Study Notes of HCF and LCM
Prime number : A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For example, 2, 3, 5, 7, 11, 13, etc. are prime numbers.
Co-Prime Number: Two numbers are said to be relatively prime, mutually prime, or co-prime to each other when they have no common factor or the only common positive factor of the two numbers is 1.In other words, two numbers are said to be co-primes if their H.C.F. is 1.
Factors: The numbers are said to be factors of a given number when they exactly divide that number.
Thus, factors of 18 are 1, 2, 3, 6, 9 and 18.
Common Factors: A common factor of two or more numbers is a number which divides each of them exactly.
Thus, each of the numbers - 2, 4 and 8 is a common factor of 8 and 24.
Multiple: When a number is exactly divisible by another number, then the former number is called the multiple of the latter number.
Thus, 45 is a multiple of 1, 3, 5, 9, 15 and 45.
Common Multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them.
For example, 12, 24 and 36 is a common multiple of 3, 4, 6 and 12.
Prime Factorisation: If a natural number is expressed as the product of prime numbers, then the factorisation of the number is called its prime factorisation.
A prime factorisation of a natural number can be expressed in the exponential form.
For example:
(1) 24 = 2 x 2 x 2 x 3 = 2^3 x 3.
(2) 420 = 2 x 2 x 3 x 5 x 7 = 2^2 x 3 x 5 x 7
Highest Common Factor (H.C.F.) or Greatest Common Divisor (G.C.D.) or Greatest Common Measure (G.C.M.) are synonymous terms:
The H.C.F of two or more than two numbers is the greatest numbers which divides each of them without any remainder.
Methods of finding the H.C.F. of a given set of numbers:
Method I: Prime Factorisation method :
Express each one of the given numbers as the product of prime factors. The product of least powers/index of common prime factors gives H.C.F.
Example I:Find the H.C.F. of 8 and 14 by Prime Factorisation method?
Solution:
8 = 2 x 2 x 2
14 = 2 x 7
Common factor of 8 and 14 = 2.
Thus, Highest Common Factor (H.C.F.) of 8 and 14 = 2.
Example II:Find the H.C.F. of 24, 36 and 72 by Prime Factorisation method?
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
72 = 2 x 2 x 2 x 3 x 3
H.C.F. of 24, 36 and 72 = Product of common factors with least powers/index = 2^2 x 3
Thus, Highest Common Factor (H.C.F.) of 24, 36 and 72 = 12
Method II: Successive Division method :
Divide the larger number 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 the required H.C.F.
Example I:Find the H.C.F. of 8 and 14 by Successive Division method?
8 | 14 | 1
8
6 | 8 | 1
6
2 | 6 | 3
6
0
Least Common Multiple (L.C.M.):
L.C.M. of two or more given numbers is the smallest number which is divisible by all the given numbers.
Methods of finding the L.C.M. of a given set of numbers:
Method I: Prime Factorisation method :
Express each one of the given numbers as the product of prime factors. The product of greatest powers/index of common prime factors gives L.C.M.
Example I:Find the L.C.M. of 8 and 14 by Prime Factorisation method?
Solution:
8 = 2 x 2 x 2
14 = 2 x 7
L.C.M. of 8 and 14 = Product of all the prime factors of each of the given number with greatest index of common prime factors
= 2^3 x 7 = 56.
Thus, L.C.M. of 8 and 14 = 56.
Method II: Division method :
Find the L.C.M. of 8 and 14 by using Division method?
2 | 8, 14
|4, 7
L.C.M. of the given numbers = product of divisors and the remaining numbers = 2 x 4 x 7 = 56.
Some important formula related to H.C.F. and L.C.M.:
(1) H.C.F. of given fractions = H.C.F. of numerator / L.C.M. of denominator
(2) L.C.M. of given fractions = L.C.M. of numerator / H.C.F. of denominator
(3) Product of two numbers (First number x Second Number) = H.C.F. X L.C.M.
(4) H.C.F. of a given number always divides its L.C.M.
(5) Largest number which divides x, y, z to leave remainder R in each case = H.C.F. of (x-R), (y-R), (z-R).
(6) Largest number which divides x, y, z to leave same remainder = H.C.F. of (y-x), (z-y), (z-x).
(7) Largest number which divides x, y, z to leave remainder a,b,c = H.C.F. of (x-a), (y-b), (z-c).
(8) Least number which when divided by x, y, z and leaves a remainder R in each case = (L.C.M. of x, y, z) + R
Some Questions based on above concept and Tricks.
