(a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 928389 → Odd places = 9 + 8 + 8 = 25 Even places = 2 + 3 + 9 = 14 Difference = 25 – 14 = 11 (b) We know that a number is divisible by 11Read more
(a) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 928389 → Odd places = 9 + 8 + 8 = 25
Even places = 2 + 3 + 9 = 14
Difference = 25 – 14 = 11
(b) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 869484 → Odd places = 8 + 9 + 8 = 25
Even places = 6 + 4 + 4 = 14
Difference = 25 – 14 = 11
Write all the numbers less than 100 which are common multiples of 3 and 4.
Multiple of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 Multiple of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100 Common multiples of 3 and 4 = 12, 24, 36,Read more
Multiple of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99
Multiple of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100
Common multiples of 3 and 4 = 12, 24, 36, 48, 60, 72, 84, 96
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
See lessFind the first three common multiples of: (a) 6 and 8 (b) 12 and 18
(a) Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72, ………… Multiple of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, ……………………. Common multiples of 6 and 8 = 24, 48, 72 (b) Multiple of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ……… Multiple of 18 = 18, 36, 54, 72, 90, 108, ……………………………… Common mRead more
(a) Multiple of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72, …………
Multiple of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, …………………….
Common multiples of 6 and 8 = 24, 48, 72
(b) Multiple of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ………
Multiple of 18 = 18, 36, 54, 72, 90, 108, ………………………………
Common multiples of 12 and 18 = 36, 72, 108
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See lessFind the common factors of: (a) 4, 8 and 12 (b) 5, 15 and 25
(a) Factors of 4 = 1, 2, 4 Factors of 8 = 1, 2, 4, 8 Factors of 12 = 1, 2, 3, 4, 6, 12 Common factors of 4, 8 and 12 = 1, 2, 4 (b) Factors of 5 = 1, 5 Factors of 15 = 1, 3, 5, 15 Factors of 25 = 1, 5, 25 Common factors of 5, 15 and 25 = 1, 5 https://www.tiwariacademy.com/ncert-solutions/class-6/mathRead more
(a) Factors of 4 = 1, 2, 4
Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 3, 4, 6, 12
Common factors of 4, 8 and 12 = 1, 2, 4
(b) Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors of 5, 15 and 25 = 1, 5
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See lessFind the common factors of: (a) 20 and 28 (b) 15 and 25 (c) 35 and 50 (d) 56 and 120
(a) Factors of 20 = 1, 2, 4, 5, 10, 20 Factors of 28 = 1, 2, 4, 7, 14, 28 Common factors = 1, 2, 4 (b) Factors of 15 = 1, 3, 5, 15 Factors of 25 = 1, 5, 25 Common factors = 1, 5 (c) Factors of 35 = 1, 5, 7, 35 Factors of 50 = 1, 2, 5, 10, 25, 50 Common factors = 1, 5 (d) Factors of 56 = 1, 2, 4, 7,Read more
(a) Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factors = 1, 2, 4
(b) Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors = 1, 5
(c) Factors of 35 = 1, 5, 7, 35
Factors of 50 = 1, 2, 5, 10, 25, 50
Common factors = 1, 5
(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 60, 120
Common factors = 1, 2, 4, 8
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See lessWrite the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 11: (a) 92 __________ 389 (b) 8 __________ 9484
(a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 928389 → Odd places = 9 + 8 + 8 = 25 Even places = 2 + 3 + 9 = 14 Difference = 25 – 14 = 11 (b) We know that a number is divisible by 11Read more
(a) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 928389 → Odd places = 9 + 8 + 8 = 25
Even places = 2 + 3 + 9 = 14
Difference = 25 – 14 = 11
(b) We know that a number is divisible by 11 if the difference of the sum of the digits
at odd places and that of even places should be either 0 or 11.
Therefore, 869484 → Odd places = 8 + 9 + 8 = 25
Even places = 6 + 4 + 4 = 14
Difference = 25 – 14 = 11
https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/
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