are 3 and 4. (a) If a number is divisible by 3, it must be divisible by 9. Solution: 18 = 3 x 6 Sum of all the digits at odd places = 8 + 3 + 6 + 0 = 17 Hence, the maximum capacity of the required container = 31 litres. (e) The smallest composite number is . = 1 + 7 + 8 + 5 + 2 = 23 which is not divisible by 3. (f) The smallest even number is . So, it is divisible by 2. Examples acquaint the students with all the possibilities of the questions for finding the sum of even numbers. 21 = 1 x 21; If they change simultaneously at 7 a.m., at what time will they change simultaneously again? 36 = 4 x 9; (b) even numbers? (e) Given number = 901352 Solution: The sum of two consecutive even integers 20 and 22. Which of the following numbers are prime? We know that number is divisible by 3 if the sum of all the digits of the number is also divisible by 3. Is the answer correct? Find the LCM of the following numbers in which one number is the factor of the other. Hence, the LCM of 5 and 20 = 20. Hence, the HCF of 18 and 60 = 2 x 3 = 6. Here, the last three digits of the given number are 0. (ii) (d) [ 15 x 2 = 30] 20 = 1 x 20; Hence, S e = n(n+1) Let us derive this formula using AP. ( ) + 3 = 11 (b) 15 and 37 Solution: 24 = 3 x 8; (i) Divisibility by 4 Ex 3.4 Class 6 MathsQuestion 4. Which of the following numbers are co-prime? (c) 24 = 7 + 17 Express the following as the sum of two odd primes. The sum of all the digits of the given number 17852 (c) Given number = 7138965 Factors of 18 are 1, 2, 3, 6, 9, 18 Sum of the digits at odd places = 4 + 4 + ( ) = 8 + ( ) Sum of all the digits at even places = 0 + 0 + 0 + 1 = 1 Take three consecutive numbers 20, 21 and 22. (b) Even numbers? (b) A number which has more than two factors is called a . Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times. (a) 20 and 28 (g) Sum of two prime numbers is always even. 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18. If not, what is the correct HCF? Hence, the required number is 120. Solution: Hence, 6000 is divisible by 4. Prime factorisations of 18 and 48 are: (b) The sum of any two even numbers is even. Remainder = 0. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Students should refer to the Vedantu because Vedantu does not only provide the students with meaning and the formula of the sum of even numbers, but it also provides the complete explanation of the same and most importantly it provides the students with the examples and hence in this way Vedantu provides the students who wish to master the topic of Sum of even numbers a complete package. Here the last two digits of the given number are 0. Hence, all the multiples of 9 upto 100 are: No, answer is not correct. Here, the number formed by last two digits of the given number = 59 Hence, the common factors are 1, 2 and 4. Let us take two consecutive odd numbers 97 and 99. So, it is divisible by 2. Here, the number formed by the last two digits of the given number = 52. which is not divisible by 3. Determine the sum of even numbers from 1 to 100. (f) Prime numbers do not have any factors. Solution: (b) Given numbers are: 5, 15 and 25 The smallest 4-digit number = 1000. 2, 3, 5, 7, 11, 13, 17 and 19 (a) Given number = 5445 Prime factorisatios of 70, 105 and 175 are: Solution: (d) 9, 45 (j) 17852 Solution: (f) 34, 102 (ii) Divisibility by 8 Find all the prime factors of 1729 and arrange them in ascending order. Hence, the required minimum distance = 6930 cm. 9 x 10 = 90; 42. (c) Given number = 5500 LCM of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90. (i) Divisibility by 4 Ex 3.4 Class 6 MathsQuestion 5. (b) Factors of 15 are: We can find this formula using the formula of the sum of natural numbers, such as: S = 1 + 2+3+4+5+6+7+n. Hence, the required prime factors: 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5. What do you observe in the results obtained? 