EXPLANATION ON YOUTUBE

Common Factor
A common factor of two or more numbers is a number
which divides each of them exactly.
For example, 4 is a common factor of 8 and 12.
Highest Common Factor
Highest common factor of two or more numbers is
the greatest number that divides each one of them
exactly. For example, 6 is the highest common factor
of 12, 18 and 24. Highest Common Factor is also
called Greatest Common Divisor or Greatest Common
Measure.
Symbolically, these can be written as H.C.F. और G.C.D. or G.C.M., respectively.

Methods of Finding H.C.F.

I. Method of Prime Factors
Step 1 Express each one of the given numbers as the
product of prime factors.
[A number is said to be a prime number if
it is exactly divisible by 1 and itself, but not
by any other number, e.g., 2, 3, 5, 7, etc. are
prime numbers]
Step 2 Choose common factors.
Step 3 Find the product of these common factors. This
is the required H.C.F. of given numbers.
Illustration 1: Find the H.C.F. of 70 and 90.
Solution: 70 = 2 × 5 × 7
90 = 2 × 5 × 9
Common factors are 2 and 5.
∴ H.C.F. = 2 × 5 = 10

Illustration 2: Find the H.C.F. of 3332, 3724 and 4508.
Solution: 3332 = 2 × 2 × 7 × 7 × 17
3724 = 2 × 2 × 7 × 7 × 19
4508 = 2 × 2 × 7 × 7 × 23
∴ H.C.F. = 2 × 2 × 7 × 7 = 196.

II. Method of Division
A. For two numbers:
Step 1 Greater number is divided by the smaller one.
Step 2 Divisor of (1) is divided by its remainder.
Step 3 Divisor of (2) is divided by its remainder. This
is continued until no remainder is left.
H.C.F. is the divisor of last step.
Illustration 5: Find the H.C.F. of 3556 and 3444

B. For more than two numbers:
Step 1 Any two numbers are chosen and their H.C.F.
is obtained.
Step 2 H.C.F. of H.C.F. (of (1)) and any other number
is obtained.
Step 3 H.C.F. of H.C.F. (of (2)) and any other number
(not chosen earlier) is obtained.
This process is continued until all numbers
have been chosen. H.C.F. of last step is the
required H.C.F.


Common Multiple
A common multiple of two or more numbers is a number
which is exactly divisible by each one of them.
For example, 32 is a common multiple of 8 and 16.
18 × 4 = 32
16 × 2 = 32.
Least Common multiple
The least common multiple of two or more given numbers
is the least or lowest number which is exactly divisible
by each of them.
For example, consider the two numbers 12 and 18.
Multiples of 12 are 12, 24, 36, 48, 60, 72, …
Multiples of 18 are 18, 36, 54, 72, …
Common multiples are 36, 72, …
∴ Least common multiple, i.e., L.C.M. of 12 and
18 is 36

METHODS OF FINDING L.C.M

I. Method of Prime Factors
Step 1 Resolve each given number into prime factors.
Step 2 Take out all factors with highest powers that
occur in given numbers.
Step 3 Find the product of these factors. This product
will be the L.C.M.

II. Method of Division
Step 1 The given numbers are written in a line separated
by common.
Step 2 Divide by any one of the prime numbers
2, 3, 5, 7, 11, … which will divide at least any
two of the given numbers exactly. The quotients
and the undivided numbers are written in a line
below the fi rst.
Step 3 Step 2 is repeated until a line of numbers (prime
to each other) appears.
Step 4 Find the product of all divisors and numbers
in the last line, which is the required L.C.M

 SHORT-CUT METHODS

01 H.C.F. and L.C.M. of Decimals
Step 1 Make the same number of decimal places
in all the given numbers by suffixing
zero(s) if necessary.
Step 2 Find the H.C.F./L.C.M. of these numbers
without decimal.
Step 3 Put the decimal point (in the H.C.F./L.C.M. of
Step 2) leaving as many digits on its right as
there are in each of the numbers.
Illustration 10: Find the L.C.M. of 1.2, 0.24 and 6.
Solution: The given numbers can be written as
1.20, 0.24 and 6.00.
Now, ignoring the decimal we fi nd the L.C.M. of
120, 24 and 600

∴ H.C.F. of 616 and 1300 is 4.
Thus, the required H.C.F. = 0.04.

