Real Numbers — NCERT Solutions

Concept: Prime factorisation — divide the number successively by the smallest prime factor until the quotient is 1.

(i) 140
$$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$$
$$\boxed{140 = 2^2 \times 5 \times 7}$$

(ii) 156
$$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$$
$$\boxed{156 = 2^2 \times 3 \times 13}$$

(iii) 3825
$$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3 \times 3 \times 5 \times 85 = 3 \times 3 \times 5 \times 5 \times 17$$
$$\boxed{3825 = 3^2 \times 5^2 \times 17}$$

(iv) 5005
$$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$$
$$\boxed{5005 = 5 \times 7 \times 11 \times 13}$$

(v) 7429
$$7429 = 17 \times 437 = 17 \times 19 \times 23$$
$$\boxed{7429 = 17 \times 19 \times 23}$$

Concept: Express each number as a product of prime factors. HCF = product of the smallest powers of common prime factors. LCM = product of the greatest powers of all prime factors. Verification: LCM × HCF = product of the two numbers.

(i) 26 and 91

Prime factorisations:
$$26 = 2 \times 13, \quad 91 = 7 \times 13$$

Common prime factor: $13$
$$\text{HCF}(26, 91) = 13$$
$$\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$$

Verification:
$$\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$$
$$26 \times 91 = 2366 \quad \checkmark$$

(ii) 510 and 92

Prime factorisations:
$$510 = 2 \times 255 = 2 \times 3 \times 85 = 2 \times 3 \times 5 \times 17$$
$$92 = 2 \times 46 = 2 \times 2 \times 23 = 2^2 \times 23$$

Common prime factor: $2$ (smallest power $= 2^1$)
$$\text{HCF}(510, 92) = 2$$
$$\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$$

Verification:
$$\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920$$
$$510 \times 92 = 46920 \quad \checkmark$$

(iii) 336 and 54

Prime factorisations:
$$336 = 2 \times 168 = 2^4 \times 3 \times 7$$
$$54 = 2 \times 27 = 2 \times 3^3$$

Common prime factors: $2$ and $3$ (smallest powers $2^1$ and $3^1$)
$$\text{HCF}(336, 54) = 2^1 \times 3^1 = 6$$
$$\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$$

Verification:
$$\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144$$
$$336 \times 54 = 18144 \quad \checkmark$$

Concept: HCF = product of smallest powers of common prime factors; LCM = product of greatest powers of all prime factors present.

(i) 12, 15 and 21

Prime factorisations:
$$12 = 2^2 \times 3, \quad 15 = 3 \times 5, \quad 21 = 3 \times 7$$

Common prime factor to all three: $3$ (smallest power $= 3^1$)
$$\text{HCF}(12, 15, 21) = 3$$

Greatest powers of all prime factors: $2^2,\ 3^1,\ 5^1,\ 7^1$
$$\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$$

(ii) 17, 23 and 29

All three numbers are prime, so they have no common factor other than 1.
$$17 = 17, \quad 23 = 23, \quad 29 = 29$$
$$\text{HCF}(17, 23, 29) = 1$$
$$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$$

(iii) 8, 9 and 25

Prime factorisations:
$$8 = 2^3, \quad 9 = 3^2, \quad 25 = 5^2$$

There is no prime factor common to all three numbers.
$$\text{HCF}(8, 9, 25) = 1$$

Greatest powers of all prime factors: $2^3,\ 3^2,\ 5^2$
$$\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800$$

Given: $\text{HCF}(306, 657) = 9$

Formula used:
$$\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}$$

Calculation:
$$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$$
$$= \frac{201042}{9} = 22338$$

$$\boxed{\text{LCM}(306, 657) = 22338}$$

Concept: A number ends with the digit 0 if and only if it has both 2 and 5 as prime factors (i.e., 10 is a factor of the number).

Prime factorisation of $6^n$:
$$6^n = (2 \times 3)^n = 2^n \times 3^n$$

The only prime factors of $6^n$ are 2 and 3.

For $6^n$ to end in 0, it must be divisible by 10, which requires 5 as a prime factor. But 5 does not appear in the prime factorisation of $6^n$ for any natural number $n$.

By the Fundamental Theorem of Arithmetic, the prime factorisation of a number is unique. Since 5 is never a factor of $6^n$, it can never be divisible by 10.

Conclusion: $6^n$ cannot end with the digit 0 for any natural number $n$.