how to find hcf anf lcm of a number
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Hello Mate,
The highest common factor is found by multiplying all the factors which appear in both lists..
The lowest common multiple is found by multiplying all the factors which appear in either list.
Hope this helps you
The highest common factor is found by multiplying all the factors which appear in both lists..
The lowest common multiple is found by multiplying all the factors which appear in either list.
Hope this helps you
chino43:
ya that's correct my brother s
ok bhaiya
hmm
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H.C.F., a.k.a. highest common factor, can be found using the Euclidean algorithm.
(I tried to find a pic on the Internet, but none meet my requirements - I didn’t learn it that way. So I tried to “write” it on Quora.)
E.g. Find the H.C.F. of 2938 and 3497. (Random numbers)
Remainders……………Quotient……Extended part
3497……………………………………………1…………0
2938……………………….1…………………0………..1
559…………………………5…………………1………..-1
143…………………………3…………………-5………..6
130…………………………1…………………16………-19
13…………………………..10……………….-21………25
From this Euclidean theorem, we can see that we are continuously dividing numbers. We obtain the quotient (if not the extended part, it is useless) and the remainder, and use the previous remainder to divide the new remainder to get another quotient and another remainder. That is repeated until the remainder is 0 (because you cannot divide by 0). The final answer will be 13 in the above case, or the last remainder in a general case.
The extended part is a side note, so I will not explain much here. But you can see that
3491(−21)+2938(25)=133491(−21)+2938(25)=13 = HCF of the numbers
Then use the following identity
HCF * LCM = product of two numbers
That means the LCM of the two numbers above will be 3497×2938÷13=790322
(I tried to find a pic on the Internet, but none meet my requirements - I didn’t learn it that way. So I tried to “write” it on Quora.)
E.g. Find the H.C.F. of 2938 and 3497. (Random numbers)
Remainders……………Quotient……Extended part
3497……………………………………………1…………0
2938……………………….1…………………0………..1
559…………………………5…………………1………..-1
143…………………………3…………………-5………..6
130…………………………1…………………16………-19
13…………………………..10……………….-21………25
From this Euclidean theorem, we can see that we are continuously dividing numbers. We obtain the quotient (if not the extended part, it is useless) and the remainder, and use the previous remainder to divide the new remainder to get another quotient and another remainder. That is repeated until the remainder is 0 (because you cannot divide by 0). The final answer will be 13 in the above case, or the last remainder in a general case.
The extended part is a side note, so I will not explain much here. But you can see that
3491(−21)+2938(25)=133491(−21)+2938(25)=13 = HCF of the numbers
Then use the following identity
HCF * LCM = product of two numbers
That means the LCM of the two numbers above will be 3497×2938÷13=790322
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