Number Systems-3
This article focuses on H.C.F/G.C.D and L.C.M
H.C.F. stands for highest common factor
G.C.D. stands for grand common divisor
Both are one and the same thing
L.C.M. stands for least common multiple
Always express a number in terms of prime numbers for finding L.C.M. and H.C.F.
Consider 2 numbers
a=2^13.It basically refers to thirteenth power of 2.(Notation used)
b=2^17
So H.C.F. is????
2^13 because the definition suggests highest common factor of the two and highest common factor of a and b is 2^13
And L.C.M. is 2^17
Concept is for H.C.F. u need to choose minimum of the powers and for L.C.M. u need to choose maximum power
Let x=2^13.3^14.5^15.7^13
y=2^6.3^56.5^17
So now H.C.F. is minimum power of(2).minimum power of (3).mimimum power of (5).minimum power of(7)
So.HCF(a,b)=2^6.3^14.5^15
L.C.M.=maximum of each power multiplied=2^13.3^56.5^17.7^13
Let x=2^13.3^23.4^5
y=2^11.3^34.4^11
So.H.C.F.=2^11.3^23.4^5
L.C.M.=2^13.3^34.4^11
Nope!!!!!!!
If you got this as answer u dint pay heed to statement that you need to express x and y in terms of prime numbers and 4 is not a prime number :)
Now express 4 in terms of 2 and get the answer.
Answers are
H.C.F.=2^21.3^23
L.C.M=2^35.3^34
Important Property
If and b are two numbers then
a * b =H.C.F. * L.C.M
Problems
1.If L.C.M of numbers upto 120 is k,find out L.C.M. of numbers upto 128?
Let their.L.C.M. be k
then k=2^6.3^4.5^2.7^2.11.13.17......remaining prime numbers upto 120 will be multiplied
Do you find it diffficult?
Writting expression for k.If yes,Please read the concept again which states that for L.C.M. you need to choose maximum powers .So for 2 ,maximum power present is in 2^6=64.Similarly other powers have been chosen
So now you have understood the concept.What i have done below is that i have basically taken care of additional prime numbers and additional powers in the extended range(maximum power concept)
L.C.M.(1,2,3....127)=k*11*5*127*2
127 is the new prime number which is added
one power of 5 comes(from 125=5^3)
one power of 11 comes(from 121=11^2)
one power of 2 comes (from 128=2^7)
So in questions like these you just have to think of additional prime numbers and additional power within the extended range
2.Find the ratio of L.C.M.(1,2,3..............200)/L.C.M.(105.106,.................200)
Now see that all the prime numbers from 1 to 100 (in numerator) are also present in denominator(in form of doubles or triples or other power).Similarly all the prime numbers in the range of 105-200 are also present in numerator and denominator.The ones which are missing are 101 and 103
And hence ratio is 101 *103
3.Given that a=6^6 ,b=8^8,c=unknown.And their L.C.M. is 12^12.Find possible values of c
Now a=2^3.3^3
b=2^24
c=...........
L.C.M.=2^24.3^12
This means highest power of 3 should come from c where there is no such restriction on power of 2 as it is already present in b
So general expression for c is c=2^t.3^12 where t can have values from 0 to 24.So total possible values of c are 25
In next post i will be discussing more problems on above concept.Happy Learning!!
This article focuses on H.C.F/G.C.D and L.C.M
H.C.F. stands for highest common factor
G.C.D. stands for grand common divisor
Both are one and the same thing
L.C.M. stands for least common multiple
Always express a number in terms of prime numbers for finding L.C.M. and H.C.F.
Consider 2 numbers
a=2^13.It basically refers to thirteenth power of 2.(Notation used)
b=2^17
So H.C.F. is????
2^13 because the definition suggests highest common factor of the two and highest common factor of a and b is 2^13
And L.C.M. is 2^17
Concept is for H.C.F. u need to choose minimum of the powers and for L.C.M. u need to choose maximum power
Let x=2^13.3^14.5^15.7^13
y=2^6.3^56.5^17
So now H.C.F. is minimum power of(2).minimum power of (3).mimimum power of (5).minimum power of(7)
So.HCF(a,b)=2^6.3^14.5^15
L.C.M.=maximum of each power multiplied=2^13.3^56.5^17.7^13
Let x=2^13.3^23.4^5
y=2^11.3^34.4^11
So.H.C.F.=2^11.3^23.4^5
L.C.M.=2^13.3^34.4^11
Nope!!!!!!!
If you got this as answer u dint pay heed to statement that you need to express x and y in terms of prime numbers and 4 is not a prime number :)
Now express 4 in terms of 2 and get the answer.
Answers are
H.C.F.=2^21.3^23
L.C.M=2^35.3^34
Important Property
If and b are two numbers then
a * b =H.C.F. * L.C.M
Problems
1.If L.C.M of numbers upto 120 is k,find out L.C.M. of numbers upto 128?
Let their.L.C.M. be k
then k=2^6.3^4.5^2.7^2.11.13.17......remaining prime numbers upto 120 will be multiplied
Do you find it diffficult?
Writting expression for k.If yes,Please read the concept again which states that for L.C.M. you need to choose maximum powers .So for 2 ,maximum power present is in 2^6=64.Similarly other powers have been chosen
So now you have understood the concept.What i have done below is that i have basically taken care of additional prime numbers and additional powers in the extended range(maximum power concept)
L.C.M.(1,2,3....127)=k*11*5*127*2
127 is the new prime number which is added
one power of 5 comes(from 125=5^3)
one power of 11 comes(from 121=11^2)
one power of 2 comes (from 128=2^7)
So in questions like these you just have to think of additional prime numbers and additional power within the extended range
2.Find the ratio of L.C.M.(1,2,3..............200)/L.C.M.(105.106,.................200)
Now see that all the prime numbers from 1 to 100 (in numerator) are also present in denominator(in form of doubles or triples or other power).Similarly all the prime numbers in the range of 105-200 are also present in numerator and denominator.The ones which are missing are 101 and 103
And hence ratio is 101 *103
3.Given that a=6^6 ,b=8^8,c=unknown.And their L.C.M. is 12^12.Find possible values of c
Now a=2^3.3^3
b=2^24
c=...........
L.C.M.=2^24.3^12
This means highest power of 3 should come from c where there is no such restriction on power of 2 as it is already present in b
So general expression for c is c=2^t.3^12 where t can have values from 0 to 24.So total possible values of c are 25
In next post i will be discussing more problems on above concept.Happy Learning!!
