HCF and LCM

Important

How to find HCF?

Step 1: Express the given number in its prime factorisation form

Step 2: Identify only the prime factors that are common to all the numbers given.

Step 3: Take the lowest power of each common prime factor and multiply them.

Example: Find the HCF of 18 and 120

18 = 2 x 3$$^2$$

120 = 2$$^3$$ x 3 x 5

Common primes → 2 and 3 → Lowest powers are 2$$^1$$ and 3$$^1$$

HCF = 2 x 3 = 6

How to find LCM?

Step 1: Express the given number in its prime factorisation form

Step 2: Identify all the prime factors that appear in any of the factorisation

Step 3: Take the highest power of every prime factor present and multiply them.

Example: Find the HCF of 60 and 90

60 = 2$$^2$$ x 3 x 5

90 = 2 x 3$$^2$$ x 5

Highest power of all primes → 2$$^2$$ , 3$$^2$$, 5

HCF = 2$$^2$$ x 3$$^2$$ x 5= 180

Basic formulas:


  • HCF * LCM of two numbers = Product of two numbers

  • The greatest number dividing a, b and c leaving remainders of $$x_1$$, $$x_2$$ and $$x_3$$ is the HCF of (a-$$x_1$$), (b-$$x_2$$) and (c-$$x_3$$).

  • The greatest number dividing a, b and c (a<b<c) leaving the same remainder each time is the HCF of (c-b), (c-a), (b-a).

  • LCM of fractions = LCM of numerators / HCF of denominators 

  • HCF of fractions = HCF of numerators / LCM of denominators

  • HCF of $$(a^m - 1)$$ and $$(a^n - 1)$$ is $$(a^{gcd(m, n)} - 1)$$

Formula Video


Question 1

Let A be the largest positive integer that divides all the numbers of the form $$3^k + 4^k + 5^k$$, and B be the largest positive integer that divides all the numbers of the form $$4^k + 3(4^k) + 4^{k + 2}$$ , where k is any positive integer. Then (A + B) equals

Question 2

A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

Question 3

The number of coins collected per week by two coin-collectors A and B are in the ratio 3 : 4. If the total number of coins collected by A in 5 weeks is a multiple of 7, and the total number of coins collected by B in 3 weeks is a multiple of 24, then the minimum possible number of coins collected by A in one week is

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