HCF & LCM - Basics Video Lecture - Banking Exams

hello students today we will learn
highest common factor and least common
multiple this is a very important
concept of number system you can see
many questions based on these two
concepts in many of the exams
let's have overview of the session first
of all we will learn how to calculate
highest common factor then we will
practice some remainder questions based
on highest common factor then we will
learn how to calculate the least common
multiple then we will practice some
remainder questions based on least
common multiple then we will learn how
to calculate the least common multiple
and fzf of fractions then we will do
applications of least common multiple
and xcf' before taking the questions let
us have a smaller example let my numbers
are 10 and 15 first of all we will learn
xcf
by long division method see first of all
the meaning of f c/f see this is f means
factor so I'm writing factors of 10 here
they are 1 2 5 10 in case of 15 they are
1 3 5 and 50 these are the factors of 10
and 15 then I'll take the second word
which is common it means we are looking
for common factors and they are 1 in 5
then I am taking the third word which is
highest so we are interested only in
highest common factor and that is 5 in
this case so what is yet CF of 10 and 15
it will be 5 see in this case 10 in 15
they are smaller numbers and they are
only 2 in case of bigger numbers on in
case of many of the numbers we can go by
long debate
method what is long division method this
is nothing but conventional division we
will take the bigger number inside and
the smaller number outside I mean we
will take the bigger number as dividend
and the smaller number as divisor
whether perform the conventional
division I am writing quotient here and
we are getting five as the remainder now
we will perform the division once again
in this case the old divisor will be the
new dividend two will be the quotient 0
will be the in remainder whenever you
are getting 0 as in a mender you stop
there
the latest divisor will be the xef that
will be the final final xef of these
numbers so in this case what would be
the h ZF of 10 and 15 that is 5 now we
will learn how to calculate least common
multiple I am taking the same numbers
they can be two approaches one is called
factorization method
and another thing is called division
method first of all I am going by
factorization you write the prime
factorization of 10 which is 2 into 5
you will write prime factorization of 15
that is 3 into 5 now what will be the
least common multiple we will select all
the primes and we will take its highest
power in case of 2 what is the highest
power of 2 that is 1 what is the highest
power of 3 that is also 1 what is the
highest power of 5 that is also 1 so
what will be the least common multiple
that will be 30 in this case now we will
go by long division maker write the
numbers here 10 and 15 now here we will
take a divisor of 10 15 I mean we will
take that number which is dividing the
maximum number of numbers so I can write
5 here we will write quotients here 2
and 3 now we are divided by 2 then we
will take 3 at the end we are getting 1
and 1 so we are left with no number
which is dividing 1 comma 1
except one we will stop here then we
will take multiplication of these
numbers so what will be the HCF of these
numbers it will be 5 into 2 into 3 that
is 30 now I am taking various examples
that is a very first question find HCF
of 18 and 120 again we will go by long
division method we will write the bigger
number inside and the smaller number
outside we will write the quotient here
now we will write the remainder once
again the division thing the old divisor
will be the new dividend the old divisor
will be the new division once again
since we are getting a remainder of 0 we
will stop here now the latest divisor
will be the xef so what will be the HCF
of 18 and 20 that will be 6 that is the
answer in this case now the next
question that is xef of 2 0 4 3 60 and
120 we will go by the long division
method you can select any of two numbers
I am selecting 120 and 360 putting 360
inside and taking 120 outside performing
the conventional division since we are
getting remainder as 0 so 120 will be
the HCF of these two numbers 120 and 360
so this is the HCF now we will perform
the division once again we will take the
latest FCF outside and we will take the
remaining number inside and will perform
the conventional division
once again the old divisor will be the
new dividend once again the old divisor
will be the new dividend putting 84
inside once again the division thing
since we are getting the meander a zero
that's why the latest divisor will be
the X here so in this case X here will
be 12 now see the next question first of
all we will try to understand the
meaning of this question it says the
numbers 2 6 0 4 1 0 2 0 & 4 8 1 2 when
divided by a common number n remainders
are 12 in every case we are getting a
remainder of 12 so if we subtract 12
from these numbers we will get remainder
as 0 so what is 2 6 0 4 minus 12 that is
2 5 9 2 1 0 2 0 minus 12 that is 1 0 0 8