1.The HCF of two numbers 12906 and 14818 is 478. Their LCM is
(1)400086
(2)200043
(3)600129
(4)800172
2.The H.C.F and L.C.M of two numbers are 8 and 48 respectively. If one of the numbers is 24, then the other number is
(1)48
(2)36
(3)24
(4)16
3.Find the smallest multiple of 13 such that on being divided by 4, 6, 7 and 10. It gives remainder 2 in each case.
(1)420
(2)840
(3)2470
(4)2522
4.The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is :
(1)144
(2)72
(3)36
(4)85
5. 4 bells ring at intervals of 30 minutes, 1 hour, 1 ½ hour and 1 hours 45 minutes respectively.All the bells ring simultaneously at 12 noon. They will again ring simultaneously at:
(1)12 mid night
(2)3 a.m
(3)6 a.m
(4)9 a.m
6.Three bells ring simultaneously at 11 a.m. They rign at regular intervals of 20 minutes. 30 minutes. 40 minutes respectively. The time when all the three ring together next is
(1)2 p.m.
(2)1 p.m.
(3)1.15 p.m.
(4)1.30 p.m.
7.The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is :
(1)91
(2)910
(3)1001
(4)1911
8.A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number of times is :
(1)1 litre
(2)5 litres
(3)15 litres
(4)25 litres
9.The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :
(1)12
(2)6
(3)8
(4)10
10.The HCF (GCD) of a, b is 12, a. b are positive integers and a > b > 12. The smallest values of (a,b) are respectively .
(1)12, 24
(2)24, 12
(3)24, 36
(4)36, 24
ANSWERS AND SOLUTION
1.(1)
Product of two numbers = HCF * LCM
= 12906 * 14818 = LCM * 478
LCM = (12906 * 148)/478
= 400086
2.(4) first number x second number
= HCF * LCM
= 24 * second number = 8 * 48
so Second Number = (8 * 48)/24 = 16
3.(4)
4 = 2^2
6 = 2 * 3
7 = 7
10 = 2 * 5
LCM = 4 * 3 * 7 * 5 = 420
Number = 420 k + 2
= 416 k + (4k + 2)
When K = 6 then 4k + 2
= 4 * 6 + 2 = 26 which is divisible by 13.
so Required number
= 420 k + 2
= 420 * 6 + 2 = 2522
4.(2)
5.(4)
6.(2) LCM of 20, 30 and 40 minutes = 120 minutes
Hence, the bells will toll together again after 2 hours i.e. at 1 p.m.
7.(1) Maximum number of students
= The greatest common divisor
= HCF of 1001 and 910 = 91
8.(3) Required maximum capacity of container
= HCF of 75 l and 45 l
Now, 75 = 5 * 5 * 3
45 = 5 * 3 * 3
so HCF = 15 litres
9.(3) Let the numbers be xH and yH where H is the HCF and yH > xH.
so LCM = xyH
xyH = 2yH
x = 2
Again, xH = H = 4
2H – H = 4
H = 4
Smaller number = xH = 8
10.(4) HCF of a and b = 12
so Numbers = 12x and 12y
Where x and y are prime to each other.
A > b > 12
a = 36; b = 24
ANSWERS AND SOLUTION
1.(1)
Product of two numbers = HCF * LCM
= 12906 * 14818 = LCM * 478
LCM = (12906 * 148)/478
= 400086
2.(4) first number x second number
= HCF * LCM
= 24 * second number = 8 * 48
so Second Number = (8 * 48)/24 = 16
3.(4)
4 = 2^2
6 = 2 * 3
7 = 7
10 = 2 * 5
LCM = 4 * 3 * 7 * 5 = 420
Number = 420 k + 2
= 416 k + (4k + 2)
When K = 6 then 4k + 2
= 4 * 6 + 2 = 26 which is divisible by 13.
so Required number
= 420 k + 2
= 420 * 6 + 2 = 2522
4.(2)
5.(4)
6.(2) LCM of 20, 30 and 40 minutes = 120 minutes
Hence, the bells will toll together again after 2 hours i.e. at 1 p.m.
7.(1) Maximum number of students
= The greatest common divisor
= HCF of 1001 and 910 = 91
8.(3) Required maximum capacity of container
= HCF of 75 l and 45 l
Now, 75 = 5 * 5 * 3
45 = 5 * 3 * 3
so HCF = 15 litres
9.(3) Let the numbers be xH and yH where H is the HCF and yH > xH.
so LCM = xyH
xyH = 2yH
x = 2
Again, xH = H = 4
2H – H = 4
H = 4
Smaller number = xH = 8
10.(4) HCF of a and b = 12
so Numbers = 12x and 12y
Where x and y are prime to each other.
A > b > 12
a = 36; b = 24
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