APTITUDE PREPARATION: L.C.M & H.C.F


Least Common Multiple (LCM):

A number which can be divided without reminder by two or more numbers is called as the LCM of those numbers.

Highest Common Factor (HCF):

The greatest number that will divide each of the two or more numbers without a reminder is called as the HCF of those numbers.
HCF is also called as Greatest Common Divisor (GCD).
If two numbers ‘a’ and ‘b’ are considered;
LCM x HCF = a x b
Example: The HCF of 6 and 8 is 2. Find the LCM.
Solution:
2 x LCM = 6 x 8 = 48
LCM = 24.

LCM by prime factorization method:

Find the LCM for the numbers 20, 25 and 30.

Solution:

(i) Start dividing by the least prime number which can divide without reminder at least one of the given numbers. If any number is not divisible then keep as it is and proceed for the next step.
(ii) Take the next prime number.
(iii) Continue this procedure till we get all ‘ones’ (number ‘1’) in the last row.
(iv) At each step, continue dividing by the same prime number till we have no multiples of that prime number.
(v)LCM = Product of all the prime numbers.
The prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29……

Here 2 is the least prime number. But the number 25 is not divisible by 2 so write 25 as in it is in the next row.
Similarly, 25 is not divisible by 3 and hence, kept as it is in the next row.
Thus, LCM of 20, 25 and 30 is = 2 x 2 x 3 x 5 x 5 = 300.

HCF by prime factorization method:

Find the H.C.F. of 24, 36 and 48.

Solution:

(i) Individually find the prime factors for each number
(ii) Select the common prime factors present in all numbers
(iii) HCF = product of all such prime numbers.
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
The common prime factors = 2, 2, 3
H.C.F. = 2 × 2 × 3 = 12.


Euclid’s simple method to find HCF of any two numbers::

Find the HCF of 765 and 65.

Solution:

(i) Divide the larger number by the smaller one;
765 ÷ 65 = 11, reminder 50
(ii) Divide the smaller number by the reminder obtained;
65 ÷ 50 = 1, reminder 15
(iii) Divide the remainder from the step (i) by the remainder from the step (ii);
50 ÷ 15 = 3, Reminder 5
(iv) In the similar way, divide the remainder from the step (ii) by the remainder from the step (iii);
15 ÷ 5 = 3, reminder 0.
At this step, the remainder is zero, so we stop.
HCF is the smallest non-zero reminder
HCF = 5.
(Euclid’s method can be used for more than two numbers also.)

LCM and HCF of Fractions:

LCM of proper and improper fractions = (LCM of Numerators)/(HCF of Denominators)
HCF of proper and improper fractions = (HCF of Numerators)/(LCM of Denominators)

Example 1:

Find the LCM of 1/( 2), 3/5, 4/( 7) and 5/21.

Solution:

LCM = (LCM of 1,3,4 and 5)/(HCF of 2,5,7 and 21) = 60/1 = 60.

Example 2:

Find the HCF of 1/( 2), 3/5, 4/( 7) and 5/21.

Solution:

HCF = (HCF of 1,3,4 and 5)/(LCM of 2,5,7 and 21) = 1/210


Example 3:

Find the greatest number that will divide 398, 436 and 542 leaving 7, 11 and 15 as the reminders.

Solution:

(i)Subtract the reminders from the respective numbers.
398-7=391
436-11=425 and,
542-14=527.
(ii) Find the HCF of the obtained numbers.
HCF of 391, 425 and 527 is 17.

Example 4:

What is the least number which when divided by 9, 12, 16 and 30 leaves a reminder of 3 in each case?

Solution:

(i) Find the LCM of given numbers.
LCM of 9, 12, 16 and 30 = 720
(ii) Add 3 to this number.
The answer is: 720+3=723.


Example 5:

Find the least number which when divided by 35 leaves a reminder of 20, when divided by 45 leaves a reminder of 30, and when divided by 55 leaves a reminder of 40.

Solution:

In each case, the difference between the divisor and reminder is 15.
(i) Find the LCM of all divisors.
LCM of 35, 45 and 55 = 3465.
(ii)Subtract the reminder from it.
The least number = 3465-15=3450.
(Since the required number when increased by 15 becomes divisible by 35, 45 and 55)

About Author

Dr. BASKAR .A,

a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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