and the third number will become 4,800
so my question will become if we divide
two five nine two one zero zero eight
and 4800 by a common number n they will
get remainder as zero it means n should
be a common factor for all these three
numbers and in question and should be
highest
so we will find HCF of that is highest
common factors of two five nine two one
zero zero eight and 4,800 we will go by
a long division method since all the
numbers are very big first of all we
will take any two numbers out of these
three I am taking first two numbers
writing bigger number inside and the
smaller number outside we will perform
the conventional division now the next
thing take the old divisor as new
dividend and old remainder as new
divisor and the normal division
again we will take all divisor as the
new dividend once again we will do
now in this case we are getting 0 as the
remainder so latest divisor will be the
f c/f so what will be the XE f that is
144
now as take the third number which was
4,800 we will find at CF of 4,800 and
the number 144
now the normal division we are getting
48 as the remainder once again the own
divisor will become the new dividend
since we are getting a remainder of zero
so the latest divisor which was 48 that
is the answer in this case so we have
performed division twice now the next
question now we will learn how to
calculate least common multiple of these
numbers they are 18 24 31 20 and 54 we
will do this concept in two approaches
one is called factorization and another
one is called division Delta first of
all we will do factorization
we will find prime factorization for
every number now 18 can be written as 2
raised to power 1 into 3 square 24 can
be written as 2 cube into 3 30 can be
written as 2 into 3 into 5 1 20 can be
written as 2 cube into 3 into 5 and 50 4
can be written as 2 into 3 cube so what
will be the least common multiple of
these numbers right all the primes which
are 2 into 3 into 5 in these case and we
will take the highest part see in case
of 2 we are getting these powers they
are 1 3 1 then 3 then 1 we will select
the highest one which is 3 in this case
then we will take power of 3 they are 2
1 1 1 & 3 so we will select 3 in this
case then we will take highest power of
5 which you can see is 1 so what will be
the least common multiples that is 2
cube into 3 cube into 5 which is 1 0 8 0
okay now we will find least common
multiple by division method see in case
of division method we will write all
these numbers 18 24 31 20 and 54 here we
will write a divisor which is dividing
many of these numbers as you can see
here 6 we can take you could take 2 as
well as 3 we will write quotients here
then again we will write a divisor here
such that we will take that number which
is dividing at least two numbers so we
can take five here again writing the
quotient if in any case a number is not
divisible by five we will write the
number as it is now we will write three
here since three is dividing three and
nine since four is not divisible by
three that is why I am writing four as
it is then I can take forward now we
will stop here since we are unable to
find a number which is dividing at least
two of the remaining numbers now what
will be the least common multiple least
common multiple will be nothing but
multiplication of these numbers and they
are 6 5 4 3 & 3 that's why and this
comes out to be 1 0 8 0 that's why this
is the least common multiple of given
numbers now let us see the next question
a nice question based on least common
multiple first of all again we will
understand the meaning of this question
see it says five bells start hauling at
the same time and they tow and after
five six 15 30 25 seconds respectively
then will they tall together again for
the first time again we will understand
the meaning of this question there are
five bells say they are b1 b2 b3 b4 and
b5 the first Bell is tolling after every
five seconds it means after five seconds
then after 10 seconds five more seconds
I mean then after 15 seconds and 20
seconds it is darling now similar is the
case for b2 but the time here is 6
seconds so we can see it will halt after
every 56 seconds or we can say it will
tall after 6 seconds then after 12
seconds then after 18 seconds and so on
in case of b3 time is 15 seconds so
these are the possible cases then we
will have 30 and so on then b5 is given
so we are looking for a time when all
the bells are tolling together you can
see this is nothing but files multiple
again in case of b2 these are 6 multiple
these are 15 s multiple these are 30 is
multiple and these are 25 multiple we
are looking for a time which will appear
in all these cases it means we are
looking for a number which is a multiple
of 5 6 15 30 and 25 it means we are
looking for a common multiple of these
numbers and the numbers are 5 6
fifteen thirty 25 Sims it is asked when
will all the bends wilt all together
again for the first time it means we are
looking for the least common multiple
now we can go by division method or by
vectorization so I am going by division
method we will write numbers here will
perform the division since 6 is not
