\nFind the prime factorisation of the following numbers:
\n(i) 168
\n(ii) 2304
\nSolution:
\n(i) 168
\n
![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-1.png\")
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 7<\/p>\n
Question 1.
\nWrite all factors of:
\n(i) 88
\n(ii) 105
\n(iii) 96
\nSolution:
\n(i) 88 = {1, 2, 4, 8, 11, 22, 44, 88}
\n(ii) 105 = {1, 3, 5, 7, 15, 21, 35, 105}
\n(iii) 96 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96}<\/p>\n
Question 2.
\nFind the common mutliples of 8 and 12.
\nSolution:
\nThe multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64 72, 80, 88, 96, 104,
\nThe multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108,
\nThe common multiples of 8 and 12 are 24, 48, 72, 96<\/p>\n
Question 3.
\nWhich of the following pairs of numbers are co-prime?
\n(i) 25 and 105
\n(ii) 59 and 97
\n(iii) 161 and 192
\nSolution:
\n(i) 25 and 105
\nThe factors of 25 are 1, 5, 25
\nThe factors of 105 are 1,3, 5, 7, 15, 21, 35, 105
\nThe common factors of 25 and 105 are 1, 5
\n\u2234 They are not co-prime<\/p>\n
(ii) 59 and 97
\nThe factors of 59 are 1, 59
\nThe factors of 97 are 1, 97
\nThe common factors of 59 and 97 is 1
\n\u2234 They are co-prime<\/p>\n
(iii) 161 and 192
\nThe factors of 161 are 1, 161
\nThe factors of 192 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192
\nThe common factors of 161, 192 is 1
\n\u2234 They are co-prime.<\/p>\n
Question 4.
\nUsing divisibility tests, determine which of the following numbers are divisible by 4, 6, 8, 9 or 11:
\n(i) 197244
\n(ii) 613440
\n(iii) 4100448
\nSolution:
\n197244: divisible by 4, 6, 9.
\nIt is divisible by 4 as last two digits is divisible by 4.
\nIt is divisible by 6 as last digit of given number is divisible 2 and their sum is also divisible by 3.
\nSum of digits 1 + 9 + 7 + 2 + 4 + 4 = 27 Which is divisible by 3.
\nIt is not divisible by 8 as the sum of last three digit 2 + 4 + 4 = 10, is not divisible by 8.
\nIt is divisible by 9 as the sum of its digits 27 is divisible by 9.
\nIt is not divisible by 11 as the difference of the sum of alternate number 1 + 7 + 4 = 12 and 9 + 2 + 4 = 15, (15 – 12) = 3 is not divisible by 11.
\n\u2234 197244 is divisible by 4, 6 and 9.<\/p>\n
(ii) 613440 : divisible by 4, 6, 8, 9.
\nIt is divisible by 4 as last two digit is divisible by 4.
\nIt is divisible by 6 as the sum of all digits (6 + 1 + 3 + 4 + 4 + 0)=18 is divisible by 3 and by 2 also as last digit is 0.
\nIt is divisible by 8 as the sum of last three digits (4 + 4 + 0) = 8 is divisible 8.
\nIt is also divisible by 9 as the sum of its digits 18 is divisible by 9.
\nIt is not divisible by 11 as the difference of the sum of alternate number 6 + 3 + 4 = 13 and 1 + 4 + 0 = 5, (13 – 5) = 8 is not divisible by 11.
\n613440 is divisible by 4, 6, 8 and 9.<\/p>\n
(iii) 4100448: divisible by 4, 6, 8, 11.
\nIt is divisible by 4 as last two digit is divisible by 4.
\nIt is divisible by 6 as the sum of all digits 4 + 1 + 0 + 0 + 4 + 4 + 8 = 21 is divisible by 3 and also last digit is divisible by 2.
\nIt is divisible by 8 as the sum of last three digits 4 + 4 + 8 = 16 is divisible by 8.
\nIt is not divisible by 9 as the sum of its digit 21 is not divisible by 9.
\nIt is divisible by 11 as the difference of the sum of alternate number 4 + 0 + 4 + 8= 16 and 1 + 0 + 4 = 5, (16 – 6) = 11 which is divisible by 11.
\n\u2234 4100448 is divisible by 4, 6, 8, 11.<\/p>\n
Question 5.
\nIn 92 * 389, replace * by a digit so that the number formed is divisible by 11.
\nSolution:
\nThe given number is 92 * 389
\nHere, * occur at odd place.
\nSum of digits at odd place = 9 + 8 = 17 (Except *)
\nSum of digits at even place = 2 + 3 + 9 = 14
\nTheir difference = 17 – 14 = 3
\nIf ‘*’ is replaced by 8, then sum of digits at odd place = 9 + 8 + 8 = 25
\nTheir difference (Sum of digits at odd places – Sum of digits at even places)
\n= 25 – 14 = 11
\nWhich is divisible by 11
\n\u2234 ‘*’ is to be replaced by the digit 8.<\/p>\n
Question 6.
\nFind the prime factorisation of the following numbers:
\n(i) 168
\n(ii) 2304
\nSolution:
\n(i) 168
\n
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 7<\/p>\n
(ii) 2304
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-2.png\")
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3<\/p>\n
Question 7.
