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Home Maths Notes Handy Notes : HCF and LCM For All Govt. Exams

Handy Notes : HCF and LCM For All Govt. Exams

| Updated On February 12th, 2020 at 04:24 pm

TIPS FOR SOLVING QUESTIONS RELATED TO 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-prime 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 factorization of the number is called its prime factorization. Prime factorization of a natural number can be expressed in the exponential form.

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For example:
• 24 = 2 x 2 x 2 x 3 = 2² x 3
• 420 = 2 x 2 x 3 x 5 x 7 = 2² 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 divide 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:

Q.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, the Highest Common Factor (H.C.F.) of 8 and 14 = 2.

Example II:

Q.Find the H.C.F. of 24, 36 and 72 by Prime Factorisation method?
Solution:
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:

Q.Find the H.C.F. of 8 and 14 by Successive Division method?
Solution:
8 | 14 | 1
8
6 | 8 | 1
6
2 | 6 | 3
6
0

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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 the greatest powers/index of common prime factors gives L.C.M.

Example I:

Q.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 :

Q.Find the L.C.M. of 8 and 14 by using the Division method?
Solution:
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.

Other important formula related to H.C.F. and L.C.M.

• H.C.F. of given fractions = H.C.F. of numerator / L.C.M. of the denominator
• L.C.M. of given fractions = L.C.M. of numerator / H.C.F. of the denominator
• Product of two numbers (First number x Second Number) = H.C.F. X L.C.M.
• H.C.F. of a given number always divides its L.C.M.
• The largest number which divides x, y, z to leave remainder R in each case = H.C.F. of (x-R), (y-R), (z-R).
• Largest number which divides x, y, z to leave same remainder = H.C.F. of (y-x), (z-y), (z-x).
• Largest number which divides x, y, z to leave remainder a,b,c = H.C.F. of (x-a), (y-b), (z-c).
• 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.

For More Questions From Advance Maths Click Here

Q1.The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers?
Sol:
Let the numbers be 13x and 13y (? HCF of the numbers = 13)
13x × 13y = 2028
=> xy = 12
co-primes with product 12 are (1, 12) and (3, 4) (? we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers
Hence numbers are 39 and 52

Q2.Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Sol:
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 30 minutes, they will toll together 30/2 + 1 = 16 times.

Q3.N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N?
Sol:
If the remainder is the same in each case and the remainder is not given, HCF of the differences of the numbers is the required greatest number
6905 – 1305 = 5600
6905 – 4665 = 2240
4665 – 1305 = 3360
Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120
Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4

Q4.Which greatest possible length can be used to measure exactly 15 meters 75 cm, 11 meters 25 cm and 7 meters 65 cm?
Sol:
Convert first all terms into cm.
i.e. 1575 cm, 1125cm, 765cm.
Now whenever we need to calculate this type of question, we need to find the HCF. HCF of above terms is 255.

Q5.A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point?
Sol:
LCM of 252, 308 and 198 = 2772
Hence they all will be again at the starting point after 2772 seconds
or 46 minutes 12 seconds

Q6. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest is
Sol:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.

Q7.What is the LCM of 2/3, 5/6 and 4/9?
Sol:
LCM for fractions = LCM of Numerators / HCF of Denominators
LCM of 2/3, 5/6 and 4/9=LCM (2, 5, 4) / HCF (3, 6, 9) = 20/3

Q8.The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
Sol:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

Q9.What is the HCF of 9/10,12/15, 18/35 and 21/40 is :
Sol.
HCF for fractions = HCF of Numerators / LCM of Denominators
HCF of 9/10,12/15, 18/35 and 21/40=HCF (9,12,18,21) / LCM (10,15,35,40) = 3/840=1/280

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