CBSE Class 10 Real Numbers Sure Shot Questions Set H

Real Numbers

• Rational no.
  • If remainder = 0: Terminating & Non repeating. Eg : \( \frac{18}{5} = 3.6 \)
  • If remainder \(\neq\) 0 & rem. = dividend: Non terminating & repeating (recurring). Eg : \( \frac{1}{3} = 0.33.... = 0.\overline{3} \)
• Irrational no.
  • If remainder \(\neq\) 0 & rem. \(\neq\) any dividend: Non terminating non repeating. Eg : \( 0.671234….. \), \( 1.343634003908…... \)

Question. Insert a rational and an irrational number between 2 and 3.
Answer: If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then \( \sqrt{ab} \) is an irrational number lying between a and b. Also, if a,b are rational numbers, then \( \frac{a + b}{2} \) is a rational number between them.
\(\therefore\) A rational number between 2 and 3 is \( \frac{2 + 3}{2} = 2.5 \)
An irrational number between 2 and 3 is \( \sqrt{2 \times 3} = \sqrt{6} \)

Question. Find two irrational numbers between 2 and 2.5.
Answer: If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then \( \sqrt{ab} \) is an irrational number lying between a and b.
\(\therefore\) Irrational number between 2 and 2.5 is \( \sqrt{2 \times 2.5} = \sqrt{5} \)
Similarly, irrational number between 2 and \( \sqrt{5} \) is \( \sqrt{2 \times \sqrt{5}} \)
So, required numbers are \( \sqrt{5} \) and \( \sqrt{2 \times \sqrt{5}} \).

Question. Find two irrational numbers lying between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: We know that, if a and b are two distinct positive irrational numbers, then \( \sqrt{ab} \) is an irrational number lying between a and b.
\(\therefore\) Irrational number between \( \sqrt{2} \) and \( \sqrt{3} \) is \( \sqrt{\sqrt{2} \times \sqrt{3}} = \sqrt{\sqrt{6}} = 6^{1/4} \)
Irrational number between \( \sqrt{2} \) and \( 6^{1/4} \) is \( \sqrt{\sqrt{2} \times 6^{1/4}} = 2^{1/4} \times 6^{1/8} \).
Hence required irrational number are \( 6^{1/4} \) and \( 2^{1/4} \times 6^{1/8} \).

Question. Find two irrational numbers between 0.12 and 0.13.
Answer: Let a = 0.12 and b = 0.13. Clearly, a and b are rational numbers such that a < b.
We observe that the number a and b have a 1 in the first place of decimal. But in the second place of decimal a has a 2 and b has 3. So, we consider the numbers
c = 0.1201001000100001 ......
and, d = 0.12101001000100001.......
Clearly, c and d are irrational numbers such that a < c < d < b.

Theorem : Let p be a prime number. If p divides \( a^2 \), then p divides a, where a is a positive integer.
Proof : Let the prime factorisation of a be as follows :
a = \( p_1 p_2 \dots p_n \), where \( p_1, p_2, \dots p_n \) are primes, not necessarily distinct.
Therefore,
\( a^2 = (p_1 p_2 \dots p_n) (p_1 p_2 \dots p_n) = p_1^2 p_2^2 \dots p_n^2 \).
Now, we are given that p divides \( a^2 \). Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of \( a^2 \). However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of \( a^2 \) are \( p_1, p_2, \dots p_n \). So p is one of \( p_1, p_2, \dots, p_n \). Now, since a = \( p_1 p_2 \dots p_n \), p divides a.

