Mixing
Fastest way to find LCM and HCF of multiple numbers?
$begingroup$
Is there any shortcut approach to find LCM and HCF of multiple numbers apart from prime factorization and hit and trial method (writing down all the multiples of respective numbers and comparing them for finding common multiple)?
greatest-common-divisor least-common-multiple
edited Jan 25 at 14:40
lioness99a
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
$endgroup$
add a comment |
$begingroup$
Is there any shortcut approach to find LCM and HCF of multiple numbers apart from prime factorization and hit and trial method (writing down all the multiples of respective numbers and comparing them for finding common multiple)?
greatest-common-divisor least-common-multiple
edited Jan 25 at 14:40
lioness99a
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
$endgroup$
add a comment |
$begingroup$
Is there any shortcut approach to find LCM and HCF of multiple numbers apart from prime factorization and hit and trial method (writing down all the multiples of respective numbers and comparing them for finding common multiple)?
greatest-common-divisor least-common-multiple
edited Jan 25 at 14:40
lioness99a
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
$endgroup$
Is there any shortcut approach to find LCM and HCF of multiple numbers apart from prime factorization and hit and trial method (writing down all the multiples of respective numbers and comparing them for finding common multiple)?
greatest-common-divisor least-common-multiple
greatest-common-divisor least-common-multiple
edited Jan 25 at 14:40
lioness99a
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
edited Jan 25 at 14:40
lioness99a
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
edited Jan 25 at 14:40
lioness99a
3,8612727
edited Jan 25 at 14:40
lioness99a
3,8612727
edited Jan 25 at 14:40
lioness99a
3,8612727
3,8612727
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
asked Jan 25 at 14:30
Yash PanchalYash Panchal
123
123
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
$$
operatorname{HCF}(a, b, c) = operatorname{HCF}(operatorname{HCF}(a, b), c)
$$
and similarily for LCM. So using an algorithm which is efficient for two numbers (say, the Euclidean algorithm for HCF, or multiplying them and dividing them by the HCF to get the LCM) will be relatively efficient for more than two numbers as well, just applied repeatedly.
Example: $a = 6, b = 10$ and $c = 15$. Then their HCF is $1$, and their LCM is $30$ (this is easily confirmed with prime factorisations, for instance). Following my desciption above, we get
$$
operatorname{HCF}(6, 10, 15) = operatorname{HCF}(operatorname{HCF}(6, 10), 15)
$$
so first we find $operatorname{HCF}(6, 10)$ with the Euclidean algorithm:
$$
operatorname{HCF}(6, 10) = operatorname{HCF}(4, 6)\
= operatorname{HCF}(2, 4) = operatorname{HCF}(0, 2) = 2
$$
which then gives
$$
operatorname{HCF}(operatorname{HCF}(6, 10), 15) = operatorname{HCF}(2, 15)\
= operatorname{HCF}(1, 2) = 1
$$
As for LCM, we get
$$
operatorname{LCM}(6, 10, 15) = operatorname{LCM}(operatorname{LCM}(6, 10), 15)
$$
so first we find $operatorname{LCM}(6, 10)$. We know from above that their HCF is $2$, so the LCM is $6cdot 10div 2 = 30$. Thus
$$
operatorname{LCM}(operatorname{LCM}(6, 10), 15) = operatorname{LCM}(30, 15)
$$
The Euclidean algorithm gives $operatorname{HCF}(30, 15) = 15$, so $operatorname{LCM}(30, 15) = 30cdot 15div 15 = 30$.
answered Jan 25 at 14:46
ArthurArthur
118k7118202
$endgroup$
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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$begingroup$
Note that
$$
operatorname{HCF}(a, b, c) = operatorname{HCF}(operatorname{HCF}(a, b), c)
$$
and similarily for LCM. So using an algorithm which is efficient for two numbers (say, the Euclidean algorithm for HCF, or multiplying them and dividing them by the HCF to get the LCM) will be relatively efficient for more than two numbers as well, just applied repeatedly.
