Principles of ArithmeticDeighton, 1885 - 220 pages |
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Page 2
... third group . For it is clear from the definition that two groups will contain the same number of objects when each contains the same number as a third group . Those objects which in the former groups correspond with the same object in ...
... third group . For it is clear from the definition that two groups will contain the same number of objects when each contains the same number as a third group . Those objects which in the former groups correspond with the same object in ...
Page 62
... third given number . Thus , if we want to find the G.C.M. of 150 , 45 and 63 , we find first the G.C.M. of 45 and 150 . 45 ) 150 ( 3 135 15 ) 45 ( 3 45 It is 15. We next find the G.C.M. of 15 and 63 . 15 ) 63 ( 4 60 3 ) 15 ( 5 15 It is ...
... third given number . Thus , if we want to find the G.C.M. of 150 , 45 and 63 , we find first the G.C.M. of 45 and 150 . 45 ) 150 ( 3 135 15 ) 45 ( 3 45 It is 15. We next find the G.C.M. of 15 and 63 . 15 ) 63 ( 4 60 3 ) 15 ( 5 15 It is ...
Page 73
... third number 7 for exam- ple , prime to each of the preceding numbers , then we may shew that any number divisible by 8 , 15 and 7 is divisible by their product 840. For any number divisible by 8 and 15 is divisible by 120 , and since 8 ...
... third number 7 for exam- ple , prime to each of the preceding numbers , then we may shew that any number divisible by 8 , 15 and 7 is divisible by their product 840. For any number divisible by 8 and 15 is divisible by 120 , and since 8 ...
Page 74
... third number and find their L.C.M. and so on . A more convenient method in prac- tice is that indicated below . 6 , 14 , 18 , 24 , 30 , 56 2 | 3 3 7 9 12 15 28 7 I 7 3 4 5 28 4 I I 3 4 5 4 I I 3 I 5 I Here we wish to find the L.C.M. of ...
... third number and find their L.C.M. and so on . A more convenient method in prac- tice is that indicated below . 6 , 14 , 18 , 24 , 30 , 56 2 | 3 3 7 9 12 15 28 7 I 7 3 4 5 28 4 I I 3 4 5 4 I I 3 I 5 I Here we wish to find the L.C.M. of ...
Page 75
... third number 7 for exam- ple , prime to each of the preceding numbers , then we may shew that any number divisible by 8 , 15 and 7 is divisible by their product 840. For any number divisible by 8 and 15 is divisible by 120 , and since 8 ...
... third number 7 for exam- ple , prime to each of the preceding numbers , then we may shew that any number divisible by 8 , 15 and 7 is divisible by their product 840. For any number divisible by 8 and 15 is divisible by 120 , and since 8 ...
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Common terms and phrases
2nd Edition 4th Edition addition arithmetical bers centimetres consecutive numbers contain Crown 8vo cube decimal fractions decimal places decimetre difference divisible by 9 divisors English equal number example F. A. Paley Fcap figures find the number finding the G.C.M. five four numbers fourth geometrical geometrical progression George Bell given number greater Greek Hence hundred hundredths Latin least common multiple length M.A. 3rd Edition means millimetres multiply Nicomachus of Gerasa notation number divisible number of combinations number of objects number of permutations number of variations number which divides numbers is equal obtain odd numbers perfect number place further places of decimals Post 8vo prime factors prime numbers prove pyramidal number quotient ratio remainder second number seven shew square inches square root subtract suppose symbols tens tenths thousand thousandths three numbers tion triangular number unaltered unit of measure vulgar fraction whole number words ΙΟ
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