LCM and HCF - Important Points
Factors and Multiples
If a number a divides another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.
Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.)
The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:
Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
Example 1: Find the H.C.F. of 72, 126 and 270.
Solution: Using Prime factorisation
72 = 2 x 2 x 2 x 3 x 3 = 23 x 32
126 = 2 x 3 x 3 x 7 = 21 x 32 x 71
270 = 2 x 3 x 3 x 3 x 5 =21 x 33 x 51
H.C.F. of the given numbers = the product of common factors with least index = 21 x 32 = 18
Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
Example 2: Find HCF of 12 and 18.
Solution:
12 ) 18 ( 1
-12
-------------
6 ) 12 ( 2
-12
---------------
0
SO, HCF = 6
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.
Similarly, the H.C.F. of more than three numbers may be obtained.
Least Common Multiple (L.C.M.)
The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
There are two methods of finding the L.C.M. of a given set of numbers:
Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the prime factors.
Example3: Find the LCM of 72, 126 and 270.
Solution: Using Prime factorisation
72 = 2 x 2 x 2 x 3 x 3 = 23 x 32
126 = 2 x 3 x 3 x 7 = 21 x 32 x 71
270 = 2 x 3 x 3 x 3 x 5 =21 x 33 x 51
L.C. M of the given numbers = Product of common factors with highest index and the non-common terms = 23 x 33 x 51 × 71
Division Method (short-cut): Arrange the given numbers in a row in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.
Example4: Find LCM of 4, 6 and 8.
Solution:
2|4,6,8
2|2,3,4
|1,3,2
LCM = 2 x 2 x 2 x 3 = 24
Some General rules for LCM and HCF
H.C.F. of Fractions: H.C.F. of Numereators / L.C.M. of Denominators
L.C.M. of Fractions: L.C.M. of Numereators / H.C.F. of Denominators
Before calculating H.C.F. or L.C.M. of Fractions, ensure that the numerator and denominator of a fraction must be co-prime to each other.
In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
Some Important Facts about LCM & HCF
- If a, b and c give remainders p, q and r respectively, when divided by the same number N, then N is HCF of (a-p), (b-q), (c-r).
- If the HCF of two numbers ‘a’ and ‘b’ is N, then, the numbers (a+b) and (a-b) are also divisible by N.
- If a number N always leaves a remainder R when divided by the numbers a, b and c, then N = LCM (or a multiple of LCM) of a, b and c + R.
- If a number when divided by a, b, c leaves remainders as x, y, z respectively and a-x = b-y = c-z = P, then the smallest number satisfying this condition is L.C.M. of (a, b, c) - P.
Applications of HCF & LCM
Example5: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 20 minutes, how many times do they toll together ?
Solution:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 20 minutes, they will toll together 20/2 + 1 = 16 times. One is added as for first time at t = 0, they tolled together.
Example6: Six racers take 2, 4, 6, 8 10 and 12 seconds respectively to run a circular field. They start together from a common point. In 20 minutes, how many times will they meet together again at the point from which they started?
Solution: L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 20 minutes, they will toll together 20/2 = 10 times .
Example7: Which is the smallest numbers which leaves a common remainder of 4 when divided by 6, 7, and 9?
Solution: All such numbers with this condition = Multiple of (LCM of 6, 7, 9) + 4
= 126 x n + 4 = 130 (By putting n=1 for smallest number)
Example8: Find the biggest 3 digit number which when divided by 6, 7 and 9 leaves a common remainder 4 in each case.
Solution: All such numbers with this condition = Multiple of (LCM of 6, 7, 9) + 4 = 126 x n + 4
Now, the biggest 3 digit number = 999. So, we have to find the number closer to 999.
By putting n = 7, The required number = 126 x 7 + 4 = 886