Real Numbers — CBSE Class 10 Maths Textbook Solutions

We factorise each number completely into primes using repeated division (the factor tree method).

(i) $140$140
$140 = 2\times70 = 2\times2\times35 = 2\times2\times5\times7$140 = 2×70 = 2×2×35 = 2×2×5×7
So $140 = 2^2\times5\times7$140 = 2^2×5×7.

(ii) $156$156
$156 = 2\times78 = 2\times2\times39 = 2\times2\times3\times13$156 = 2×78 = 2×2×39 = 2×2×3×13
So $156 = 2^2\times3\times13$156 = 2^2×3×13.

(iii) $3825$3825
$3825 = 3\times1275 = 3\times3\times425 = 3\times3\times5\times85 = 3\times3\times5\times5\times17$3825 = 3×1275 = 3×3×425 = 3×3×5×85 = 3×3×5×5×17
So $3825 = 3^2\times5^2\times17$3825 = 3^2×5^2×17.

(iv) $5005$5005
$5005 = 5\times1001 = 5\times7\times143 = 5\times7\times11\times13$5005 = 5×1001 = 5×7×143 = 5×7×11×13
So $5005 = 5\times7\times11\times13$5005 = 5×7×11×13.

(v) $7429$7429
$7429 = 17\times437 = 17\times19\times23$7429 = 17×437 = 17×19×23
So $7429 = 17\times19\times23$7429 = 17×19×23.

We find the prime factorisation of each number. The HCF is the product of the smallest power of each common prime; the LCM is the product of the greatest power of each prime involved.

(i) $26$26 and $91$91
$26 = 2\times13,\qquad 91 = 7\times13$26 = 2×13, 91 = 7×13
$\text{HCF} = 13,\qquad \text{LCM} = 2\times7\times13 = 182$HCF = 13, LCM = 2×7×13 = 182
Verification: $\text{HCF}\times\text{LCM} = 13\times182 = 2366$HCF×LCM = 13×182 = 2366, and $26\times91 = 2366$26×91 = 2366. Equal, as required.

(ii) $510$510 and $92$92
$510 = 2\times3\times5\times17,\qquad 92 = 2^2\times23$510 = 2×3×5×17, 92 = 2^2×23
$\text{HCF} = 2,\qquad \text{LCM} = 2^2\times3\times5\times17\times23 = 23460$HCF = 2, LCM = 2^2×3×5×17×23 = 23460
Verification: $\text{HCF}\times\text{LCM} = 2\times23460 = 46920$HCF×LCM = 2×23460 = 46920, and $510\times92 = 46920$510×92 = 46920. Equal.

(iii) $336$336 and $54$54
$336 = 2^4\times3\times7,\qquad 54 = 2\times3^3$336 = 2^4×3×7, 54 = 2×3^3
$\text{HCF} = 2\times3 = 6,\qquad \text{LCM} = 2^4\times3^3\times7 = 3024$HCF = 2×3 = 6, LCM = 2^4×3^3×7 = 3024
Verification: $\text{HCF}\times\text{LCM} = 6\times3024 = 18144$HCF×LCM = 6×3024 = 18144, and $336\times54 = 18144$336×54 = 18144. Equal.

For three numbers, the HCF is the product of the smallest power of each prime common to all three, and the LCM is the product of the greatest power of every prime that appears in any of them.

(i) $12, 15, 21$12, 15, 21
$12 = 2^2\times3,\qquad 15 = 3\times5,\qquad 21 = 3\times7$12 = 2^2×3, 15 = 3×5, 21 = 3×7
The only prime common to all three is $3$3 (power $1$1), so $\text{HCF} = 3$HCF = 3.
All primes involved are $2,3,5,7$2,3,5,7 with greatest powers $2^2,3^1,5^1,7^1$2^2,3^1,5^1,7^1, so $\text{LCM} = 2^2\times3\times5\times7 = 420$LCM = 2^2×3×5×7 = 420.

(ii) $17, 23, 29$17, 23, 29
All three are prime numbers, so they share no common factor other than $1$1: $\text{HCF} = 1$HCF = 1.
$\text{LCM} = 17\times23\times29 = 11339$LCM = 17×23×29 = 11339

(iii) $8, 9, 25$8, 9, 25
$8 = 2^3,\qquad 9 = 3^2,\qquad 25 = 5^2$8 = 2^3, 9 = 3^2, 25 = 5^2
These three numbers share no common prime factor, so $\text{HCF} = 1$HCF = 1.
$\text{LCM} = 2^3\times3^2\times5^2 = 1800$LCM = 2^3×3^2×5^2 = 1800

For any two positive integers, $\text{HCF}(a,b)\times\text{LCM}(a,b) = a\times b$HCF(a,b)×LCM(a,b) = a× b. This relation holds only for two numbers, so it applies directly here.

$\text{LCM}(306,657) = \frac{306\times657}{\text{HCF}(306,657)} = \frac{306\times657}{9}$LCM(306,657) = 306×657/HCF(306,657) = 306×657/9

Since $306\times657 = 201042$306×657 = 201042,
$\text{LCM}(306,657) = \frac{201042}{9} = 22338.$LCM(306,657) = 201042/9 = 22338.

A number ends in the digit $0$0 only if it is divisible by $10 = 2\times5$10 = 2×5 — which means its prime factorisation must contain both $2$2 and $5$5.

Now $6 = 2\times3$6 = 2×3, so
$6^n = (2\times3)^n = 2^n\times3^n.$6^n = (2×3)^n = 2^n×3^n.

The only primes in the factorisation of $6^n$6^n are $2$2 and $3$3; the prime $5$5 never appears. By the uniqueness part of the Fundamental Theorem of Arithmetic, this factorisation is the only possible one, so $6^n$6^n is never divisible by $10$10.

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

A number is composite if it can be written as a product of two factors, each greater than $1$1.

$7\times11\times13+13$7×11×13+13
$7\times11\times13+13 = 13\times(7\times11+1) = 13\times78$7×11×13+13 = 13×(7×11+1) = 13×78
This is a product of $13$13 and $78$78, both greater than $1$1 (and $78 = 2\times3\times13$78 = 2×3×13 factorises further), so $7\times11\times13+13$7×11×13+13 is composite.

$7\times6\times5\times4\times3\times2\times1+5$7×6×5×4×3×2×1+5
$7\times6\times5\times4\times3\times2\times1 = 5040,\qquad 5040+5 = 5045$7×6×5×4×3×2×1 = 5040, 5040+5 = 5045
Factoring out the common $5$5:
$5045 = 5\times(1008+1) = 5\times1009$5045 = 5×(1008+1) = 5×1009
This is a product of $5$5 and $1009$1009, both greater than $1$1, so $7\times6\times5\times4\times3\times2\times1+5$7×6×5×4×3×2×1+5 is composite.

In both cases the expression has the form $k\times m+k=k(m+1)$k× m+k=k(m+1) with $k>1$k>1 and $m+1>1$m+1>1, which is always composite.

Sonia completes one round every $18$18 minutes and Ravi every $12$12 minutes. They are next together at the starting point after a time that is a common multiple of $18$18 and $12$12 — and the first such time is the LCM.

$18 = 2\times3^2,\qquad 12 = 2^2\times3$18 = 2×3^2, 12 = 2^2×3
$\text{LCM}(18,12) = 2^2\times3^2 = 36$LCM(18,12) = 2^2×3^2 = 36

So Sonia and Ravi will meet again at the starting point after 36 minutes.