## The following real numbers have decimal expansions as given below. In each case, decide

[question] Question. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational, or not. If they are rational, and of the form $\frac{\mathbf{P}}{\mathbf{q}}$, what can you say about the prime factors of $\mathrm{g}$ ? (i) $43.123456789$ (ii) $0.120120012000120000 \ldots$ (iii) $43 . \overline{23456789}$ [/question] [solution] Solution: (i) $43.123456789$ Since, the decimal expansion terminates, so the given real number is rational and there...

## Write down the decimal expansions of those rational numbers in Question

[question] Question. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. [/question] [solution] Solution: (i) $\frac{13}{3125}=\frac{13}{5^{5}}=\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}=\frac{416}{10^{5}}=0.00416$ (ii) $\frac{17}{8}=\frac{17}{2^{3}}$$=\frac{17 \times 5^{3}}{\mathbf{2}^{3} \times \mathbf{5}^{3}}=\frac{\mathbf{1 7} \times \mathbf{5}^{3}}{\mathbf{1 0}^{3}}=\frac{\mathbf{2 1 2 5}}{\mathbf{1 0}^{3}}=2.125$ (iv) $...

## Without actually performing the long division, state whether the following rational numbers expansion.

[question] Question. Without actually performing the long division, state whether the following rational numbers expansion. will have a terminating decimal expansion or a non-terminating repeating decimal (i) $\frac{13}{3125}$ (ii) $\frac{17}{8}$ (iii) $\frac{64}{455}$ (iv) $\frac{15}{1600}$ (v) $\frac{29}{343}$ (vi) $\frac{23}{2^{3} 5^{2}}$ (vii) $\frac{129}{2^{2} 5^{7} 7^{5}}$ (viii) $\frac{6}{15}$ (ix) $\frac{35}{50}$ (x) $\frac{77}{210}$ [/question] [solution] Solution: (i) $\frac{13}{3125}=...

## Prove that the following are irrationals :

[question] Question. Prove that the following are irrationals : (i) $\frac{1}{\sqrt{2}}$ (ii) $7 \sqrt{5}$ (iii) $\mathbf{6}+\sqrt{\mathbf{2}}$ [/question] [solution] Solution: (i) Let us assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational. That is we can find coprime integers a and $b(b \neq 0)$ such that, $\frac{1}{\sqrt{2}}=\frac{\mathbf{P}}{q}$ Therefore, $q=\sqrt{\mathbf{z}_{\mathbf{p}}}$ Squaring on both sides, we get $\mathrm{q}^{2}=2 \mathrm{p}^{2}$ ...(1) Therefore, 2 divides...

## Prove that $3+2 \sqrt{5}$ is irrational.

[question] Question. Prove that $3+2 \sqrt{5}$ is irrational. [/question] [solution] Solution: Let us assume, to the contrary, that $3+2 \sqrt{5}$ is rational. That is, we can find coprime integers a and $b(b \neq 0)$ such that $3+2 \sqrt{5}=\frac{\mathbf{a}}{\mathbf{b}}$ Therefore, $\frac{\mathbf{a}}{\mathbf{b}}-3=2 \sqrt{\mathbf{5}}$ $\Rightarrow \frac{\mathbf{a}-\mathbf{3} \mathbf{b}}{\mathbf{b}}=2 \sqrt{\mathbf{5}}$ $\Rightarrow \frac{a-3 b}{2 b}=\sqrt{5} \Rightarrow \frac{a}{2 b}-\frac{3}{2...

## Prove that $\sqrt{\mathbf{5}}$ is irrational.

[question] Question. Prove that $\sqrt{5}$ is irrational. [/question] [solution] Solution: Let us assume, to the contrary, that $\sqrt{5}$ is rational. So, we can find coprime integers a and $b(\neq 0)$ such that $\sqrt{5}=\frac{a}{b}$ $\Rightarrow \sqrt{5} b=a$ Squaring on both sides, we get $5 b^{2}=a^{2}$ Therefore, 5 divides $\mathrm{a}^{2}$. Therefore, 5 , divides a So, we can write a = 5c for some integer c. Substituting for a, we get $5 \mathrm{~b}^{2}=25 \mathrm{c}^{2}$ $\Rightarrow \mat...

## Explain why $7 \times 11 \times 13+13$ and $7 \times 6 \times 5 \times 4$ $\times 3 \times 2 \times 1+5$ are composite numbers.

[question] Question. Explain why $7 \times 11 \times 13+13$ and $7 \times 6 \times 5 \times 4$ $\times 3 \times 2 \times 1+5$ are composite numbers. [/question] [solution] Solution: (i) $7 \times 11 \times 13+13=(7 \times 11+1) \times 13$ $=(77+1) \times 13$ $=78 \times 13=(2 \times 3 \times 13) \times 13$ So, $78=2 \times 3 \times 13$ $78 \times 13=2 \times 3 \times 13^{2}$ Since, $7 \times 11 \times 13+13$ can be expressed as a product of primes, therefore, it is a composite number. (ii) $7 \t...

