Q: Express each number as product of its prime factors:(i) 140(ii) 156(iii) 3825(iv) 5005(v) 7429

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

Q: Find the LCM and HCF of the following pairs of integers and verify that $\mathrm{LCM} \times \mathrm{HCF}=$ product of the two numbers.(i) 26 and 91(ii) 510 and 92(iii) 336 and 54

(i) 26 and 91$$\begin{aligned}& 26=2 \times 13 \\& 91=7 \times 13 \\& H C F=13 \\& L C M=2 \times 7 \times 13=182\end{aligned}$$Product of the two numbers $=26 \times 91=2366$$$\mathrm{HCF} \times \mathrm{LCM}=13 \times 182=2366$$Hence, product of two numbers $=\mathrm{HCF} \times \mathrm{LCM}$(ii) 510 and 92$$\begin{aligned}& 510=2 \times 3 \times 5 \times 17 \\& 92=2 \times 2 \times 23 \\& \mathrm{HCF}=2 \\& \mathrm{LCM}=2 \times 2 \times 3 \times 5 \times 17 \times 23=23460\end{aligned}$$Product of the two numbers $=510 \times 92=46920$$$\begin{aligned}\mathrm{HCF} \times \mathrm{LCM} & =2 \times 23460 \\& =46920\end{aligned}$$Hence, product of two numbers $=\mathrm{HCF} \times \mathrm{LCM}$(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$$$\mathrm{HCF} \times \mathrm{LCM}=6 \times 3024=18144$$Hence, product of two numbers $=\mathrm{HCF} \times \mathrm{LCM}$

Q: Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) 12,15 and 21(ii) 17, 23 and 29(iii) 8,9 and 25

(i) 12,15 and 21$$\begin{aligned}& 12=2^{2} \times 3 \\& 15=3 \times 5 \\& 21=3 \times 7 \\& H C F=3 \\& L C M=2^{2} \times 3 \times 5 \times 7=420\end{aligned}$$(ii) 17,23 and 29$$\begin{aligned}& 17=1 \times 17 \\& 23=1 \times 23 \\& 29=1 \times 29 \\& H C F=1 \\& L C M=17 \times 23 \times 29=11339\end{aligned}$$(iii) $\quad 8,9$ and 25$$\begin{aligned}& 8=2 \times 2 \times 2 \\& 9=3 \times 3 \\& 25=5 \times 5 \\& \text { HCF }=1 \\& \text { LCM }=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5=1800\end{aligned}$$

Question 1

Use Euclid's division algorithm to find the HCF of:(i) 135 and 225(ii) 196 and 38220(iii) 867 and 255

Accepted Answer

(i) 135 and 225Since $225>135$, we apply the division lemma to 225 and 135 to obtain $225=135 	imes 1+90$Since remainder $90 
eq 0$, we apply the division lemma to 135 and 90 to obtain$135=90 	imes 1+45$We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain$90=2 	imes 45+0$Since the remainder is zero, the process stops.Since the divisor at this stage is 45 ,Therefore, the HCF of 135 and 225 is 45.(ii) 196 and 38220Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain$38220=196 	imes 195+0$Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196.(iii) 867 and 255Since 867 > 255, we apply the division lemma to 867 and 255 to obtain $867=255 	imes 3+102$Since remainder $102 
eq 0$, we apply the division lemma to 255 and 102 to obtain$255=102 	imes 2+51$We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain$102=51 	imes 2+0$Since the remainder is zero, the process stops.Since the divisor at this stage is 51 , Therefore,HCF of 867 and 255 is 51.

Question 2

Show that any positive odd integer is of the form $6 q+1$, or $6 q+3$, or $6 q+5$, where $q$ is some integer.

Accepted Answer

Let $a$ be any positive integer and $b=6$.Then, by Euclid's algorithm, $a=6 q+r$ for some integer $q \geq 0$, and $r=0,1,2,3,4,5$ because $0 \leq rTherefore, $a=6 q$ or $6 q+1$ or $6 q+2$ or $6 q+3$ or $6 q+4$ or $6 q+5$Also, $6 q+1=2 	imes 3 q+1=2 k_{1}+1$, where $k_{1}$ is a positive integer $6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_{2}+1$, where $k_{2}$ is an integer $6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_{3}+1$, where $k_{3}$ is an integer Clearly,$6 q+1,6 q+3,6 q+5$ are of the form $2 k+1$, where $k$ is an integer.Therefore, $6 q+1,6 q+3,6 q+5$ are not exactly divisible by 2 .Hence, these expressions of numbers are odd numbers.And therefore, any odd integer can be expressed in the form $6 q+1$, or $6 q+3$, or $6 q+5$

Question 3

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Accepted Answer

HCF $(616,32)$ will give the maximum number of columns in which they can march.We can use Euclid's algorithm to find the HCF.$616=32 	imes 19+8$$32=8 	imes 4+0$The $\operatorname{HCF}(616,32)$ is 8.Therefore, they can march in 8 columns each.

Question 4

Use Euclid's division lemma to show that the square of any positive integer is either of form $3 m$ or $3 m+1$ for some integer $m$.[Hint: Let $x$ be any positive integer then it is of the form $3 q, 3 q+1$ or $3 q+2$. Now square each of these and show that they can be rewritten in the form $3 m$ or $3 m+1$.]

