## Free PDF Download of NCERT Exemplar for Class 10 Maths Chapter 1

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## Access NCERT Exemplar Solutions for Class 10 Mathematics Chapter 1 – Real Numbers

### SOLVED EXAMPLE

1. The decimal expansion of the rational number $\frac{{33}}{{{2^2}.5}}$ will terminate after

(A) One decimal place

(B) Two decimal place

(C) Three decimal place

(D) More than 3 decimal places

Ans: B

Express the denominator as the multiple of 10,

If we divide both numerator and denominator by $5,$ denominator can be changed to $100$,

$\frac{{33}}{{{2^2} \cdot 5}} = \frac{{33 \times 5}}{{4 \times 5 \times 5}}$

$\Rightarrow \frac{{33 \times 5}}{{100}} = \frac{{165}}{{100}}$

Convert the obtained fraction into decimal.

$\frac{{165}}{{100}} = 1 \cdot 65$

2. Euclid’s division Lemma states that for two positive integers $a$ and $b$ there exist unique integers $q$ and $r$ such that $a = bq + r$, where $r$ must satisfy.

(A) $1 < r < b$

(B) $0 < r \leqslant b$

(C) $0 \leqslant r < b$

(D) $0 < r < b$

Ans: C

Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that $a = bq + r$, where $0 \leqslant r < b$.

### EXERCISE 1.1

1. For some integer $m$, every even integer is of the form

(A) $m$

(B)$m + 1$

(C) $2m$

(D) $2m + 1$

Ans: C

Every Even integer is a multiple of $2$.

Let,

$m = {\text{Integer}} = Z$

$\Rightarrow m = .... - 3, - 2, - 1,0,1,2,3....$

$\Rightarrow 2m = .... - 6, - 4, - 2,0,2,4,6....$(all even numbers)

Hence,

Even numbers can be written in the form of 2m.

2. For some integer $q$, every odd integer is of the form

(A) $q$

(B) $q + 1$

(C) $2q$

(D) $2q + 1$

Ans: D

Every Even integer is a multiple of 2.

Let,

$q = {\text{Integer}} = Z$

$\Rightarrow q = .... - 3, - 2, - 1,0,1,2,3....$

$\Rightarrow 2q = .... - 6, - 4, - 2,0,2,4,6....$(all are even numbers)

$\Rightarrow 2q + 1 = .... - 5, - 3, - 1,1,3,5,7....$(all are odd numbers)

Hence, every odd integer can be written in the form of $2q + 1.$

3. ${n^2} - 1$ is divisible by 8, if $n$ is

(A) an integer

(B)a natural number

(C) an odd integer

(D) More than 3 decimal places

Ans: C

Let $x = {n^2} - 1$

Since, ‘n’ is any number. Therefore, it can be even or odd.

Condition I,

When ‘n’ is an even integer.

Let, $n = 2k$ (where $'k'$ is an integer)

$\Rightarrow x = {n^2} - 1$

$\Rightarrow x = {\left( {2k} \right)^2} - 1$

$\Rightarrow x = 4{k^2} - 1$

When,

$k = - 1$

$\Rightarrow x = 4{k^2} - 1$

$\Rightarrow x = 4{\left( { - 1} \right)^2} - 1 = 3$(which is not divisible by $8$)

$k = 0$

$\Rightarrow x = 4{k^2} - 1$

$\Rightarrow x = 4{\left( 0 \right)^2} - 1 = - 1$ (which is not divisible by $8$)

Condition II,

When ‘n’ is an odd integer.

$n = 2k + 1$ (where $'k'$ is an integer)

$\Rightarrow x = {\left( {2k + 1} \right)^2} - 1$

$\Rightarrow x = 4{k^2} + 1 + 4k - 1$

$\Rightarrow x = 4{k^2} + 4k$

$\Rightarrow x = 4k\left( {k + 1} \right)$

When,

$k = - 1$

$\Rightarrow x = 4k\left( {k + 1} \right)$

$\Rightarrow x = 4\left( { - 1} \right)\left( { - 1 + 1} \right) = 0$ (which is divisible by $8$)

$k = 0$

$\Rightarrow x = 4k\left( {k + 1} \right)$

$\Rightarrow x = 4\left( 0 \right)\left( {0 + 1} \right) = 0$ (which is divisible by $8$)

$k = 1$

$\Rightarrow x = 4k\left( {k + 1} \right)$

$\Rightarrow x = 4\left( 1 \right)\left( {1 + 1} \right) = 8$ (which is divisible by $8$)

Hence, we can conclude from the above two cases that if n is odd, then ${n^2} - 1$ is divisible by $8$.

4. If the HCF of $65$ and $117$ is expressible in the form $65 m -- 117$, then the value of $m$ is

(A)$4$

(B) $2$

(C) $1$

(D) $3$

Ans: B

HCF of $65$ and $117$ by prime factorisation method.

$65 = 5 \times 13$

$117 = 3 \times 3 \times 13$

As, $13$ is the common factor between $65$ and $117$.

Therefore, ${\text{HCF}}\left( {65,{\text{ }}117} \right) = 13$

$\because 65m - 117 = {\text{HCF}}$

$\therefore 65m - 117 = 13$

$\Rightarrow 65m = 13 + 117$

$\Rightarrow 65m = 130$

$\Rightarrow m = \frac{{130}}{{65}} = 2$

5. The largest number which divides $70 and 125$, leaving remainder $5 and 8$ respectively is

(A) 13

(B) 65

(C) 875

(D) 1750

Ans: A

Since, 5 and 8 are the remainder of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers $65 = \left( {70 - 5} \right)$ and$117 = \left( {125--8} \right)$, which is divisible by the required number.