Example: $a = 6, b = 10$ and $c = 15$. Then their HCF is $1$, and their LCM is $30$ (this is easily confirmed with prime factorisations, for instance). Following my desciption above, we get
$$
operatorname{HCF}(6, 10, 15) = operatorname{HCF}(operatorname{HCF}(6, 10), 15)
$$
so first we find $operatorname{HCF}(6, 10)$ with the Euclidean algorithm:
$$
operatorname{HCF}(6, 10) = operatorname{HCF}(4, 6)\
= operatorname{HCF}(2, 4) = operatorname{HCF}(0, 2) = 2
$$
which then gives
$$
operatorname{HCF}(operatorname{HCF}(6, 10), 15) = operatorname{HCF}(2, 15)\
= operatorname{HCF}(1, 2) = 1
$$
As for LCM, we get
$$
operatorname{LCM}(6, 10, 15) = operatorname{LCM}(operatorname{LCM}(6, 10), 15)
$$
so first we find $operatorname{LCM}(6, 10)$. We know from above that their HCF is $2$, so the LCM is $6cdot 10div 2 = 30$. Thus
$$
operatorname{LCM}(operatorname{LCM}(6, 10), 15) = operatorname{LCM}(30, 15)
$$
The Euclidean algorithm gives $operatorname{HCF}(30, 15) = 15$, so $operatorname{LCM}(30, 15) = 30cdot 15div 15 = 30$.
answered Jan 25 at 14:46
ArthurArthur
118k7118202
$endgroup$
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
add a comment |
$begingroup$
Note that
$$
operatorname{HCF}(a, b, c) = operatorname{HCF}(operatorname{HCF}(a, b), c)
$$
and similarily for LCM. So using an algorithm which is efficient for two numbers (say, the Euclidean algorithm for HCF, or multiplying them and dividing them by the HCF to get the LCM) will be relatively efficient for more than two numbers as well, just applied repeatedly.
Example: $a = 6, b = 10$ and $c = 15$. Then their HCF is $1$, and their LCM is $30$ (this is easily confirmed with prime factorisations, for instance). Following my desciption above, we get
$$
operatorname{HCF}(6, 10, 15) = operatorname{HCF}(operatorname{HCF}(6, 10), 15)
$$
so first we find $operatorname{HCF}(6, 10)$ with the Euclidean algorithm:
$$
operatorname{HCF}(6, 10) = operatorname{HCF}(4, 6)\
= operatorname{HCF}(2, 4) = operatorname{HCF}(0, 2) = 2
$$
which then gives
$$
operatorname{HCF}(operatorname{HCF}(6, 10), 15) = operatorname{HCF}(2, 15)\
= operatorname{HCF}(1, 2) = 1
$$
As for LCM, we get
$$
operatorname{LCM}(6, 10, 15) = operatorname{LCM}(operatorname{LCM}(6, 10), 15)
$$
so first we find $operatorname{LCM}(6, 10)$. We know from above that their HCF is $2$, so the LCM is $6cdot 10div 2 = 30$. Thus
$$
operatorname{LCM}(operatorname{LCM}(6, 10), 15) = operatorname{LCM}(30, 15)
$$
The Euclidean algorithm gives $operatorname{HCF}(30, 15) = 15$, so $operatorname{LCM}(30, 15) = 30cdot 15div 15 = 30$.
answered Jan 25 at 14:46
ArthurArthur
118k7118202
$endgroup$
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
add a comment |
$begingroup$
Note that
$$
operatorname{HCF}(a, b, c) = operatorname{HCF}(operatorname{HCF}(a, b), c)
$$
and similarily for LCM. So using an algorithm which is efficient for two numbers (say, the Euclidean algorithm for HCF, or multiplying them and dividing them by the HCF to get the LCM) will be relatively efficient for more than two numbers as well, just applied repeatedly.
