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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