Real Numbers
Mathematics Class Tenth
NCERT Exercise 1.2
The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic states that Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. And it says that given any composite number it can be factorised as a product of prime numbers in a unique way, except for the order in which the primes occur. This 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.
The prime factorization of a natural number is unique, except for the order of its factors.
In general, given a composite number, x, we factorise 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 the powers of primes.
According to The Fundamental Theorem of Arithmetic,
HCF = Product of the smallest power of each common prime factor in the numbers.
And LCM = Product of the greatest power of each prime factor, involved in the numbers.
Thus,
For any two positive integers a and b
HCF (a, b) × LCM (a, b) = a × b.
This result is used to find the LCM of two positive integers if we have already found the HCF of the two positive integers.
NCERT Exercise 1.2 solution class ten mathematics
Question (1) Express each number as a product of its prime factors:
(i) 140
Solution:
140 = 2 × 2 × 5 × 7
=22 × 5 × 7 Answer
(ii) 156
Solution:
156 = 2 × 2 × 3 × 13
= 22 × 2 × 13 Answer
(iii) 3825
Solution:
3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17 Answer
(iv) 5005
Solution:
5005 = 5 × 7 × 11 × 13 Answer
(v) 7429
Solution:
7429 = 17 × 19 × 23 Answer
Question (2) 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
Solution:
Prime factor of 26 = 2 × 13
Prime factor of 91 = 7 × 13
Thus, HCF = common in prime factors = 13
And LCM = 2 × 7 × 13 = 182
Now, product of given two numbers = 26 × 91 = 2366
And, HCF × LCM = 13 × 182 = 2366
Thus, the product of the given two numbers = LCM × HCF of the given two numbers. Proved
(ii) 510 and 92
Solution:
Prime factor of 510 = 2 × 3 × 17
And prime factor of 92 = 2 × 2 × 23
Therefore, HCF = 2
And LCM = 2 × 2 × 3 × 5 × 17 × 23
=23460
Now, product of two numbers = 510 × 92 = 46920
And HCF × LCM of given two numbers
= 2 × 23460 = 46920
Thus, HCF × LCM = product of two given numbers proved
(iii) 336 and 54
Solution:
Prime factors of 336
=2 × 2 × 2 × 2 × 3 × 7
=24 × 3 × 7
Prime factor of 54
=2 × 3 × 3 × 3
=2 × 33
Thus, HCF = 2 × 3 = 6
And LCM
= 24 × 33 × 7 = 3024
Now the product of the given two numbers =336 × 54 = 18144
And HCF × LCM = 6 × 3024 = 18144
Thus, HCF × LCM = product of two given numbers proved
Question (3) Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
Solution:
Prime factor of 12 = 2 × 2 × 3
= 22 × 3
Prime factor of 15 = 3 × 5
Prime factor of 21 = 3 × 7
Thus, HCF = 3
And LCM = 22 × 3 × 5 × 7 = 420
Thus, HCF = 3 and LCM = 420 Answer
(ii) 17, 23 and 29
Solution:
Prime factor of 17 = 17 × 1
Prime factor of 23 = 23 × 1
Prime factor of 29 = 1 × 29
Thus, HCF = 1
And LCM = 17 × 23 × 29 = 11339
Thus, HCF = 1 and LCM = 11339 Answer
(iii) 8, 9 and 25
Solution:
Prime factor of 8 = 2 × 2 × 2 = 23
Prime factor of 9 = 3 × 3 = 32
Prime factor of 25 = 5 × 5 = 52
Thus, HCF = 1
And LCM = 23 × 32 × 52 = 1800
Thus, HCF = 1 and LCM = 1800 Answer
Question (4) Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given, HCF (306, 657) = 9
∴ LCM = ?
Now, we know that, LCM × HCF = product of two numbers
∴ LCM × HCF = 306 × 657
=> LCM × 9 = 306 × 657
=> LCM = 306 34 × 657/9
=> LCM = 34 × 657
=> LCM = 22338 Answer
Question (5) Check whether 6n can end with the digit 0 for any natural number n.
Solution:
If any number ends with the digit 0, it will be divisible by 10, 5, and/or 2
Such as, 10, which is ends with 0, is divisible by 10, 5 and 2 because prime factor of 10 = 2 × 5
Now, prime factors of 6n = (2 × 3)n
As 5 is not in the prime factorization of 6n
Thus, for any value of n, 6n will not be divisible by 5.
Thus, 6n cannot end with the digit 0 for any natural number n. Proved
Question (6) Explain why 7 × 11 × 13 + 3 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Numbers can be categorized into two types. These are Prime numbers and Composite numbers.
Numbers which has only two factors, i.e. number itself and one (1) are called prime numbers. And numbers that have factors other than 1 and themselves are called composite numbers.
Thus, as given in the question
One
7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 13 × 6
Since, the given expression has 13 and 6 as its factors, thus it is a composite number.
Two
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 ( 7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × 1009
Since, the given expression has 5 and 1009 as factors, thus it is a composite number.
Thus, given expressions are composite numbers.
Question (7) 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 straight point?
Solution: It is clear that Ravi takes lesser time than Sonia to complete one round of sports field. Thus, when they start at the same point and in the same direction, LCM of their timing to complete one round will give the time for their meeting point.
Now,
Prime factor of 18 = 2 × 3 × 3
And Prime factor of 12 = 2 × 2 × 3
Thus, LCM of 18 and 12
= 2 × 2 × 3 × 3 = 36
Thus, Sonia and Ravi will meet together after 36 minutes of starting. Answer
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