divisible by 5 so I'm writing 6 as it is
now taking 3 as the divisor since we are
unable to find a number which is
dividing at least two numbers so we will
stop here now we will take
multiplication of these numbers and
these are 5 3 2 & 5 that is 150 so after
every 150 seconds there is wealth all
together so this is the answer in this
case we could do one more thing
see we are interested in least common
multiple of 5 6 15 30 25 since we are
interested only in multiples that's why
we can strike out all these factors see
in case of 30 you can see 5 6 15 they
all are factors of 30 that's why we can
strike it out so we can find least
common multiple of only 30 and 25 which
will be same in this case that will be
150 only so in case of least common
multiple we can neglect all the factors
of any of the numbers and in case of
highest common factor we can neglect the
multiple of any of the numbers now come
to the next question first of all we
will understand the meaning of this
question it says find the smallest
number which when divided by 2015 18 and
60 let the number be n remainder should
be 5 on base first of all we will find a
number instead of taking this remainder
we will take remainder as zero it means
when n is divided by all these numbers
remainder should be zero that implies n
should be a common multiple of these
numbers since we are interested only in
least number that's why we will try to
find least common multiples of 12 15 18
and 60 so we will find least common
multiple of 12 15 18 and 60 you can go
by long division vector as well as by
prime factorization method in this
question I am going by prime
factorization method now 12 can be
written as
two square into 3 raise to power 115 can
be written as 3 raise to power 1 into 5
raised to power 118 can be written as 2
raised to power 1 into 3 squared and 60
can be written as 2 square into 3 raise
to power 1 into 5 raised to power 1 now
what is least common multiple it will be
all the primes with highest powers so in
case of 2 this is 2 in case of 3 this is
2 and in case of 5 it will be 1 so this
is nothing but 4 into 9 into 5 which is
180 so 180 is the least number which is
exactly divisible by 12 15 18 and 60 but
coming back to original question it says
the remainder should be 5 always 180 is
a number which is exactly divisible by
these numbers so if we add 5 to this
number we will get a number which will
leave a remainder of 5 when divided by
these numbers so I am adding 5 to 180
that is 185 so this is the answer in
this case 185 now come to next question
a very nice question based on least
common multiple difficult to say it says
find the smallest number which when
divided by 9 15 18 m21 remainders should
be 5 11 14 and 17 let that number be in
in previous question remainders were
same but in this question all the
Dementors are different but you can
observe one thing what does the
remainder
between the divisor and what is the
difference between the divisor and the
remainder that is for in this case again
for in this case for in this case and
for in the last case whenever you are
getting this type of question where the
difference between the demanded and the
divisor is always same what is the
formula to calculate smallest such
number it will be least common multiple
of all the divisors minus the difference
which is four in this case so first of
all we will find the divisor least
common multiple of nine 15 18 and 21 we
can go by prime factorization as well as
division method
we can put three here
now since we are not able to find a
number which is dividing at least two of
the numbers we will stop here then we
will take multiplication of these
numbers that we the LCM this is 3 into 3
into 5 into 2 into 7 that is 630
so what would be the answer to this
question it will be 630 minus 4 that is
626 seconds so up 626 is the such number
which when divided by 9 15 18 and 21
remainders will be 5 11 14 and 17
respectively
now this is a formula how to calculate
least common multiple of flexions it
will be least common multiple of the
numerator divided by highest common
vector of the denominators similarly how
to calculate highest common factor
this is highest common factor of the
numerator divided by least common
multiple of the denominator let's have
an example let's see F of 15 by 18 20 by
15 and 25 by 12 will be HC f of the
numerators they are 15 20 and 25 and LCM
of 18 15 and 12 now what is HC over 15
20 25 I think you are comfortable now
that will be 5 what will be the least
common multiple of 18 15 and 12 you can
go by either of the approaches it will
be 180 in this case so this will be the
answer in this case now moving to LCM
part
what is least common multiple of these
numbers 15 by 18 20 by 15 and 25 by 12
this will be least common multiple of
numerators which are 15 20 and 25 in
this case divided by HC F of the
denominators so what is the least common
multiple of 15 20 and 25 it will be 300
in this case and what will be the HCF
that is 3 so what is the answer in this
case it will be 100 now you can see
practice questions there are 4 practice
questions and these are the answers
thank you so much happy learning