Question. Prove that (i) \( \sqrt{2} \) is irrational number (ii) \( \sqrt{3} \) is irrational number. Similarly \( \sqrt{5}, \sqrt{7}, \sqrt{11} \dots \) are irrational numbers.
Answer:
(i) Let us assume, to the contrary, that \( \sqrt{2} \) is rational. So, we can find integers r and s (\( s \neq 0 \)) such that \( \sqrt{2} = \frac{r}{s} \). Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get \( \sqrt{2} = \frac{a}{b} \), where a and b are coprime.
So, \( b\sqrt{2} = a \). Squaring on both sides and rearranging, we get \( 2b^2 = a^2 \). Therefore, 2 divides \( a^2 \). Now, by Theorem it following that 2 divides a.
So, we can write a = 2c for some integer c. Substituting for a, we get \( 2b^2 = 4c^2 \), that is, \( b^2 = 2c^2 \). This means that 2 divides \( b^2 \), and so 2 divides b (again using Theorem with p = 2). Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that \( \sqrt{2} \) is rational. So, we conclude that \( \sqrt{2} \) is irrational.
(ii) Let us assume, to contrary, that \( \sqrt{3} \) is rational. That is, we can find integers a and b (\( b \neq 0 \)) such that \( \sqrt{3} = \frac{a}{b} \). Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, \( b\sqrt{3} = a \). Squaring on both sides, and rearranging, we get \( 3b^2 = a^2 \). Therefore, \( a^2 \) is divisible by 3, and by Theorem, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get \( 3b^2 = 9c^2 \), that is, \( b^2 = 3c^2 \). This means that \( b^2 \) is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3). Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime. This contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that \( \sqrt{3} \) is rational. So, we conclude that \( \sqrt{3} \) is irrational.

Question. Prove that \( 7 - \sqrt{3} \) is irrational.
Answer:
Method I: Let \( 7 - \sqrt{3} \) is rational number. \(\therefore 7 - \sqrt{3} = \frac{p}{q} \) (p, q are integers, \( q \neq 0 \)).
\(\therefore 7 - \frac{p}{q} = \sqrt{3} \Rightarrow \frac{7q - p}{q} = \sqrt{3} \).
Here p, q are integers \(\therefore \frac{7q - p}{q} \) is also integer. \(\therefore\) LHS = \( \sqrt{3} \) is also integer but this is contradiction that \( \sqrt{3} \) is irrational so our assumption is wrong that \( (7 - \sqrt{3}) \) is rational. \(\therefore 7 - \sqrt{3} \) is irrational proved.
Method II: Let \( 7 - \sqrt{3} \) is rational. we know sum or difference of two rationals is also rational.
\(\therefore 7 - (7 - \sqrt{3}) = \sqrt{3} = \text{rational} \). But this is contradiction that \( \sqrt{3} \) is irrational.
\(\therefore (7 - \sqrt{3}) \) is irrational proved.

Question. Prove that : (i) \( \frac{\sqrt{5}}{3} \) (ii) \( 2\sqrt{7} \) are irrationals.
Answer:
(i) Let \( \frac{\sqrt{5}}{3} \) is rational. \(\therefore \left( \frac{\sqrt{5}}{3} \right) \times 3 = \sqrt{5} \) is rational (\(\because\) product of two rationals is also rational). But this is contradiction that \( \sqrt{5} \) is irrational. \(\therefore \frac{\sqrt{5}}{3} \) is irrational proved.
(ii) Let \( 2\sqrt{7} \) is rational. \(\therefore (2\sqrt{7}) \times \frac{1}{2} = \sqrt{7} \) (\(\because\) division of two rational no. is also rational). \(\therefore \sqrt{7} \) is rational. But this is contradiction that \( \sqrt{7} \) is irrational. \(\therefore 2\sqrt{7} \) is irrational proved.

Theorem 1 :

Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form \( \frac{p}{q} \), where p and q are coprime and the prime factorization of q is of the form \( 2^n 5^m \), where n, m are non-negative integers.
(A) Numbers are terminating (remainder = zero)
Eg : \( \frac{32}{125} = \frac{2^5}{5^3} = \frac{2^5 \times 2^3}{(2 \times 5)^3} = \frac{256}{10^3} = 0.256 \)
Eg : \( \frac{9}{25} = \frac{9 \times 2^2}{5^2 \times 2^2} = \frac{36}{(2 \times 5)^2} = \frac{36}{(10)^2} = 0.36 \)
So we can convert a rational number of the form \( \frac{p}{q} \), where q is of the form \( 2^n 5^m \) to an equivalent rational number of the form \( \frac{a}{b} \) where b is a power of 10. These are terminates.

