application of fundamental theorem of maths

Application Of The Fundamental Theorem Of Arithmetic To Find The HCF And LCM Of Numbers

All composite numbers can be written as the product of two or more prime numbers. For example, 20 can be written as 2² × 5; 54 can be written as 2 × 3³, and so on.

Note that if we do not consider the way in which the prime factors are written, then we can prime factorise every number in only one way. This applies to other numbers as well.

This leads to the fundamental theorem of arithmetic, which states that:

Every composite number can be factorised as the product of certain prime numbers and this factorisation is unique, although the order in which the prime factors occur may be changed.

Even though we did not notice it before, whenever we prime factorise a number, we use the fundamental theorem of arithmetic to do so.

For example, the prime factorisation of 980 is represented as

980 = 2 × 490

= 2 × 2 × 245

= 2 × 2 × 5 × 49

= 2 × 2 × 5 × 7 × 7

= 2² × 5¹ ´ 7²

Hence, 2² × 5¹ ´ 7² is the prime factorisation of 980; and 2, 5, and 7 are its prime factors.

By applying the fundamental theorem of arithmetic to the prime factorized numbers, we can also find their HCF and LCM.

This is known as the prime factorisation method, which states that:

For any two positive integers a and b: HCF (a, b) = Product of the smallest power of each common prime factor in the prime factorisation of numbers LCM (a, b) = Product of the greatest power of each prime factor in the prime factorisation of numbers

Let us understand this method with the help of some examples.

Example 1:

Find the LCM and the HCF of 432 and 676 using the prime factorization method.

Solution:

We can write these numbers as

432 = 2⁴ × 3³

676 = 2² × 13²

To calculate the HCF

We observe that the only common prime factor is 2 and the smallest power of this prime factor is also 2.

Thus, HCF (432, 676) = 2² = 4

To calculate the LCM

We observe that the prime factors of 432 and 676 are 2, 3, and 13. The greatest powers of these factors are 4, 3, and 2 respectively.

LCM is the product of the greatest power of each prime factor.

Thus, LCM (432, 676) = 2⁴ × 3³ × 13² = 73008

Example 2:

Find the HCF and the LCM of 28, 42, and 64 using the prime factorization method.

Solution:

We can write these numbers as

28 = 2² × 7¹

42 = 2 × 3¹ × 7¹

64 = 2⁶

HCF is the product of the smallest power of each common prime factor.

Here, the only common prime factor is 2 and its power is 1.

Thus, HCF (28, 42, 64) = 2¹ = 2

LCM is the product of the greatest power of each prime factor.

Thus, LCM (28, 42, 64) = 2⁶ × 3¹ × 7¹ = 1344

Example 3:

Find the HCF and the LCM of 1080 and 900 using the prime factorization and show that HCF ´ LCM = Product of two numbers.

Solution:

1080 = 2³ × 3³ × 5

900 = 2² × 3² × 5²

Hence, HCF (1080, 900) = 2² × 3² × 5 = 180

LCM (1080, 900) = 2³ × 3³ × 5² = 5400

HCF × LCM = 180 × 5400 = 972000

Product of numbers = 1080 × 900 = 972000

Hence, HCF × LCM = Product of two numbers

Example 4:

The HCF of 273 and another number is 7, while their LCM is 3003. Find the other number.

Solution:

Let the first number (a) be 273 and the second number be b.

It is given that HCF (a, b) = 7 and LCM (a, b) = 3003.

We know that HCFLCM = Product of two numbers.

⇒ HCF (a, b) × LCM (a, b) = a × b

⇒ 7 × 3003 = 273 × b

⇒ b = 77

Hence, the other number is 77.

Example 5:

Anurag takes 6 minutes to complete one round of jogging around the circular track of a park, while Twinkle takes 8 minutes to do the same. If both of them start jogging at the same time from the same point, then how much time will they take before they meet at the point from which they started?

Solution:

Since Anurag and Twinkle take 6 minutes and 8 minutes respectively to complete one round of the circular track, the time after which they will meet at the starting point will be the lowest multiple of 6 and 8, i.e., their LCM.

6 = 2 × 3

8 = 2³

∴ LCM (6, 8) = 2³ × 3 = 24

Thus, they will meet at the starting point after 24 minutes.

Example 6:

There are 120 students in a class. When the students were arranged according to their roll numbers, it was observed that every second student got distinction in Mathematics, every third student got distinction in Science, and every fifth student got distinction in English. How many students got distinction in all three subjects?

Solution:

As every second, third, and fifth student got distinction in Math, Science, and English respectively, the roll numbers of the students who got distinction in all three subjects will be equal to the multiples of the LCM of 2, 3, and 5.

LCM (2, 3, 5) = 2 × 3 × 5 = 30

Thus, every 30^th student got distinction in all three subjects.

Thus, a total of  students got distinction in all three subjects.