For finding maximum weight, we have to find H.C.F. of 75 and 69. Factors of 75 = 3 x 5 x 5 Factors of 69 = 3 x 69 H.C.F. = 3 Therefore the required weight is 3 kg. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
For finding maximum weight, we have to find H.C.F. of 75 and 69.
Factors of 75 = 3 x 5 x 5
Factors of 69 = 3 x 69
H.C.F. = 3
Therefore the required weight is 3 kg.
(a) H.C.F. of two consecutive numbers be 1. (b) H.C.F. of two consecutive even numbers be 2. (c) H.C.F. of two consecutive odd numbers be 1. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
(a) H.C.F. of two consecutive numbers be 1.
(b) H.C.F. of two consecutive even numbers be 2.
(c) H.C.F. of two consecutive odd numbers be 1.
(a) Factors of 18 = 2 x 3 x 3 Factors of 48 = 2 x 2 x 2 x 2 x 3 H.C.F. (18, 48) = 2 x 3 = 6 (b) Factors of 30 = 2 x 3 x 5 Factors of 42 = 2 x 3 x 7 H.C.F. (30, 42) = 2 x 3 = 6 (c) Factors of 18 = 2 x 3 x 3 Factors of 60 = 2 x 2 x 3 x 5 H.C.F. (18, 60) = 2 x 3 = 6 (d) Factors of 27 = 3 x 3 x 3 FactorRead more
(a) Factors of 18 = 2 x 3 x 3
Factors of 48 = 2 x 2 x 2 x 2 x 3
H.C.F. (18, 48) = 2 x 3 = 6
(b) Factors of 30 = 2 x 3 x 5
Factors of 42 = 2 x 3 x 7
H.C.F. (30, 42) = 2 x 3 = 6
(c) Factors of 18 = 2 x 3 x 3
Factors of 60 = 2 x 2 x 3 x 5
H.C.F. (18, 60) = 2 x 3 = 6
(d) Factors of 27 = 3 x 3 x 3
Factors of 63 = 3 x 3 x 7
H.C.F. (27, 63) = 3 x 3 = 9
(e) Factors of 36 = 2 x 2 x 3 x 3
Factors of 84 = 2 x 2 x 3 x 7
H.C.F. (36, 84) = 2 x 2 x 3 = 12
(f) Factors of 34 = 2 x 17
Factors of 102 = 2 x 3 x 17
H.C.F. (34, 102) = 2 x 17 = 34
(g) Factors of 70 = 2 x 5 x 7
Factors of 105 = 3 x 5 x 7
Factors of 175 = 5 x 5 x 7
H.C.F. = 5 x 7 = 35
(h) Factors of 91 = 7 x 13
Factors of 112 = 2 x 2 x 2 x 2 x 7
Factors of 49 = 7 x 7
H.C.F. = 1 x 7 = 7
(i) Factors of 18 = 2 x 3 x 3
Factors of 54 = 2 x 3 x 3 x 3
Factors of 81 = 3 x 3 x 3 x 3
H.C.F. = 3 x 3 = 9
(j) Factors of 12 = 2 x 2 x 3
Factors of 45 = 3 x 3 x 5
Factors of 75 = 3 x 5 x 5
H.C.F. = 1 x 3 = 3
The smallest four prime numbers are 2, 3, 5 and 7. Hence, the required number is 2 x 3 x 5 x 7 = 210 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
The smallest four prime numbers are 2, 3, 5 and 7.
Hence, the required number is 2 x 3 x 5 x 7 = 210
The prime factorization of 45 = 5 x 9 25110 is divisible by 5 as ‘0’ is at its unit place. 25110 is divisible by 9 as sum of digits is divisible by 9. Therefore, the number must be divisible by 5 x 9 = 45 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
The prime factorization of 45 = 5 x 9
25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
Therefore, the number must be divisible by 5 x 9 = 45
3 + 5 = 8 and 8 is divisible by 4. 5 + 7 = 12 and 12 is divisible by 4. 7 + 9 = 16 and 16 is divisible by 4. 9 + 11 = 20 and 20 is divisible by 4. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6. Example: (i) 2 x 3 x 4 = 24 (ii) 4 x 5 x 6 = 120 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example: (i) 2 x 3 x 4 = 24
(ii) 4 x 5 x 6 = 120
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.
For finding maximum weight, we have to find H.C.F. of 75 and 69. Factors of 75 = 3 x 5 x 5 Factors of 69 = 3 x 69 H.C.F. = 3 Therefore the required weight is 3 kg. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
For finding maximum weight, we have to find H.C.F. of 75 and 69.
Factors of 75 = 3 x 5 x 5
Factors of 69 = 3 x 69
H.C.F. = 3
Therefore the required weight is 3 kg.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessH.C.F. of co-prime numbers 4 and 15 was found as follows by factorization: 4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F.?
