NCERT Class 10 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2; 4 = 2 × 2; 253 = 11 × 23; and so on. Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying some prime numbers? Let us study.

Take any collection of prime numbers; say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many). Let us list a few:

7 × 11 × 23 = 1771
3 × 7 × 11 × 23 = 5313
2 × 3 × 7 × 11 × 23 = 10626
2^3 × 3 × 7^3 = 8232
2^2 × 3 × 7 × 11 × 23 = 21252
and so on.

Now, let us suppose your collection of prime numbers includes all the possible prime numbers. What is your guess about the size of this collection? Does it contain only a finite number of integers, or infinitely many? In fact, there are infinitely many prime numbers. So, if we combine all these primes in all possible ways, we will get an infinite collection of numbers, all the primes and all possible products of primes.

The question is can we produce all the composite numbers this way?
What do you think?
Do you think that there may be a composite number which is not the product of powers of prime numbers?

Before we answer this, let us factorize positive integers, that is, do the opposite of what we have done so far.

We are going to use the factor tree with which you are all familiar. Let us take some large number, say, 32760, and factorize it as shown:

32760
=2×16380
=2×2×8190
=2×2×2×4095
=2×2×2×3×1365
= 2×2×2×3×3×455
= 2×2×2×3×3×5×91
= 2×2×2×3×3×5× 7×13

So, we have factorized 32760 as 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 as a product of primes,
i.e., 32760 = 2^3 × 3^2 × 5 × 7 × 13 as a product of powers of primes.

Let us try another number, say, 123456789.
This can be written as 3^2 × 3803 × 3607. Of course, you have to check that 3803 and 3607 are primes!

(Try it out for several other natural numbers yourself.)

This leads us to a conjecture that every composite number can be written as the product of powers of primes.

In fact, this statement is true, and is called the Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers.

Let us now formally state this theorem.

Theorem 1.2 (Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

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Start of Blue Box [An equivalent version of Theorem 1.2 was probably first recorded as Proposition 14 of Book 9 in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic. However, the first correct proof was given by Carl Friedrich Gauss in his Disquisitiones Arithmeticae.

Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science.

Carl Friedrich Gauss (1777 - 1855)]

End of Blue Box

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The Fundamental Theorem of Arithmetic says that every composite number can be factorized as a product of primes. Actually, it says more. It says that given any composite number it can be factorized as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur.

So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written. This fact is also stated in the following form:

The prime factorization of a natural number is unique, except for the order of its factors.

In general, given a composite number x, we factorize it as x = p1p2 ... pn, where p1, p2,..., pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤ . . . ≤ pn. If we combine the same primes, we will get powers of primes. For example,

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2^3 × 3^2 × 5 × 7 × 13

Once we have decided that the order will be ascending, and then the way the number is factorized, is unique.

Applications of Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. Let us look at some examples.

Example 5

Consider the numbers 4^n, where n is a natural number. Check whether there is any value of n for which 4^n ends with the digit zero.

Solution:

If the number 4^n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorization of 4^n would contain the prime 5. This is not possible because 4^n = (2) ^ (2n); so the only prime in the factorization of 4^n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4^n. So, there is no natural number n for which 4^n ends with the digit zero.

You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realizing it! This method is also called the prime factorization method. Let us recall this method through an example.

Example 6

Find the LCM and HCF of 6 and 20 by the prime factorization method.

Solution:

We have:

\[6\ =\ 2^1\ ×\ 3^1\]

and

\[20\ =\ 2\ ×\ 2\ ×\ 5\]

\[=\ 2^2\ ×\ 5^1.\]

You can find
HCF (6, 20) = 2 and
LCM(6, 20) = 2 × 2 × 3 × 5 = 60,
as done in your earlier classes.

Note that HCF (6, 20) = 2^1 = Product of the smallest power of each common prime factor in the numbers.

LCM (6, 20) = 2^2 × 3^1 × 5^1 = Product of the greatest power of each prime factor, involved in the numbers.

From the example above, you might have noticed that HCF (6, 20) × LCM (6, 20) = 6 × 20.

In fact, we can verify that for any two positive integers α and β, HCF (α, β) × LCM (α, β) = α × β.

We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.

Example 7

Find the HCF of 96 and 404 by the prime factorization method. Hence, find their LCM.

Solution:

The prime factorization of 96 and 404 gives:

96 = 2^5 × 3,and
404 = 2^2 × 101

Therefore, the HCF of these two integers is 2^2 = 4.

Also, LCM (96, 404) = (96×404)/HCF (96, 404) = 96×404 /4 = 9696

Example 8

Find the HCF and LCM of 6, 72 and 120, using the prime factorization method.

Solution:

We have :
6 = 2 × 3,
72 = 2^3 × 3^2,
120 = 2^3 × 3 × 5
Here, 2^1 and 3^1 are the smallest powers of the common factors 2 and 3, respectively.

So, HCF (6, 72, 120) =2^1 × 3^1 = 2 × 3 = 6

2^3, 3^2 and 5^1 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM (6, 72, 120) = 2^3 × 3^2 × 5^1 = 360

Remarks: Notice, (6 × 72 × 120) ≠ [HCF (6, 72, 120) × LCM (6, 72, 120)].
So, the product of three numbers is not equal to the product of their HCF and LCM.

EXERCISE 1.2

Q1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

A1. The prime factors of the numbers are:
(i) 2^2 × 5 × 7
(ii) 2^2 × 3 × 13
(iii) 3^2 × 5^2 × 17
(iv) 5 × 7 × 11 × 13
(v) 17 × 19 × 23

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

A2.
(i) LCM = 182; HCF = 13
(ii) LCM = 23460; HCF = 2
(iii) LCM = 3024; HCF = 6

Q3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25

A3.
(i) LCM = 420; HCF = 3
(ii) LCM = 11339; HCF = 1
(iii) LCM = 1800; HCF = 1

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

A4. 22338

Q5. Check whether 6^n can end with the digit 0 for any natural number n.

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

A7. 36 minutes