Question 3

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

\n

$$15 = 3\times 5$$

\n

$$21 = 3\times 7$$

The only prime appearing in all three factorizations is $3$, and its lowest exponent is $1$. Hence $\text{HCF}=3^{1}=3$.

Collect the highest exponent of each prime present:

• $2$: highest exponent $2$ (from $12$)\n- $3$: highest exponent $1$ (appears in all)\n- $5$: exponent $1$ (from $15$)\n- $7$: exponent $1$ (from $21$)\nThus $\text{LCM}=2^{2}\times 3^{1}\times 5^{1}\times 7^{1}=\dfrac{2^{2}\cdot 3\cdot 5\cdot 7}{1}=420$.

All three numbers are prime, so

$$17 = 17$$

\n

$$23 = 23$$

\n

$$29 = 29$$

There is no prime common to all three numbers, so $\text{HCF}=1$.

When numbers share no common factors, the LCM is the product of the numbers:

$$\text{LCM}=17\times 23\times 29=\dfrac{17\cdot 23\cdot 29}{1}=11339$$

$$8 = 2^{3}$$

\n

$$9 = 3^{2}$$

\n

$$25 = 5^{2}$$

No prime appears in all three factorizations, therefore $\text{HCF}=1$.

Take the highest exponent of each prime present:\n- $2^{3}$ (from $8$)\n- $3^{2}$ (from $9$)\n- $5^{2}$ (from $25$)\nThus

$$\text{LCM}=2^{3}\times 3^{2}\times 5^{2}=\dfrac{2^{3}\cdot 3^{2}\cdot 5^{2}}{1}=1800$$