No. The correct H.C.F. is 1. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
No. The correct H.C.F. is 1.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessWhat is the H.C.F. of two consecutive: (a) numbers? (b) even numbers? (c) odd numbers?
(a) H.C.F. of two consecutive numbers be 1. (b) H.C.F. of two consecutive even numbers be 2. (c) H.C.F. of two consecutive odd numbers be 1. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
(a) H.C.F. of two consecutive numbers be 1.
(b) H.C.F. of two consecutive even numbers be 2.
(c) H.C.F. of two consecutive odd numbers be 1.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessFind the H.C.F. of the following numbers: (a) 18, 48 (b) 30, 42 (c) 18, 60 (d) 27, 63 (e) 36, 84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75
(a) Factors of 18 = 2 x 3 x 3 Factors of 48 = 2 x 2 x 2 x 2 x 3 H.C.F. (18, 48) = 2 x 3 = 6 (b) Factors of 30 = 2 x 3 x 5 Factors of 42 = 2 x 3 x 7 H.C.F. (30, 42) = 2 x 3 = 6 (c) Factors of 18 = 2 x 3 x 3 Factors of 60 = 2 x 2 x 3 x 5 H.C.F. (18, 60) = 2 x 3 = 6 (d) Factors of 27 = 3 x 3 x 3 FactorRead more
(a) Factors of 18 = 2 x 3 x 3
Factors of 48 = 2 x 2 x 2 x 2 x 3
H.C.F. (18, 48) = 2 x 3 = 6
(b) Factors of 30 = 2 x 3 x 5
Factors of 42 = 2 x 3 x 7
H.C.F. (30, 42) = 2 x 3 = 6
(c) Factors of 18 = 2 x 3 x 3
Factors of 60 = 2 x 2 x 3 x 5
H.C.F. (18, 60) = 2 x 3 = 6
(d) Factors of 27 = 3 x 3 x 3
Factors of 63 = 3 x 3 x 7
H.C.F. (27, 63) = 3 x 3 = 9
(e) Factors of 36 = 2 x 2 x 3 x 3
Factors of 84 = 2 x 2 x 3 x 7
H.C.F. (36, 84) = 2 x 2 x 3 = 12
(f) Factors of 34 = 2 x 17
Factors of 102 = 2 x 3 x 17
H.C.F. (34, 102) = 2 x 17 = 34
(g) Factors of 70 = 2 x 5 x 7
Factors of 105 = 3 x 5 x 7
Factors of 175 = 5 x 5 x 7
H.C.F. = 5 x 7 = 35
(h) Factors of 91 = 7 x 13
Factors of 112 = 2 x 2 x 2 x 2 x 7
Factors of 49 = 7 x 7
H.C.F. = 1 x 7 = 7
(i) Factors of 18 = 2 x 3 x 3
Factors of 54 = 2 x 3 x 3 x 3
Factors of 81 = 3 x 3 x 3 x 3
H.C.F. = 3 x 3 = 9
(j) Factors of 12 = 2 x 2 x 3
Factors of 45 = 3 x 3 x 5
Factors of 75 = 3 x 5 x 5
H.C.F. = 1 x 3 = 3
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessI am the smallest number, having four different prime factors. Can you find me?
The smallest four prime numbers are 2, 3, 5 and 7. Hence, the required number is 2 x 3 x 5 x 7 = 210 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
The smallest four prime numbers are 2, 3, 5 and 7.
Hence, the required number is 2 x 3 x 5 x 7 = 210
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See less18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by 4 and 6. Can we say that the number must be divisible by 4 x 6 = 24? If not, give an example to justify your answer.
No. Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
No. Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessDetermine if 25110 is divisible by 45. [Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9.]
The prime factorization of 45 = 5 x 9 25110 is divisible by 5 as ‘0’ is at its unit place. 25110 is divisible by 9 as sum of digits is divisible by 9. Therefore, the number must be divisible by 5 x 9 = 45 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
The prime factorization of 45 = 5 x 9
25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
Therefore, the number must be divisible by 5 x 9 = 45
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessIn which of the following expressions, prime factorization has been done: (a) 24 = 2 x 3 x 4 (b) 56 = 7 x 2 x 2 x 2 (c) 70 = 2 x 5 x 7 (d) 54 = 2 x 3 x 9
In expressions (b) and (c), prime factorization has been done. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
In expressions (b) and (c), prime factorization has been done.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessThe sum of two consecutive odd numbers is always divisible by 4. Verify this statement with the help of some examples.
3 + 5 = 8 and 8 is divisible by 4. 5 + 7 = 12 and 12 is divisible by 4. 7 + 9 = 16 and 16 is divisible by 4. 9 + 11 = 20 and 20 is divisible by 4. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessThe product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6. Example: (i) 2 x 3 x 4 = 24 (ii) 4 x 5 x 6 = 120 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example: (i) 2 x 3 x 4 = 24
(ii) 4 x 5 x 6 = 120
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See less