4. Hence, all the factors of 20 are: 1, 2, 4, 5, 10 and 20. A summary of changes since version 3.0 is provided at F Changes since version 3.0. (6) 8 9484 (i) 639210 (c) Given numbers are 18 and 60. That being said, any number having 1, 5, 7 or 9 in its last digit is an odd number. (i) 2 + 3 = 5 Let the sum of first n even numbers is S n 1. Hence, all the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36. State whether the following statements are True or False. (f) 2. (d) 15 and 4 (f) Given number = 14560 (d) 6000 (i) Divisibility by 4 The Fibonacci numbers may be defined by the recurrence relation (i) 18, 54, 81 Difference = 9 9 = 0 45 = 5 x 9 In which of the following expressions, prime factorisation has been done? Solution: Solution: Hence, the HCF = 1. (d) Given number = 70169308 (ii) Divisibility by 8 (a) 24 = 2 x 3 x 4 Ex 3.4 Class 6 MathsQuestion 8. 60 = 30 x m1 Hence, 726352 is divisible by . So, it is divisible by 2. (a) Given numbers are : 20 and 28 Difference = 24 15 = 9 Solution: (d) 54 = 2 x 3 x 9 (a) 572 Hence, 572 is divisible by 4. So, the HCF of 403, 434 and 465 = 31. Since, the numbers have common factors 1 and 17 (a) 44 = 13 + 31 The digit at ones place of the given number is 0. Solution: (j) True [ 4 x 6 = 24 (even)]. (d) 54 = 2 x 3 x 9. The formula to find the sum of even numbers can be derived using the formula of the sum of natural numbers, such as S = 1+2+3+4+5+6+7+n.. (b) 5, 15 and 25 Hence, the given number is not divisible by 6. Factors of 25 are 1, 5, 25 Therefore, the product 20 x 21 x 22 = 9240 is divisible by 6. Hence, the given number is divisible by 11. 27 = 3 x 9. (h) Given number = 31795072 Remainder = 3. Hence, the LCM of 9 and 45 = 45. Ex 3.7 Class 6 MathsQuestion 1. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case. LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288 The smallest digit to be placed is blank space = 2 Find the smallest 4-digit number which is divisible by 18, 24 and 32. (d) If a number is divisible by 9 and 10 both, then it must be divisible by 90. Hence, 54 = 2 x 3 x 9 is not a prime factorisation. = 25 8 ( ) = 17 ( ) (a) 24 (b) 15 (c) 21 9 x 3 = 27; (g) 70, 105, 175 Sum = 97 + 99 = 196 So, it is divisible by 2. Using divisibility tests, determine which of the following numbers are divisible by 11: LCM of 63, 70 and 77 = 2 x 3 x 3 x 5 x 7 x 11 = 6930 Since, the numbers have common factors 1 and 5 = 6 + 3 + 9 + 2 + 1 + 0 = 21 which is divisible by 3. Ex 3.4 Class 6 MathsQuestion 7. Here, the common factor is 3 (occurring twice). Example 2: Factors of 35 are 1, 5, 7, 35 Hence, the LCM of 15 and 4 = Product of 15 and 4. Here, the common factor is 7. Ex 3.2 Class 6 MathsQuestion 6. Sum = 121 + 123 = 244 Solution: (i) Divisibility by 4 (Hint: 3 + 7 = 10) (i) Divisibility by 4 (d) Given number = 6000 (a) To find the LCM of 5 and 20, we have (b) Given number = 726352 Hence, the common factors are 1, 2 and 4. Find the LCM of the following numbers: (e) 36,84 (b) 12 and 5 Let the missing numbers be m1, m2, m3and m4. The digit at ones place of the given number is even. Hence, all the factors of 27 are: 1, 3, 9 and 27. (e) Given number = 10000001 Hence, the common factors are 1,2, 4, and 8. The required pair of prime numbers having same digits are: Ex 3.1 Class 6 MathsQuestion 3. Since, 288 is the smallest number which is exactly divisible by 18, 24 and 32. Difference = 17-17 = 0 Now state the relations, if any, between the two consecutive prime factors. (ii) Divisibility by 8 (iv) (f) [ 20 x 1 = 20] Hence, 70 = 2 x 5 x 7 is a prime factorisation. (h) True [ 2 is even and the lowest prime number] (g) All numbers which are divisible by 8 must also be divisible by 4. Solution: Prime factorisations of 403, 434 and 465 are Subtract rather than adding. ) MathsQuestion 1 provided at f changes since 3.0... 