Theorem 2 :

Let x = \( \frac{p}{q} \) be a rational number, such that the prime factorization of q is of the form \( 2^n 5^m \), where n, m are non-negative integers. Then x has a decimal expansion which terminates.
(B) Non terminating & recurring
Eg : \( \frac{1}{7} = 0.\overline{142857} = 0.142857142857..... \)
Since denominator 7 is not of the form \( 2^n 5^m \) so we zero (0) will not show up as a remainder.

Theorem 3 :

Let x = \( \frac{p}{q} \) be a rational number, such that the prime factorization of q is not of the form \( 2^n 5^m \), where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).
From the discussion above, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.

Question. From given rational numbers check terminating or non terminating.
(1) \( \frac{13}{3125} = \frac{13}{(5)^5} = \frac{13 \times 2^5}{2^5 \times 5^5} = \frac{(13 \times 32)}{(10)^5} \) = terminating.
(2) \( \frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{(2 \times 5)^3} = \frac{17 \times 125}{(10)^3} \) = terminating.
(3) \( \frac{64}{455} = \frac{2^6}{5 \times 7 \times 13} \) (\(\because\) we can not remove 7 & 13 from denominator) non-terminating repeating (\(\because\) no. is rational \(\therefore\) it is always repeating or recurring).
(4) \( \frac{15}{1600} = \frac{3 \times 5}{2^4 \times 10^2} = \frac{3 \times 5^5}{(2 \times 5)^4 \times 10^2} = \frac{3 \times 5^5}{10^6} \) = terminating.
(5) \( \frac{29}{343} = \frac{29}{(7)^3} \) = non terminating.
(6) \( \frac{23}{2^3 5^2} = \frac{23 \times 5}{(2 \times 5)^3} = \frac{23 \times 5}{(10)^3} \) = terminating.
(7) \( \frac{129}{2^5 \times 5^7 \times 7^5} = \frac{3 \times 43 \times 2^2}{(2 \times 5)^7 \times 7^5} \) = non terminating (\(\because\) 7 cannot remove from denominator).
(8) \( \frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5} = \frac{2 \times 2}{10} \) = terminating.
(9) \( \frac{35}{50} = \frac{35 \times 2}{100} \) = terminating.
(10) \( \frac{77}{210} = \frac{7 \times 11}{7 \times 30} = \frac{11}{2 \times 5 \times 3} \) = non terminating.

EUCLID’S DIVISION LEMMA OR EUCLID’S DIVISION ALGORITHM

For any two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where \( 0 \le r < b \).
For Example:
(i) Consider number 23 and 5, then: 23 = 5 \(\times\) 4 + 3. Comparing with a = bq + r; we get: a = 23, b = 5, q = 4, r = 3 and \( 0 \le r < b \) (as \( 0 \le 3 < 5 \)).
(ii) Consider positive integers 18 and 4. 18 = 4 \(\times\) 4 + 2 \(\Rightarrow\) For 18 (= a) and 4(= b) we have q = 4, r = 2 and \( 0 \le r < b \).
In the relation a = bq + r, where \( 0 \le r < b \) is nothing but a statement of the long division of number a by number b in which q is the quotient obtained and r is the remainder.
Thus, dividend = divisor \(\times\) quotient + remainder \(\Rightarrow\) a = bq + r.

H.C.F. (Highest Common Factor)

The H.C.F. of two or more positive integers is the largest positive integer that divides each given positive number completely. i.e., if positive integer d divides two positive integers a and b then the H.C.F. of a and b is d.
For Example:
(i) 14 is the largest positive integer that divides 28 and 70 completely; therefore H.C.F. of 28 and 70 is 14.
(ii) H.C.F. of 75, 125 and 200 is 25 as 25 divides each of 75, 125 and 200 completely and so on.