NOTE-. If the given set of numbers includes fractions as
well as whole numbers, treat whole number too
as fraction with 1 in its denominator.
2. The H.C.F. of a number of fractions is always a
fraction, but the L.C.M. may be a fraction or an integer.
03 Product of two numbers
= L.C.M. of the numbers × H.C.F. of the numbers
Illustration 14: The H.C.F. and the L.C.M. of any two
numbers are 63 and 1260, respectively. If one of the two
numbers is 315, find the other number

Solution: The required number
= × = L.C.M. H.C.F. ×
First Number
1260 63
315
= 252.
04 To fi nd the greatest number that will exactly divide
x, y and z.
Required number = H.C.F. of x, y and z.
Illustration 15: Find the greatest number that will
exactly divide 200 and 320.
Solution: The required greatest number
= H.C.F. of 200 and 320 = 40.
05 To fi nd the greatest number that will divide x, y
and z leaving remainders a, b and c, respectively.
Required number = H.C.F. of (x – a), (y – b) and
(z – c).
Illustration 16: Find the greatest number that will
divide 148, 246 and 623 leaving remainders 4, 6 and
11, respectively.
Solution: The required greatest number
= H.C.F. of (148 – 4), (246 – 6) and (623 – 11),
i.e., H.C.F. of 144, 240 and 612 = 12.
06 To fi nd the least number which is exactly divisible
by x, y and z.
Required number = L.C.M. of x, y and z.
Illustration 17: What is the smallest number which is
exactly divisible by 36, 45, 63 and 80?
Solution: The required smallest number
= L.C.M. of 36, 45, 63 and 80

= 5040.
07 To fi nd the least number which when divided by
x, y and z leaves the remainders a, b and c,
respectively. It is always observed that (x – a)
= (y – b) = (z – c) = k (say)
∴ Required number = (L.C.M. of x, y and z) – k.
Illustration 18: Find the least number which when
divided by 36, 48 and 64 leaves the remainders 25, 37
and 53, respectively.
Solution: Since, (36 – 25) = (48 – 37) = (64 – 53) = 11,
therefore, the required smallest number
= (L.C.M. of 36, 48 and 64) – 11
= 576 – 11 = 565.
08 To fi nd the least number which when divided by
x, y and z leaves the same remainder r in each
case.
Required number = (L.C.M. of x, y and z) + r.

Illustration 19: Find the least number which when
divided by 12, 16 and 18, will leave in each case a
remainder 5.
Solution: The required smallest number
= (L.C.M. of 12, 16 and 18) + 5
= 144 + 5 = 149.
09 To fi nd the greatest number that will divide x, y
and z leaving the same remainder in each case.
(a) When the value of remainder r is given:
Required number = H.C.F. of (x – r), (y – r)
and (z – r).
(b) When the value of remainder is not given:
Required number = H.C.F. of | (x – y) |, | (y – z) |
and | (z – x) |
Illustration 20: Find the greatest number which will
divide 772 and 2778 so as to leave the remainder 5 in
each case.
Solution: The required greatest number
= H.C.F. of (772 – 5) and (2778 – 5)
= H.C.F. of 767 and 2773
= 59

Illustration 21: Find the greatest number which on
dividing 152, 277 and 427 leaves equal remainder.
Solution: The required greatest number
= H.C.F. of |(x – y)|, |(y – z)| and |(z – x)|
= H.C.F. of |(152 – 277)|, |(277 – 427)| and
|(427 – 152)|
= H.C.F. of 125, 150 and 275
= 25

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