\nFind the G.C.D. of the given numbers by prime factorisation method :
\n(i) 24,45
\n(ii) 180, 252, 324
\nSolution:
\n(i) 24, 45
\n24 = 2 \u00d7 2 \u00d7 2 \u00d7 3
\n45 = 3 \u00d7 3 \u00d7 5
\nThe greatest common factor is 3.
\nG.C.D = 3
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-3.png\")
<\/p>\n
(ii) 180, 252, 324
\n180 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-4.png\")
\n252 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 7
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-5.png\")
\n324 = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-6.png\")
\nG.C.D = 2 \u00d7 2 \u00d7 3 \u00d7 3 = 36
\nWe notice that 2 and 3 both occurs as the common factor in the given numbers two time each.<\/p>\n
Question 8.
\nFind the H.C.F of the given numbers by division method.
\n(i) 54, 82
\n(ii) 84, 120, 156
\nSolution:
\n(i) 54, 82
\nH.C.F = 2
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-7.png\")
<\/p>\n
(ii) 84, 120, 156
\nSolution:
\nH.C.F = 12
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-8.png\")
<\/p>\n
Question 9.
\nFind the L.C.M of the given numbers by prime factorisation method.
\n(i) 27, 90
\n(ii) 36, 48, 210
\nSolution:
\n(i) 27, 90
\n= 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 = 270
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-9.png\")
<\/p>\n
(ii) 36, 48, 210
\n= 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 7 = 5040
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-10.png\")
<\/p>\n
Question 10.
\nFind the L.C.M of the given numbers by division method:
\n(i) 48, 60
\n(ii) 112, 168, 266
\nSolution:
\n(i) 48, 60
\n= 2 \u00d7 2 \u00d7 3 \u00d7 4 \u00d75 = 240
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-11.png\")
<\/p>\n
(ii) 112, 168, 266
\n= 2 \u00d7 2 \u00d7 2 \u00d7 7 \u00d7 2 \u00d7 3 \u00d7 19 = 6384
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-12.png\")
<\/p>\n
Question 11.
\nFind the greatest number which divides 2706, 7041 and 8250 leaving remainder 6, 21 and 42 respectively.
\nSolution:
\nWhen 2706 is divided by the required number, 6 is left as a remainder. So, 2706 – 6 = 2700 i.e. 2700 is exactly divisible by that number.
\nSimilarly, 7041 – 21 = 7020 is exactly divisible by that number.
\nSimilarly, also, 8250 – 42 = 8208 is exactly divisible by that number.
\nTherefore, 2700, 7020 and 8208 are divisible by that number.
\nThus, the required number is the H.C.F. of 2700, 7020 and 8208.
\nFirst, we find H.C.F. of 2700 and 7020.
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-13.png\")
\n\u2234 The H.C.F. of 2700, 7020 and 8208 is 108.
\nHence the required number is 108<\/p>\n
Question 12.
\nFind the least number which on decreasing by 20 is exactly divisible by 18, 21, 28 and 30.
\nSolution:
\nFirst, we find the least number which is exactly divisible by the numbers 18, 21, 28 and 30. For this, we find the L.C.M. of 18, 21, 28 and 30.
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-14.png\")
\n\u2234 L.C.M. = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 7 = 1260
\nAccording to given, the required number will be 20 more than 1260.
\nThe required number = 1260 + 20 = 1280<\/p>\n
Question 13.
\nThere are three heaps of rice weighing 120 kg, 144 kg and 204 kg. Find the maximum capacity of a bag so that the rice of each heap can be packed in exact number of bags.
\nSolution:
\nWeights of three heaps = 120 kg, 144 kg and 204 kg
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-15.png\")
\n\u2234 Maximum capacity of a bag, which exactly divides the heaps in exact number HCF of 120, 144, 204 = 12
\n\u2234\u00a0 Required capacity of bag = 12 kg<\/p>\n
Question 14.
\nThree bells are ringing continuously at intervals of 30, 36 and 45 minutes respectively. At what time will they ring together again if they ring simultaneously at 8 a.m.
\nSolution:
\nL.C.M = 2 \u00d7 3 \u00d7 3 \u00d7 2 \u00d7 5 = 180.
\nAfter 180 minute at 11: 00 a.m.
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-16.png\")
<\/p>\n
Question 15.
\nTwo numbers are co-prime and their L.C.M. is 4940. If one of the numbers is 65, find the other number.
\nSolution:
\nOne number = 65
\nand let the other number = x
\nWe know that,
\nTwo numbers are co-prime if their HCF is 1
\nNow, H.C.F. \u00d7 L.C.M. of two numbers = Product of given two numbers
\n1 \u00d7 4940 = 65 \u00d7 x
\n\u21d2 4940 = 65 \u00d7 x
\n\u21d2 65 \u00d7 x = 4940
\n\u21d2 x = 4940 \u00f7 65 = 76
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-6-Solutions-for-ICSE-Maths-Chapter-4-Playing-with-Numbers-Check-Your-Progress-17.png\")
\n\u2234 The other number is 76<\/p>\n