Example: $a = 6, b = 10$ and $c = 15$. Then their HCF is $1$, and their LCM is $30$ (this is easily confirmed with prime factorisations, for instance). Following my desciption above, we get
$$
operatorname{HCF}(6, 10, 15) = operatorname{HCF}(operatorname{HCF}(6, 10), 15)
$$
so first we find $operatorname{HCF}(6, 10)$ with the Euclidean algorithm:
$$
operatorname{HCF}(6, 10) = operatorname{HCF}(4, 6)\
= operatorname{HCF}(2, 4) = operatorname{HCF}(0, 2) = 2
$$
which then gives
$$
operatorname{HCF}(operatorname{HCF}(6, 10), 15) = operatorname{HCF}(2, 15)\
= operatorname{HCF}(1, 2) = 1
$$
As for LCM, we get
$$
operatorname{LCM}(6, 10, 15) = operatorname{LCM}(operatorname{LCM}(6, 10), 15)
$$
so first we find $operatorname{LCM}(6, 10)$. We know from above that their HCF is $2$, so the LCM is $6cdot 10div 2 = 30$. Thus
$$
operatorname{LCM}(operatorname{LCM}(6, 10), 15) = operatorname{LCM}(30, 15)
$$
The Euclidean algorithm gives $operatorname{HCF}(30, 15) = 15$, so $operatorname{LCM}(30, 15) = 30cdot 15div 15 = 30$.
answered Jan 25 at 14:46
ArthurArthur
118k7118202
$endgroup$
Note that
$$
operatorname{HCF}(a, b, c) = operatorname{HCF}(operatorname{HCF}(a, b), c)
$$
and similarily for LCM. So using an algorithm which is efficient for two numbers (say, the Euclidean algorithm for HCF, or multiplying them and dividing them by the HCF to get the LCM) will be relatively efficient for more than two numbers as well, just applied repeatedly.
Example: $a = 6, b = 10$ and $c = 15$. Then their HCF is $1$, and their LCM is $30$ (this is easily confirmed with prime factorisations, for instance). Following my desciption above, we get
$$
operatorname{HCF}(6, 10, 15) = operatorname{HCF}(operatorname{HCF}(6, 10), 15)
$$
so first we find $operatorname{HCF}(6, 10)$ with the Euclidean algorithm:
$$
operatorname{HCF}(6, 10) = operatorname{HCF}(4, 6)\
= operatorname{HCF}(2, 4) = operatorname{HCF}(0, 2) = 2
$$
which then gives
$$
operatorname{HCF}(operatorname{HCF}(6, 10), 15) = operatorname{HCF}(2, 15)\
= operatorname{HCF}(1, 2) = 1
$$
As for LCM, we get
$$
operatorname{LCM}(6, 10, 15) = operatorname{LCM}(operatorname{LCM}(6, 10), 15)
$$
so first we find $operatorname{LCM}(6, 10)$. We know from above that their HCF is $2$, so the LCM is $6cdot 10div 2 = 30$. Thus
$$
operatorname{LCM}(operatorname{LCM}(6, 10), 15) = operatorname{LCM}(30, 15)
$$
The Euclidean algorithm gives $operatorname{HCF}(30, 15) = 15$, so $operatorname{LCM}(30, 15) = 30cdot 15div 15 = 30$.
answered Jan 25 at 14:46
ArthurArthur
118k7118202
edited Jan 25 at 14:52
answered Jan 25 at 14:46
ArthurArthur
118k7118202
answered Jan 25 at 14:46
ArthurArthur
118k7118202
answered Jan 25 at 14:46
ArthurArthur
118k7118202
118k7118202
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
add a comment |
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
Actually, for LCF you can just use the Euclidean algorithm directly on the triple of numbers: For instance, go from $(6, 10, 15)$ to $(6, 10, 5)$ to $(1, 10, 5)$ and once you see the $1$ there you're done. The advantage of this is that you can be clever with which order you do subtractions, and there is no downside because if you don't immediately spot something obvious and simple, you can just take the first two numbers of the tuple, and you're doing exactly what you would be doing with the pairwise algorithm described above.
$endgroup$
– Arthur
Jan 25 at 15:02
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
$begingroup$
I was about to suggest thst! And once you see $6$ with $5$ you know there's no common factor but $1$.
$endgroup$
– timtfj
Jan 25 at 15:05
add a comment |
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