Accepted Answer

Let $a$ be any positive integer and $b=3$.Then $a=3 q+r$ for some integer $q \geq 0$And $r=0,1,2$ because $0 \leq rTherefore, $a=3 q$ or $3 q+1$ or $3 q+2$ Or,$$\begin{aligned}a^{2} & =(3 q)^{2} 	ext { or }(3 q+1)^{2} 	ext { or }(3 q+2)^{2} \& =(3 q)^{2} 	ext { or } 9 q^{2}+6 q+1 	ext { or } 9 q^{2}+12 q+4 \& =3 	imes\left(3 q^{2}ight) 	ext { or } 3 	imes\left(3 q^{2}+2 qight)+1 	ext { or } 3 	imes\left(3 q^{2}+4 q+1ight)+1 \& =3 k_{1} 	ext { or } 3 k_{2}+1 	ext { or } 3 k_{3}+1\end{aligned}$$Where $k_{1}, k_{2}$, and $k_{3}$ are some positive integersHence, it can be said that the square of any positive integer is either of the form $3 m$ or $3 m+1$.

Question 5

Use Euclid's division lemma to show that the cube of any positive integer is of the form $9 m, 9 m+1$ or $9 m+8$.

Accepted Answer

Let $a$ be any positive integer and $b=3$$a=3 q+r$, where $q \geq 0$ and $0 \leq r$a=3 q$ or $3 q+1$ or $3 q+2$Therefore, every number can be represented as these three forms.There are three cases.Case 1: When $a=3 q$,$a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m$Where $m$ is an integer such that $m=3 q^{3}$Case 2: When $a=3 q+1$,$a^{3}=(3 q+1)^{3}$$a^{3}=27 q^{3}+27 q^{2}+9 q+1$$a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1$$a^{3}=9 m+1$Where $m$ is an integer such that $m=\left(3 q^{3}+3 q^{2}+q\right)$Case 3: When $a=3 q+2$,$a^{3}=(3 q+2)^{3}$$a^{3}=27 q^{3}+54 q^{2}+36 q+8$$a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8$$a^{3}=9 m+8$Where $m$ is an integer such that $m=\left(3 q^{3}+6 q^{2}+4 q\right)$Therefore, the cube of any positive integer is of the form $9 m, 9 m+1$, or $9 m+8$.\section*{Mathematics}(Chapter - 1) (Real Numbers)(Class X)\section*{Exercise 1.2}

Question 6

Express each number as product of its prime factors:(i) 140(ii) 156(iii) 3825(iv) 5005(v) 7429

Accepted Answer

(i) $140=2 	imes 2 	imes 5 	imes 7=2^{2} 	imes 5 	imes 7$(ii) $156=2 	imes 2 	imes 3 	imes 13=2^{2} 	imes 3 	imes 13$(iii) $3825=3 	imes 3 	imes 5 	imes 5 	imes 17=3^{2} 	imes 5^{2} 	imes 17$(iv) $5005=5 	imes 7 	imes 11 	imes 13$(v) $7429=17 	imes 19 	imes 23$

Question 7

Find the LCM and HCF of the following pairs of integers and verify that $\mathrm{LCM} 	imes \mathrm{HCF}=$ product of the two numbers.(i) 26 and 91(ii) 510 and 92(iii) 336 and 54

Accepted Answer

(i) 26 and 91$$\begin{aligned}& 26=2 	imes 13 \& 91=7 	imes 13 \& H C F=13 \& L C M=2 	imes 7 	imes 13=182\end{aligned}$$Product of the two numbers $=26 	imes 91=2366$$$\mathrm{HCF} 	imes \mathrm{LCM}=13 	imes 182=2366$$Hence, product of two numbers $=\mathrm{HCF} 	imes \mathrm{LCM}$(ii) 510 and 92$$\begin{aligned}& 510=2 	imes 3 	imes 5 	imes 17 \& 92=2 	imes 2 	imes 23 \& \mathrm{HCF}=2 \& \mathrm{LCM}=2 	imes 2 	imes 3 	imes 5 	imes 17 	imes 23=23460\end{aligned}$$Product of the two numbers $=510 	imes 92=46920$$$\begin{aligned}\mathrm{HCF} 	imes \mathrm{LCM} & =2 	imes 23460 \& =46920\end{aligned}$$Hence, product of two numbers $=\mathrm{HCF} 	imes \mathrm{LCM}$(iii) 336 and 54$$\begin{aligned}& 336=2 	imes 2 	imes 2 	imes 2 	imes 3 	imes 7 \& 336=2^{4} 	imes 3 	imes 7 \& 54=2 	imes 3 	imes 3 	imes 3 \& 54=2 	imes 3^{3} \& H C F=2 	imes 3=6 \& L C M=2^{4} 	imes 3^{3} 	imes 7=3024\end{aligned}$$Product of the two numbers $=336 	imes 54=18144$$$\mathrm{HCF} 	imes \mathrm{LCM}=6 	imes 3024=18144$$Hence, product of two numbers $=\mathrm{HCF} 	imes \mathrm{LCM}$

Question 8

Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) 12,15 and 21(ii) 17, 23 and 29(iii) 8,9 and 25

Accepted Answer

(i) 12,15 and 21$$\begin{aligned}& 12=2^{2} 	imes 3 \& 15=3 	imes 5 \& 21=3 	imes 7 \& H C F=3 \& L C M=2^{2} 	imes 3 	imes 5 	imes 7=420\end{aligned}$$(ii) 17,23 and 29$$\begin{aligned}& 17=1 	imes 17 \& 23=1 	imes 23 \& 29=1 	imes 29 \& H C F=1 \& L C M=17 	imes 23 	imes 29=11339\end{aligned}$$(iii) $\quad 8,9$ and 25$$\begin{aligned}& 8=2 	imes 2 	imes 2 \& 9=3 	imes 3 \& 25=5 	imes 5 \& 	ext { HCF }=1 \& 	ext { LCM }=2 	imes 2 	imes 2 	imes 3 	imes 3 	imes 5 	imes 5=1800\end{aligned}$$

Chapter 1

Exercise 1.1

Exercise 1.2

Exercise 1.3