Using Euclid’s Division Lemma For Finding H.C.F.

Consider positive integers 418 and 33.
Step-1: Taking bigger number (418) as a and smaller number (33) as b express the numbers as a = bq + r \(\Rightarrow\) 418 = 33 \(\times\) 12 + 22.
Step-2: Now taking the divisor 33 and remainder 22; apply the Euclid’s division algorithm to get: 33 = 22 \(\times\) 1 + 11 [Expressing as a = bq + r].
Step-3: Again with new divisor 22 and new remainder 11; apply the Euclid’s division algorithm to get: 22 = 11 \(\times\) 2 + 0.
Step-4: Since, the remainder = 0 so we cannot proceed further.
Step-5: The last divisor is 11 and we say H.C.F. of 418 and 33 = 11.

Verification :
(i) Using factor method: \(\therefore\) Factors of 418 = 1, 2, 11, 19, 22, 38, 209 and 418 and, Factor of 33 = 1, 3, 11 and 33. Common factors = 1 and 11 \(\Rightarrow\) Highest common factor = 11 i.e., H.C.F. = 11.
(ii) Using prime factor method: Prime factors of 418 = 2, 11 and 19. Prime factors of 33 = 3 and 11. \(\therefore\) H.C.F. = Product of all common prime factors = 11. For any two positive integers a and b which can be expressed as a = bq + r, where \( 0 \le r < b \), the, H.C.F. of (a, b) = H.C.F. of (q, r) and so on. For number 418 and 33: 418 = 33 \(\times\) 12 + 22; 33 = 22 \(\times\) 1 + 11 and 22 = 11 \(\times\) 2 + 0 \(\Rightarrow\) H.C.F. of (418, 33) = H.C.F. of (33, 22) = H.C.F. of (22, 11) = 11.

Question. Using Euclid’s division algorithm, find the H.C.F. of [NCERT] (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255.
Answer:
(i) Starting with the larger number i.e., 225, we get: 225 = 135 \(\times\) 1 + 90. Now taking divisor 135 and remainder 90, we get 135 = 90 \(\times\) 1 + 45. Further taking divisor 90 and remainder 45, we get 90 = 45 \(\times\) 2 + 0. \(\therefore\) Required H.C.F. = 45.
(ii) Starting with larger number 38220, we get: 38220 = 196 \(\times\) 195 + 0. Since, the remainder is 0 \(\Rightarrow\) H.C.F. = 196.
(iii) Given number are 867 and 255 \(\Rightarrow\) 867 = 255 \(\times\) 3 + 102 (Step-1); 255 = 102 \(\times\) 2 + 51 (Step-2); 102 = 51 \(\times\) 2 + 0 (Step-3). \(\Rightarrow\) H.C.F. = 51.

Question. Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer.
Answer: According to Euclid’s division lemma, if a and b are two positive integers such that a is greater than b; then these two integers can be expressed as a = bq + r; where \( 0 \le r < b \).
Now consider b = 2; then a = bq + r will reduce to a = 2q + r; where \( 0 \le r < 2 \), i.e., r = 0 or r = 1.
If r = 0, a = 2q + r \(\Rightarrow\) a = 2q i.e., a is even. and, if r = 1, a = 2q + r \(\Rightarrow\) a = 2q + 1 i.e., a is odd; as if the integer is not even; it will be odd. Since, a is taken to be any positive integer so it is applicable to the every positive integer that when it can be expressed as a = 2q \(\therefore\) a is even and when it can expressed as a = 2q + 1; a is odd.

Question. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Answer: Let a and b be two positive integers in which a is greater than b. According to Euclid’s division algorithm; a and b can be expressed as a = bq + r, where q is quotient and r is remainder and \( 0 \le r < b \).
Taking b = 4, we get: a = 4q + r, where \( 0 \le r < 4 \) i.e., r = 0, 1, 2 or 3.
r = 0 \(\Rightarrow\) a = 4q, which is divisible by 2 and so is even.
r = 1 \(\Rightarrow\) a = 4q + 1, which is not divisible by 2 and so is odd.
r = 2 \(\Rightarrow\) a = 4q + 2, which is divisible by 2 and so is even.
and r = 3 \(\Rightarrow\) a = 4q + 3, which is not divisible by 2 and so is odd.
\(\therefore\) Any positive odd integer is of the form 4q + 1 or 4q + 3; where q is an integer.

Question. Show that one and only one out of n; n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Answer: Consider any two positive integers a and b such that a is greater than b, then according to Euclid’s division algorithm: a = bq + r; where q and r are positive integers and \( 0 \le r < b \). Let a = n and b = 3, then a = bq + r \(\Rightarrow\) n = 3q + r; where \( 0 \le r < 3 \).
r = 0 \(\Rightarrow\) n = 3q + 0 = 3q. r = 1 \(\Rightarrow\) n = 3q + 1 and r = 2 \(\Rightarrow\) n = 3q + 2.
If n = 3q; n is divisible by 3.
If n = 3q + 1; then n + 2 = 3q + 1 + 2 = 3q + 3; which is divisible by 3 \(\Rightarrow\) n + 2 is divisible by 3.
If n = 3q + 2; then n + 4 = 3q + 2 + 4 = 3q + 6; which is divisible by 3 \(\Rightarrow\) n + 4 is divisible by 3.
Hence, if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3.

Question. Show that any positive integer which is of the form 6q + 1 or 6q + 3 or 6q + 5 is odd, where q is some integer.
Answer: If a and b are two positive integers such that a is greater than b; then according to Euclid’s division algorithm; we have a = bq + r; where q and r are positive integers and \( 0 \le r < b \). Let b = 6, then a = bq + r \(\Rightarrow\) a = 6q + r; where \( 0 \le r < 6 \).
When r = 0 \(\Rightarrow\) a = 6q + 0 = 6q; which is even integer.
When r = 1 \(\Rightarrow\) a = 6q + 1; which is odd integer.
When r = 2 \(\Rightarrow\) a = 6q + 2 which is even.
When r = 3 \(\Rightarrow\) a = 6q + 3 which is odd.
When r = 4 \(\Rightarrow\) a = 6q + 4 which is even.
When r = 5 \(\Rightarrow\) a = 6q + 5 which is odd.
This verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number.

Question. Use Euclid’s Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer: Let a and b are two positive integers such that a is greater than b; then: a = bq + r; where q and r are also positive integers and \( 0 \le r < b \). Taking b = 3, we get: a = 3q + r; where \( 0 \le r < 3 \). \(\Rightarrow\) The value of positive integer a will be 3q + 0, 3q + 1 or 3q + 2 i.e., 3q, 3q + 1 or 3q + 2.
Now we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m.
\(\therefore\) Square of 3q = \( (3q)^2 = 9q^2 = 3(3q^2) = 3m \); where m is some integer.
Square of 3q + 1 = \( (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1 \) for some integer m.
Square of 3q + 2 = \( (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1 \) for some integer m.
\(\therefore\) The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question. Use Euclid’s Division Algorithm to show that the cube of any positive integer is either of the 9m, 9m + 1 or 9m + 8 for some integer m.
Answer: Let a and b be two positive integers such that a is greater than b; then: a = bq + r; where q and r are positive integers and \( 0 \le r < b \). Taking b = 3, we get: a = 3q + r; where \( 0 \le r < 3 \).
\(\Rightarrow\) Different values of integer a are 3q, 3q + 1 or 3q + 2.
Cube of 3q = \( (3q)^3 = 27q^3 = 9(3q^3) = 9m \); where m is some integer.
Cube of 3q + 1 = \( (3q + 1)^3 = (3q)^3 + 3(3q)^2 \times 1 + 3(3q) \times 1^2 + 1^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1 \); where m is some integer.
Cube of 3q + 2 = \( (3q + 2)^3 = (3q)^3 + 3(3q)^2 \times 2 + 3 \times 3q \times 2^2 + 2^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8 \); where m is some integer.
\(\therefore\) Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.