## Check whether $6^{\mathrm{n}}$ can end with the digit 0 for any natural number $\mathrm{n}$.

[question] Question. Check whether $6^{\mathrm{n}}$ can end with the digit 0 for any natural number $\mathrm{n}$. [/question] [solution] Solution: If the number $6^{\mathrm{n}}$, for any natural number $\mathrm{n}$, ends with digit 0 , then it would be divisible by 5 . That is, the prime factorisation of $6^{\mathrm{n}}$ would contain the prime number 5 . This is not possible because $6^{\mathrm{n}}=(2 \times 3)^{\mathrm{n}}=2^{\mathrm{n}} \times 3^{\mathrm{n}} ;$ so the only primes in the facto...

## Given that $\operatorname{HCF}(306,657)=9$, find $\operatorname{LCM}(306,657)$.

[question] Question. Given that $\operatorname{HCF}(306,657)=9$, find $\operatorname{LCM}(306,657)$. [/question] [solution] Solution: $\operatorname{LCM}(306,657)$ $=\frac{\mathbf{3 0 6} \times \mathbf{6 5 7}}{\mathbf{H C F}(\mathbf{3 0 6}, \mathbf{6 5 7})}=\frac{\mathbf{3 0 6} \times \mathbf{6 5 7}}{\mathbf{9}}=22338$ [/solution]...

## Find the LCM and HCF of the following integers by applying the prime factorisation method

[question] Question. 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 [/question] [solution] Solution: (i) 12,15 and 21 So, $12=2 \times 2 \times 3=2^{2} \times 3$ So, $15=3 \times 5$ So, $21=3 \times 7$ Therefore $\mathrm{HCF}(12,15,21)=3 ;$ $\mathrm{LCM}=(12,15,21)=2^{2} \times 3 \times 5 \times 7=420$ (ii) $17,23,29$ $17=1 \times 17$ $23=1 \times 23$ $29=1 \times 29$ $\mathrm{LCM}=1 \times 17 \time...

## Find the $\mathrm{LCM}$ and HCF of the following pairs of integers and verify that LCM $\times \mathrm{HCF}=$ product of two numbers.

[question] Question. Find the LCM and HCF of the following pairs of integers and verify that LCM $\times \mathrm{HCF}=$ product of two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 [/question] [solution] Solution: (i) 26 and 91 So, $26=2 \times 13$ So. $91=7 \times 13$ Therefore, $\operatorname{LCM}(26,91)=2 \times 7 \times 13=182$ $\operatorname{HCF}(26,91)=13$ Verification : LCM $\times$ HCF $=182 \times 13=2366$ and $26 \times 91=2366$ i.e., $\mathrm{LCM} \times \mathrm{HCF}=$ produ...

## Express each number as product of its prime factors :

[question] Question. Express each number as product of its prime factors : (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 [/question] [solution] Solution: (i) 140 So, $140=2 \times 2 \times 5 \times 7=2^{2} \times 5 \times 7$ (ii) 156 So, $156=2 \times 2 \times 3 \times 13=2^{2} \times 3 \times 13$ So, $3825=3^{2} \times 5^{2} \times 17$ (iv) 5005 So, $5005=5 \times 7 \times 11 \times 13$ (v) 7429 So, $7429=17 \times 19 \times 23$ [/solution]...

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

[question] Question. Use Euclid's division lemma to show that the cube of any positive integer is of the form $9 \mathrm{~m}, 9 \mathrm{~m}$ $+1$ or $9 m+8$. [solution] Solution: Any positive integer is of the form $3 q, 3 q+1,3 q+2$ Case1: Let, $\mathrm{n}=3 \mathrm{q}$ Cube of this will be $n^{3}=27 q^{3}$ $n^{3}=9\left(3 q^{3}\right)$ So, $\mathrm{n}^{3}=9 \mathrm{~m}$, where $\mathrm{m}=3 \mathrm{q}^{3}$ Case2: $\mathrm{n}=3 \mathrm{q}+1$ So, $n^{3}=(3 q+1)^{3}$ $\mathrm{n}^{3}=27 \mathrm{q}...

## Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

[question] Question. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [/question] [solution] Solution: Let a be any odd positive integer. We apply the division lemma with a and $b=3$. Since $0 \leq r<3$, the possible remainders are 0,1 and 2 . That is, a can be $3 \mathrm{q}$, or $3 \mathrm{q}+1$, or $3 \mathrm{q}+2$, where $\mathrm{q}$ is the quotient. Now, $\quad(3 q)^{2}=9 q^{2}$ which can be written in the form...

## An army contingent of 616 members is to march behind an army band of 32 members in a parade

[question] Question. 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? [/question] [solution] Solution: 616 and 32 $616=32 \times 19+8$ $32=4 \times 8$ $\operatorname{HCF}$ of $(616,32)=8$ [/solution]...

## Show that any positive odd integer is of the form

[question] Question. 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. [/question] [solution] Solution: Let us start with taking a, where a is any positive odd integer. We apply the divisionalgorithm, with a and b = 6. Since $0 \leq \mathrm{r}<6$, the possible remainders are 0, 1, 2, 3, 4, 5. That is, a can be 6q or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient. However, since a is odd, we do not consider th...

## Use Euclid's division algorithm to find the HCF of :

[question] Question. Use Euclid's division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and $225 .$ [/question] [solution] Solution: (i) 135 and $225 .$ Start with the larger integer, that is, 225. Apply the division lemma to 225 and 135, to get. $225=135 \times 1+90$ Since the remainder $90 \neq 0$, we apply the division lemma to 135 and 90 to get $135=90 \times 1+45$ We consider the new divisior 90 and the new remainder 45, and apply the division lemma to get $90...