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Common factors
↪ Factors in common between two or more numbers are called common factors of those numbers. e.g.,Factors of 4 are 1, 2 and 4
Factors of 12 are 1, 2, 3, 4, 6 and 12
Factors of 16 are 1, 2, 4, 8 and 16
∴ Common factors of 4, 12 and 16 are 1, 2 and 4.
Among these common factors, 4 is the highest or greatest (4 > 2 > 1), so we can say that the highest common factor of 4, 12 and 16 is 4.
"The Highest Common factor (HCF) or Greatest Common Divisor of two or more given numbers is the highest (or greatest ) of their common factors. "
↪ Then the prime factors in common with their least occurrence is taken and multiplied to get the required HCF.
e.g.,Prime factorisation method to find HCF
↪ First the numbers are factorised into their prime factors.↪ Then the prime factors in common with their least occurrence is taken and multiplied to get the required HCF.
HCF of 12, 45 and 75 is found as follows -
12= 2×2×3,
45= 3×5×5,
75= 3×5×5,
The common factor of 12, 45, and 75 is 3 (occuring only once). Thus, the HCF of 12, 45 and 75 is 3.
Common multiples
Multiples in common between two or more numbers are called common multiples. e.g.,Multiples of 4 are - 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, ...
Multiples of 7 are - 7, 14, 21, 28, 35, 42, 49, 56, 63, ...
Common multiples of 4 and 7 are 28, 56, ...
Among all the infinite common multiples, 28 is the lowest (28 < 56 <...), so we can say that the Lowest common multiple of 4 and 7 is 28..
"The Lowest Common factor (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. "
Prime factorisation method to find LCM
↪ First the numbers are factorised into their prime factors.↪ Then the each prime factors with their maximum occurrence is taken and multiplied to get the required LCM.
e.g.,
LCM of 40, 48 and 45 is found as follows -
40 = 2×2×2×5
48 = 2×2×2×2×3
45 = 3×3×5
Maximum occurrence of prime factors 2, 3 and 5 is four, two and one respectively.
∴ Required LCM = 2×2×2×2×3×3×5 = 720
Time to think -
1) What is the HCF of two consecutive-(a) numbers (b) even numbers (c) odd numbers.
2) HCF of two co-prime numbers (4 = 2×2 & 15 = 3×5) found to be 0 using prime factorisation method, since there is no common prime factor. Is the answer correct? If not, what is the correct HCF?
Answer -
1) (a) 1 (b) 2 (c) 12) No, correct answer is 1 which is only common factor co-primes have but it is not a prime number that's why we don't find it in prime factorisation method.
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