(i) $140 = 2^2 \times 5 \times 7$
(ii) $156 = 2^2 \times 3 \times 13$
(iii) $3825 = 3^2 \times 5^2 \times 17$
(iv) $5005 = 5 \times 7 \times 11 \times 13$
(v) $7429 = 17 \times 19 \times 23$
(i) $26 = 2 \times 13$, $91 = 7 \times 13$, so HCF $= 13$ and LCM $= 2 \times 7 \times 13 = 182$.
(ii) $510 = 2 \times 3 \times 5 \times 17$, $92 = 2^2 \times 23$, so HCF $= 2$ and LCM $= 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$.
(iii) $336 = 2^4 \times 3 \times 7$, $54 = 2 \times 3^3$, so HCF $= 2 \times 3 = 6$ and LCM $= 2^4 \times 3^3 \times 7 = 3024$.
(i) HCF $= 13$, LCM $= 182$; check $13 \times 182 = 2366 = 26 \times 91$.
(ii) HCF $= 2$, LCM $= 23460$; check $2 \times 23460 = 46920 = 510 \times 92$.
(iii) HCF $= 6$, LCM $= 3024$; check $6 \times 3024 = 18144 = 336 \times 54$.
(i) $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$; HCF $= 3$, LCM $= 2^2 \times 3 \times 5 \times 7 = 420$.
(ii) $17, 23, 29$ are all primes; HCF $= 1$, LCM $= 17 \times 23 \times 29 = 11339$.
(iii) $8 = 2^3$, $9 = 3^2$, $25 = 5^2$; HCF $= 1$, LCM $= 2^3 \times 3^2 \times 5^2 = 1800$.
(i) HCF $= 3$, LCM $= 420$.
(ii) HCF $= 1$, LCM $= 11339$.
(iii) HCF $= 1$, LCM $= 1800$.
For two numbers, LCM $\times$ HCF $=$ product of the numbers, so LCM $= \dfrac{306 \times 657}{\text{HCF}} = \dfrac{201042}{9} = 22338$.
LCM $(306, 657) = 22338$.
If a number ends in 0, it is divisible by $10 = 2 \times 5$, so its prime factorisation must contain both 2 and 5. But $6^n = (2 \times 3)^n = 2^n \times 3^n$ has no factor of 5. By the uniqueness of prime factorisation, $6^n$ is never divisible by 5, so it can never end with the digit 0.
No. $6^n$ can never end with the digit 0 for any natural number $n$.
Both numbers are composite because each can be written as a product of two factors greater than 1.
For the first: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78$, so 13 is a factor besides 1 and itself.
For the second: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009$, so 5 is a factor besides 1 and itself.
Since each number has a factor other than 1 and itself, both are composite.
They meet at the starting point after a time that is a common multiple of 18 and 12; the earliest such time is the LCM. $18 = 2 \times 3^2$ and $12 = 2^2 \times 3$, so LCM $= 2^2 \times 3^2 = 36$. Hence they meet again after 36 minutes.
They will meet again at the starting point after 36 minutes.
Suppose, to the contrary, that $\sqrt{5}$ is rational. Then $\sqrt{5} = \dfrac{a}{b}$ for some coprime integers $a$ and $b$ with $b \neq 0$. So $\sqrt{5}\,b = a$, and squaring gives $5b^2 = a^2$. Hence 5 divides $a^2$, and since 5 is prime, 5 divides $a$. Write $a = 5c$. Then $5b^2 = 25c^2$, i.e. $b^2 = 5c^2$, so 5 divides $b^2$ and therefore 5 divides $b$. Thus 5 is a common factor of $a$ and $b$, contradicting that they are coprime. So our assumption is false, and $\sqrt{5}$ is irrational.
Suppose, to the contrary, that $3 + 2\sqrt{5}$ is rational. Then $3 + 2\sqrt{5} = \dfrac{a}{b}$ for some coprime integers $a, b$ with $b \neq 0$. Rearranging, $2\sqrt{5} = \dfrac{a}{b} - 3 = \dfrac{a - 3b}{b}$, so $\sqrt{5} = \dfrac{a - 3b}{2b}$. Since $a$ and $b$ are integers, the right-hand side is rational, which would make $\sqrt{5}$ rational. But $\sqrt{5}$ is irrational — a contradiction. Hence $3 + 2\sqrt{5}$ is irrational.
In each part we use that $\sqrt{2}$ and $\sqrt{5}$ are irrational, and argue by contradiction.
(i) If $\dfrac{1}{\sqrt{2}}$ were rational, say $\dfrac{1}{\sqrt{2}} = r$, then $\sqrt{2} = \dfrac{1}{r}$ would be rational — a contradiction. So $\dfrac{1}{\sqrt{2}}$ is irrational.
(ii) If $7\sqrt{5}$ were rational, say $7\sqrt{5} = r$, then $\sqrt{5} = \dfrac{r}{7}$ would be rational — a contradiction. So $7\sqrt{5}$ is irrational.
(iii) If $6 + \sqrt{2}$ were rational, say $6 + \sqrt{2} = r$, then $\sqrt{2} = r - 6$ would be rational — a contradiction. So $6 + \sqrt{2}$ is irrational.