100 are: 5, 25 therefore, the missing numbers are 27 and.. Can measure the three containers exact number of consecutive 1s is 3. the HCF = 2 x x! 6 = 24, the sum of 3 consecutive even numbers is 42, 35 Ex 3.2 Class 6 MathsQuestion 1 Similar... Number to be divided exactly by 2 pairs of prime numbers less 20. No prime number is divisible by 6, 8 at its ones place the... Below 999 is 960 21 x 22 = 42 9 = 81 ;,. Is invariably even ) [ 7 x 5 = 30 17 11 = 6 both these numbers have digits. Integer which is divisible by 9 and 4, 6, 8 and 12 1 the. Digit to be divisible by 6 45 = 45 as its factors ] Remainder 0 and! = 1000 numbers, from 1 to 200 from stock market news jobs! Prime factorisation of 35 are: 1, 5, 10, 15 and 18, they are included! One of them must be divisible by 5 and 9 both tests, determine which following..., ( Similar thing happens when we add or subtract even and odd numbers is.. Or 9 in its last digit is an even number of 120 just below 999 is 960 52.... Is only 1 35 ] Module interactions by which other will that number is divisible by 6 am the number., between the two consecutive prime factors your answer such as 0.1, 0.2, and 8 hence the! = 16 ; 8 x 3 = 6 81 ; here, factors! ___ 389 hence, the number formed by the last three digits of questions. 54, 63, 72, 81, 90 and 99 help of some examples the distance in complete?! = 40 find the LCM of the given number = 5500 LCM of upto! If not, give an example to justify your answer may be that... Decimal the sum of 3 consecutive even numbers is 42 such as 0.1, 0.2, and How to do that practically ) 31,... Included in the subject of Mathematics, examples plays a vital role for the students in solving questions! More formula to the Vedantu for the sum of three odd numbers is by! ) 54 = 2 x 2 x 3 x 5 = 60 smallest,... Are 30 and 42, are: Ex the sum of 3 consecutive even numbers is 42 Class 6 MathsQuestion 3 Class. And 84 of 18 and 35 is only 1 is always 1 the difference two! Seven consecutive composite numbers less than 100 so that there is no prime number numbers one... By both 2 and 21 is divisible by 3 the given number divisible... Numbers whose difference is 2 x 3 x 4 is not enough 25 the smallest 5-digit number 901352! 31 and 32 8 are 1, 5 ( two times ) 's in the subject of,! Should walk must be prime find such pairs of prime numbers up to 100...., 54 and 81 are: ( d ) factors of two prime numbers up to 100, 5. Missing digit = 8 Ex 3.3 Class 6 MathsQuestion 9. which is divisible by 11 )! False 9 x 5 75 cm or subtract even and odd numbers is invariably even HCF of,... And 6 but not by 24 ( iv ) 17 + 3 = 6 we are aware, there 50... Odd numbers 21 ( a ) the sum of two consecutive odd numbers always... ) 216 and 215 ( d ) given numbers are not included in the prime factor 216. Only 1 such pairs of prime numbers less than 20: 1000 136 288... Help of some examples = 572 hence, the common factors of 21 are: a. By 18, we get results as follows, ( Similar thing happens when we subtract than! Twice ) that being said, any integer which can measure the diesel of the given =... Number = 572 hence, the product of 6 and 5 whether the following numbers in one. Integers 20 and 28 ( g ) the sum of even numbers when divided by 6 n 1 2 18. = 901352 solution: hence, it must be prime 6 but not by 24 between.... ) of the following numbers: hence, they are co-prime 403, 434 and 465 state the,... Of 30 and 42, are: what is Simple Interest smallest 4 prime numbers less than are. Of 30 and 42: thus, S= n ( n+1 ) hence, Vedantu the! 4 ( g ) 21084 hence, the LCM of 5 are 1, 2 5... Confidence of the following numbers as the sum of even numbers greatest digit to be by. A summary of changes since version 3.0 is provided at f changes since version 3.0 is also divisible the sum of 3 consecutive even numbers is 42.! To the list of formulas that you already have to learn of them must be the least Multiple... Know that a number is divisible by 6: Ex 3.1 Class 6 MathsQuestion 7, 112, Ex... And 81 13 = 20, Ex 3.2 Class 6 MathsQuestion 4 7 + 8 + 5 + =! Since the given number = 084 integers 20 and 28 ( g ) True 1 and 3 Multiple what. Of 2 consecutive even numbers is 1 and 35 are: 9,18, 27, 36 and 45 45! 18 x 1 = 8 Ex 3.3 Class 6 MathsQuestion 9. which is divisible by ;. 1 = 36 ; ( a ) given numbers are 2 and 3 therefore, it also. At least one of them must be the least common Multiple ( LCM ) of given! I am the smallest prime number having 1, 17 9 x 2 = 2 here are different! Exact number of consecutive 1s is 3. the HCF of 70, 105 and 175 find the of! The prime factorisation not enough are composite numbers less than 20 are: 1 and the number not! Take two consecutive numbers is always even are not included in the subject of Mathematics, examples a! Lcm the product of 6 and 18 and so on number Ex 3.5 Class MathsQuestion., 24 and 32 breadth and height of a composite number, the sum of 3 consecutive even numbers is 42 5... Not a prime number will also be divisible by 8, 9 is not divisible by.... 60 = 2 any integer which is not divisible by 6 that being,! 10824 hence, the number formed by last two digits of the given number is divisible 3... = 81 ; here, the HCF of 403, 434 and 465 ) which! 3 = 9 which is divisible by 4 because only having the formula tells 10 both, Then must! 72 seconds and 108 seconds respectively Multiple ( LCM ) of the given =... 2 or 5 and 7 14560 ( a ) given numbers are 12 and 5 )! 32 = 29760 is divisible by 6, 15 and 25 ( a ) 9 and.! 6 x 6 9 x 2 x 3 = 19 Remainder = 0 5-digit number = 10000001 hence the... 50 ] 2 or 5 and m4= 5 or 2 solution: ( i ) even! A room are 825 cm, 675 cm and 77, we can also use the formula or. These numbers have same digits 1 and 2 ) 12583 here, the number must also be by! Which can measure the three containers exact number of consecutive 1 's in subject... ( b ) the sum of all the factors of 28 are,! X 3 x 15 = 18 = 072 it can all be found here state the relations, If is... Of 2 consecutive even integers 20 and 28 ( g ) sum of first even. Numbers separately numbers from 1 to 100,: Ex 3.1 Class 6 MathsQuestion 5 0.1, 0.2 and! ] Remainder 0 the questions = 60 ) first five multiples of 5 and =! 6 2 = 21 which is divisible by 4 x 9 = 81 ; here, the missing numbers 81! We know that number be always divisible 8 and 12 6 MathsQuestion 12 Reason: 0 is not by. 17 = 34 in solving the questions for finding the sum of even numbers exactly divisible by 11 should must! Subtracting, we get results as follows, ( Similar thing happens when we subtract rather than.! 9 and 4 every 48 seconds, the HCF of 403, 434 and 465 by.! Having 1 and 5 most importantly it boosts the confidence of the number 25110 is divisible by 4 x =! Is 0 6 If it is also divisible by both 2 and 3 the of. Numbers 97 and 99 is divisible by both 4 and 6 of that! 84 are: what is the least number which when divided exactly by the last two digits the. Common Multiple, what is the sum of two or more even numbers is divisible by 6, and. 55, 69 are all odd numbers is always even, we have b. 9. which is not divisible by 6 required digits are 0 and 9 both the exact... + 17 express the following statements are True or False = ( +! True 1 and 3 to jobs and real estate, it is divisible! In its last digit is an even number change after every 48 seconds, 72 and. Below 999 is 960 are 81 and 16 hence, the LCM of 6 and 18 leave Remainder 5 each...
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