Solution

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

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

(iii) $3825=3 \times 3 \times 5 \times 5 \times 17=3^{2} \times 5^{2} \times 17$

(iv) $5005=5 \times 7 \times 11 \times 13$

(v) $7429=17 \times 19 \times 23$

Solution

(i) 26 and 91

$26=2 \times 13$

$91=7 \times 13$

$HCF=13$

$LCM=2 \times 7 \times 13=182$

Product of the two numbers $=26 \times 91=2366$

$HCF \times LCM=13 \times 182=2366$

Hence, product of two numbers $=HCF \times LCM$

(ii) 510 and 92

$510=2 \times 3 \times 5 \times 17$

$92=2 \times 2 \times 23$

$HCF=2$

LCM $=2 \times 2 \times 3 \times 5 \times 17 \times 23=23460$

Product of the two numbers $=510 \times 92=46920$

$\begin{aligned} HCF \times LCM & =2 \times 23460 \\ & =46920\end{aligned}$

Hence, product of two numbers $=HCF \times LCM$

(iii) 336 and 54

$ \begin{aligned} & 336=2 \times 2 \times 2 \times 2 \times 3 \times 7 \\ & 336=2^{4} \times 3 \times 7 \\ & 54=2 \times 3 \times 3 \times 3 \\ & 54=2 \times 3^{3} \\ & H C F=2 \times 3=6 \\ & L C M=2^{4} \times 3^{3} \times 7=3024 \end{aligned} $

Product of the two numbers $=336 \times 54=18144$

$HCF \times LCM=6 \times 3024=18144$

Hence, product of two numbers $=HCF \times$ LCM

Solution

(i) 12,15 and 21

$12=2^{2} \times 3$

$15=3 \times 5$

$21=3 \times 7$

$HCF=3$

LCM $=2^{2} \times 3 \times 5 \times 7=420$

(ii) 17,23 and 29

$17=1 \times 17$

$23=1 \times 23$

$29=1 \times 29$

$HCF=1$

$LCM=17 \times 23 \times 29=11339$

(iii) 8,9 and 25

$8=2 \times 2 \times 2$

$9=3 \times 3$

$25=5 \times 5$

$HCF=1$

$LCM=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5=1800$

Solution

$HCF(306,657)=9$

We know that, $LCM \times HCF=$ Product of two numbers

$\therefore LCM \times HCF=306 \times 657$

$LCM=\dfrac{306 \times 657}{HCF}=\dfrac{306 \times 657}{9}$

$LCM=22338$

Solution

If any number ends with the digit 0 , it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as $10=2 \times 5$

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

It can be observed that 5 is not in the prime factorisation of $6^{n}$.

Hence, for any value of $n, 6^{n}$ will not be divisible by 5 .

Therefore, $6^{n}$ cannot end with the digit 0 for any natural number $n$.

Solution

Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

$7 \times 11 \times 13+13=13 \times(7 \times 11+1)=13 \times(77+1)$

$=13 \times 78$

$=13 \times 13 \times 6$

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1+5=5 \times(7 \times 6 \times 4 \times 3 \times 2 \times 1+1)$

$=5 \times(1008+1)$

$=5 \times 1009$

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Solution

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the

LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

$18=2 \times 3 \times 3$

And, $12=2 \times 2 \times 3$

LCM of 12 and $18=2 \times 2 \times 3 \times 3=36$

Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes.