Table of contents |
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HCF: Highest Common Factor |
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Methods of Finding H.C.F |
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LCM: Least Common Multiple |
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Methods of Finding L.C.M |
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Important Concepts |
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Important Results |
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Example:
Evaluate the HCF of 60 and 75.
Solution:
Write each number as a product of its prime factors.
22 x 3 x 5 = 60
3 x 52 = 75
The product of all common prime factors is the HCF.
The common prime factors in this example are 3 & 5.
The lowest power of 3 is 3 and 5 is 5.
So, HCF = 3 x 5 = 15
(b) Division Method
Step I:Here we need to divide 180 by 75.
[Divide the larger number by the smaller one].
Step II:The first divisor is 75 and the remainder is 30, so we need to divide 75 by 30.
[Divide the first divisor by the first remainder].
Step III: Now divide the second divisor 30 by the second remainder 15.
[Divide the second divisor by the second remainder].
Step IV: The remainder becomes 0.
Step V: Therefore, highest common factor = 15.
[The last divisor is the required highest common factor (H.C.F) of the given numbers].
(a) Factorisation Method
Example: Find the L.C.M of 2, 4, 6, 8, 10.
- Write the numbers as the product of primes.
- 2, 22, 2× 3, 23, 2× 5
- Take the highest powers of all the primes. i.e., 23 × 3 × 5 = 120.
Example: Find the H.C.F. of 2, 4, 6, 8, 10 using the factorization method.
- For H.C.F, take the common prime factors. i.e., 2. (This is only the prime factor common in all the numbers).
(b) LCM by Division Method
(i) Product of two numbers = L.C.M. × H.C.F.
(ii) Product of n numbers = L.C.M of n numbers × Product of the HCF of each possible pair.
Or
If the HCF of all the possible pairs taken is same then we will have:
Product of n numbers = L.C.M of n numbers × (H.C.F of each pair)(n – 1)
(iii) If the ratio of numbers is a : b and H is the HCF of the numbers. Then,
LCM of the numbers = H × a × b = HCF × Product of the ratios.
(vi) H.C.F of fractions =
(vii) L.C.M. of fractions =
(viii) If HCF (a, b) = H1 and HCF (c, d) = H2, then HCF (a, b, c, d) = HCF (H1, H2).
Tip:
1. LCM is always a multiple of HCF of the numbers.
2. The numbers can be written as the multiple of HCF of them.
Example 1: What is the greatest number which exactly divides 110, 154 and 242?
- The required number is the HCF of 110, 154 & 242.
- 110 = 2 × 5 × 11
- 154 = 2 × 7 × 11
- 242 = 2 × 11 × 11
- ∴ HCF = 2 × 11 = 22
Example 2: What is the greatest number, which when divides 3 consecutive odd numbers produces a remainder of 1.
- If x, y, z be 3 consecutive odd numbers, then the required number will be the HCF of x – 1, y – 1 and z – 1.
- Since x-1, y-1 & z-1 are 3 consecutive even integers, their HCF will be 2. So the answer is 2.
Example 3: What is the highest 3 digit number, which is exactly divisible by 3, 5, 6 and 7?
- The least no. which is exactly divisible by 3, 5, 6, & 7 is LCM (3, 5, 6, 7) = 210. So, all the multiples of 210 will be exactly divisible by 3, 5, 6 and 7.
- So, such greatest 3 digit number is 840 = (210 × 4).
Example 4: In a farewell party, some students are giving pose for photographs, if the students stand at 4 students per row, 2 students will be left if they stand 5 per row, 3 will be left and if they stand 6 per row 4 will be left. If the total number of students are greater than 100 and less than 150, how many students are there?
- If ‘N’ is the number of students, it is clear from the question that if N is divided by 4, 5, and 6, it produces a remainder of 2, 3, & 4 respectively.
- Since (4 – 2) = (5 – 3) = (6 – 4) = 2, the least possible value of N is LCM (4, 5, 6) – 2 = 60 – 2, = 58.
- But, 100 < N < 150. So, the next possible value is 58 + 60 = 118.
Example 5: There are some students in the class. Mr X brought 130 chocolates and distributed to the students equally; then he was left with some chocolates. Mr Y brought 170 chocolates and distributed equally to the students. He was also left with the same no of chocolates as MrX was left. Mr Z brought 250 chocolates, did the same thing and left with the same no of chocolates. What is the max possible no of students that were in the class?
- The question can be stated as, what is the highest number, which divides 130, 170 and 250 gives the same remainder, i.e. HCF ((170 −130),(250 −170),(250 −130)).
i.e. HCF (40, 80, 120) = 40.
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1. What is the highest common factor (HCF) and how is it calculated? | ![]() |
2. Can you explain the prime factorization method to find the HCF? | ![]() |
3. What is the least common multiple (LCM) and how is it calculated? | ![]() |
4. How can we calculate the LCM using the prime factorization method? | ![]() |
5. Can you explain the relationship between HCF and LCM? | ![]() |