THE FUNDAMENTAL THEOREM OF ARITHMETIC

Statement : Every composite number can be decomposed as a product prime numbers in a unique way, except for the order in which the prime numbers occur.
For example :
(i) 30 = 2 \(\times\) 3 \(\times\) 5, 30 = 3 \(\times\) 2 \(\times\) 5, 30 = 2 \(\times\) 5 \(\times\) 3 and so on.
(ii) 432 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 3 = \( 2^4 \times 3^3 \) or 432 = \( 3^3 \times 2^4 \).
(iii) 12600 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 5 \(\times\) 5 \(\times\) 7 = \( 2^3 \times 3^2 \times 5^2 \times 7 \).
In general, a composite number is expressed as the product of its prime factors written in ascending order of their values. e.g., (i) 6615 = 3 \(\times\) 3 \(\times\) 3 \(\times\) 5 \(\times\) 7 \(\times\) 7 = \( 3^3 \times 5 \times 7^2 \). (ii) 532400 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 5 \(\times\) 5 \(\times\) 11 \(\times\) 11 \(\times\) 11.

Question. Consider the number \( 6^n \), where n is a natural number. Check whether there is any value of \( n \in N \) for which \( 6^n \) is divisible by 7.
Answer: Since, 6 = 2 \(\times\) 3; \( 6^n = 2^n \times 3^n \). \(\Rightarrow\) The prime factorisation of given number \( 6^n \). \(\Rightarrow 6^n \) is not divisible by 7.

Question. Consider the number \( 12^n \), where n is a natural number. Check whether there is any value of \( n \in N \) for which \( 12^n \) ends with the digit zero.
Answer: We know, if any number ends with the digit zero it is always divisible by 5. \(\Rightarrow\) If \( 12^n \) ends with the digit zero, it must be divisible by 5. This is possible only if prime factorisation of \( 12^n \) contains the prime number 5. Now, 12 = 2 \(\times\) 2 \(\times\) 3 = \( 2^2 \times 3 \). \(\Rightarrow 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n \). i.e., prime factorisation of \( 12^n \) does not contain the prime number 5. \(\Rightarrow\) There is no value of \( n \in N \) for which \( 12^n \) ends with the digit zero.

USING THE FACTOR TREE

Question. Find the prime factors of : (i) 540 (ii) 21252 (iii) 8232.
Answer:
(i) 540 divided by 2 gives 270; 270 divided by 2 gives 135; 135 divided by 3 gives 45; 45 divided by 3 gives 15; 15 divided by 3 gives 5. 5 is a prime number and so cannot be further divided. \(\therefore 540 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^2 \times 3^3 \times 5 \).
(ii) 21252 = 2 \(\times\) 10626 = 2 \(\times\) 2 \(\times\) 5313 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 1771 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 \(\times\) 253 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 \(\times\) 11 \(\times\) 23. \(\therefore 21252 = 2^2 \times 3 \times 11 \times 7 \times 23 \).
(iii) 8232 = 2 \(\times\) 4116 = 2 \(\times\) 2 \(\times\) 2058 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 1029 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 343 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 \(\times\) 49 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 7 \(\times\) 7 \(\times\) 7. \(\therefore 8232 = 2^3 \times 3 \times 7^3 \).

Question. Find the missing numbers a, b and c in the following factorisation: a \(\rightarrow\) 2, b; b \(\rightarrow\) 2, c; c \(\rightarrow\) 2, 17. Can you find the number on top without finding the other?
Answer: c = 17 \(\times\) 2 = 34; b = c \(\times\) 2 = 34 \(\times\) 2 = 68 and a = b \(\times\) 2 = 68 \(\times\) 2 = 136 i.e., a = 136, b = 68 and c = 34. Yes, we can find the number on top without finding the others. Reason: The given numbers 2, 2, 2 and 17 are the only prime factors of the number on top and so the number on top = 2 \(\times\) 2 \(\times\) 2 \(\times\) 17 = 136.

USING THE FUNDAMENTAL THEOREM OF ARITHMETIC TO FIND H.C.F. AND L.C.M.

Question. Find the L.C.M. and H.C.F. of the following pairs of integers by applying the Fundamental theorem of Arithmetic method i.e., using the prime factorisation method. (i) 26 and 91 (ii) 1296 and 2520 (iii) 17 and 25.
Answer:
(i) Since, 26 = 2 \(\times\) 13 and, 91 = 7 \(\times\) 13. \(\therefore\) L.C.M. = Product of each prime factor with highest powers = 2 \(\times\) 13 \(\times\) 7 = 182. H.C.F. = Product of common prime factors with lowest powers = 13.
(ii) Since, 1296 = \( 2^4 \times 3^4 \) and, 2520 = \( 2^3 \times 3^2 \times 5 \times 7 \). \(\therefore\) L.C.M. = \( 2^4 \times 3^4 \times 5 \times 7 = 45,360 \). H.C.F. = \( 2^3 \times 3^2 = 8 \times 9 = 72 \).
(iii) Since, 17 = 17 and, 25 = \( 5^2 \). \(\therefore\) L.C.M. = 17 \(\times\) 25 = 425. H.C.F. = 1, as given numbers do not have any common prime factor.

Notes on LCM and HCF:
For any two positive integers :
Their L.C.M. \(\times\) their H.C.F. = Product of the number
(i) L.C.M. = \( \frac{\text{Product of the numbers}}{\text{H.C.F.}} \)
(ii) H.C.F. = \( \frac{\text{Product of the numbers}}{\text{L.C.M.}} \)
(iii) One number = \( \frac{\text{L.C.M. \(\times\) H.C.F.}}{\text{Other number}} \)

Question. Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657).
Answer: H.C.F. (306, 657) = 9 means H.C.F. of 306 and 657 = 9. Required L.C.M. means required L.C.M. of 306 and 657. For any two positive integers; their L.C.M. = \( \frac{\text{Product of the numbers}}{\text{Their H.C.F.}} \) i.e., L.C.M. (306, 657) = \( \frac{306 \times 657}{9} = 22,338 \).

Question. Given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100).
Answer: L.C.M. (150, 100) = 300 \(\Rightarrow\) L.C.M. of 150 and 100 = 300. Since, the product of number 150 and 100 = 150 \(\times\) 100. And, we know : H.C.F. (150, 100) = \( \frac{\text{Product of 150 and 100}}{\text{L.C.M. (150, 100)}} = \frac{150 \times 100}{300} = 50 \).

Question. The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers.
Answer: Since, the product of two numbers = Their H.C.F. \(\times\) Their L.C.M. \(\Rightarrow\) One no. \(\times\) other no. = H.C.F. \(\times\) L.C.M. \(\Rightarrow\) Other no. = \( \frac{12 \times 240}{48} = 60 \).

Question. Explain why 7 \(\times\) 11 \(\times\) 13 + 13 and 7 \(\times\) 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 + 5 are composite numbers.
Answer: Since, 7 \(\times\) 11 \(\times\) 13 + 13 = 13 \(\times\) (7 \(\times\) 11 + 1) = 13 \(\times\) 78 = 13 \(\times\) 13 \(\times\) 3 \(\times\) 2; that is, the given number has more than two factors and it is a composite number. Similarly, 7 \(\times\) 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 + 5 = 5 \(\times\) (7 \(\times\) 6 \(\times\) 4 \(\times\) 3 + 1) = 5 \(\times\) 505 = 5 \(\times\) 5 \(\times\) 101 \(\Rightarrow\) The given no. is a composite number.