1)Express each number as a product of its prime factors:
(i) 140
$$\begin{array}{r|l}
2 & 140 \\
\hline 2 & 70 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$$
(ii) 156
$$\begin{array}{r|l}
2 & 156 \\
\hline 2 & 78 \\
\hline 3 & 39 \\
\hline 13 & 13 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$$
(iii) 3825
$$\begin{array}{r|l}
3 & 3825 \\
\hline 3 & 1275 \\
\hline 5 & 425 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\begin{aligned} 3825 &=3 \times 3 \times 5 \times 5 \times 17 \\ & =3^2 \times 5^2 \times 17\end{aligned}$$
(iv) 5005
$$\begin{array}{r|l}
5 & 5005 \\
\hline 7 & 1001 \\
\hline 11 & 143 \\
\hline 13 & 13 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 5005 = 5 \times 7 \times 11 \times 13 $$
(v) 7429
$$\begin{array}{r|l}
17 & 7429 \\
\hline 19 & 437 \\
\hline 23 & 23 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 7429 = 17 \times 19 \times 23 $$
2) Find the LCM and HCF of the following pairs of integers and verify that $$\text{LCM} \times \text{HCF} = \text{product of the two numbers}$$
i) 26 and 91
Finding HCF
$$\begin{array}{r|l}
2 & 26 \\
\hline 13 & 13 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 26 = 2 \times 13$$
$$\begin{array}{r|l}
7 & 91 \\
\hline 13 & 13 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 91 = 7 \times 13$$
Now,
\begin{aligned}
26 &= 2 \times \boxed{13} \\
91 &= 7 \times \boxed{13} \\
\end{aligned}
Therefore HCF = 13
Finding LCM
$$\begin{array}{r|l}
2 & 26, 91 \\
\hline 7 & 13, 91 \\
\hline 13 & 13, 13 \\
\hline & 1, 1
\end{array}$$
Hence,
$$\begin{aligned} LCM &= 2 \times 13 \times 7 \\ & =180 \end{aligned}$$
Verification:
Now, we have to verify that,
$$ HCF \times LCM = \text{Product of two numbers}$$
$$\begin{aligned}L.H.S &= HCF \times LCM \\ &= 13 \times 182 \\ &= 2366\end{aligned}$$
Therefore, R.H.S = Product of two numbers
$$\begin{aligned}&= 26 \times 91\\ &=2366\end{aligned}$$
Therefore, L.H.S = R.H.S
Hence verified.
2) Find the LCM and HCF of the following pairs of integers and verify that $$\text{LCM} \times \text{HCF} = \text{product of the two numbers}$$
ii) 510 and 92
Finding HCF
$$\begin{array}{r|l}
2 & 510 \\
\hline 3 & 255 \\
\hline 5 & 85 \\
\hline 17 & 17 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 510 = 2 \times 3 \times 5 \times 17$$
$$\begin{array}{r|l}
2 & 92 \\
\hline 2 & 46 \\
\hline 23 & 23 \\
\hline & 1 \\
\end{array}$$
Hence,
$$\mathbf 92 = 2 \times 2 \times 23$$
Now,
\begin{aligned}
510 &= \boxed{2} \times 3 \times 5 \times 17 \\
92 &= \boxed{2} \times 2 \times 23 \\
\end{aligned}
Therefore HCF = 13
Finding LCM
$$\begin{array}{r|l}
2 & 510, 92 \\
\hline 2 & 255, 46 \\
\hline 3 & 255, 23 \\
\hline 5 & 85, 23 \\
\hline 17 & 17, 23 \\
\hline 23 & 1, 23 \\
\hline & 1, 1
\end{array}$$
Hence,
$$\begin{aligned} LCM &= 2 \times 2 \times 3 \times 5 \times 17 \times 23 \\ & =23460 \end{aligned}$$
Verification:
Now, we have to verify that,
$$ HCF \times LCM = \text{Product of two numbers}$$
$$\begin{aligned}L.H.S &= HCF \times LCM \\ &= 2 \times 23460 \\ &= 46920\end{aligned}$$
Therefore, R.H.S = Product of two numbers
$$\begin{aligned}&= 510 \times 92\\ &=46920\end{aligned}$$
Therefore, L.H.S = R.H.S
Hence verified.