The number to be obtained ${\text{ = HCF}}\left( {{\text{70,}}{\text{125}}} \right)$

$\Rightarrow 117 = 65 \times 1 + 52$ ($\therefore $ Dividend $= $ divisor $ \times $ quotient $ + $remainder)

$\Rightarrow 65 = 52 \times 1 + 13$

$\Rightarrow 52 = 13 \times 4 + 0$

Hence, 13 is the highest common factor of 70 and 125, leaving 5 and 8 as remainder receptively.

6. If two positive integers a and b are written as $a = {x^3}{y^2}$ and $b = x{y^3};x,y$are prime numbers, then HCF $\left( {a, b} \right)$ is

(A) $xy$

(B) $x{y^2}$

(C) ${x^3}{y^3}$

(D) ${x^2}{y^2}$

Ans: B

HCF is defined as the highest common factor between the two numbers and for variables it’s the smallest exponent of every common variable.

We can write,

$a = x \times x \times x \times y \times y$

$b = x \times y \times y \times y$

So,

$x$ and $y$ are both common variables in $a$ and $b$, with lowest exponents as 1 and 2 respectively.

Therefore, the HCF will be $x{y^2}$.

7. If two positive integer $p$and $q$ can be expressed as $p = a{b^2}$ and $q = {a^3}b;a,b$ being prime number then LCM $\left( {p, q} \right)$ is

(A) $ab$

(B) ${a^2}{b^2}$

(C) ${a^3}{b^2}$

(D) ${a^3}{b^3}$

Ans: C

LCM is defined as the least common multiple of integers $a$ and $b$. For variables it’s the highest exponent of every common variable.

We can write,

$p = a \times b \times b$

$q = a \times a \times a \times b$

So,

a and b are both common variables in p and q, with highest exponent as 3 and 2 respectively.

Therefore, the LCM will be ${a^3}{b^2}$.

8. The product of a non-zero rational and an irrational no. is

(A) Always irrational

(B) Always rational Two decimal place

(C) Rational or Irrational

(D) One

Ans: A

Irrational number can be defined as a non- repeating, non-terminating decimal number unlike the case of rational number which is a terminating number.

When a non-repeating and non-terminating number is multiplied by an irrational number. The result number is always an irrational number.

Example:

$\frac{2}{5} \times \sqrt 3 = \frac{{2\sqrt 3 }}{5}$

9. The least number that is divisible by all the numbers from 1-10 (both inclusive) is

(A) 10

(B) 100

(C) 504

(D)2520

Ans: D

Since, the required number is divisible from all the numbers from 1 to 10. Therefore, we need to find the LCM.

Factors of numbers from 1 to 10.

$1 = 1$

$2 = 1 \times 2$

$3 = 1 \times 3$

$4 = 2 \times 2$

$5 = 1 \times 5$

$6 = 2 \times 3$

$7 = 1 \times 7$

$8 = 2 \times 2 \times 2$

$9 = 3 \times 3$

$10 = 2 \times 5$

We know that:

LCM is defined as the least common multiple of integers $a$ and $b$. For variables it’s the highest exponent of every common variable.

Therefore,

${\text{LCM}}\left( {1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}8,{\text{ }}9,{\text{ }}10} \right) = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520$

10. The decimal expansion of the rational number $\frac{{14587}}{{1250}}$ will terminate after:

(A) One decimal place

(B) Two decimal place

(C) Three decimal place

(D) More than 3 decimal places

Ans: D

Rational number = $\frac{{14587}}{{1250}}$

Multiply the numerator and the denominator with ${2^3}$ to make the denominator as a multiple of 10.

$\frac{{14587}}{{1250}} = \frac{{14587}}{{{2^1} \times {5^4}}} \times \frac{{{2^3}}}{{{2^3}}}$

$\Rightarrow = \frac{{14587 \times 8}}{{{2^4} \times {5^4}}}$

$\Rightarrow = \frac{{116696}}{{{{\left( {10} \right)}^4}}} = \frac{{116696}}{{10000}}$

$\Rightarrow = 11.6696$

Hence, a given rational number will terminate after four decimal places.

### SOLVED EXAMPLE

1. The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer.

Ans: The given statement “The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only.” is False.

According to Euclid's division Lemma,

$a = bq + r,{\text{ }}a = 3q + r$

where b = 3

$0 \leqslant r < b$

$\Rightarrow 0 \leqslant r < 3$

Therefore, r can have 0, 1 and 2.

2. Can the number${6^n}$, n being a natural number end with the digits 5 ? Give reasons.

Ans: Consider the number ${6^n}$

On prime factorisation of 6,

${6^n} = {\left( {2 \times 3} \right)^n}$

${6^n} = {2^n} \times {3^n}$

Since, on factoring ${6^n}$ we get ${2^n} \times {3^n}$, which does not contain 5. Therefore, ${6^n}$ cannot end with the digit 5.

### EXERCISE 1.2

1. Write whether every positive integer can be of the form $4q + 2$, where q is an integer. Justify your answer

Ans: Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that $a = bq + r$, where $0 \leqslant r < b$.

Where,

a = dividend

b = divisor

q = quotient

r = remainder

According to question,

‘a’ is any positive integer.

$b = 4$

$r = 0,{\text{ }}1,{\text{ }}2,{\text{ }}3$$0 \leqslant r < b;0 \leqslant r < 4$

Therefore, ‘a’ can be in the form of $4q,4q + 1,4q + 2,4q + 3$ i.e., $a = bq + r$.

Hence, it is not necessary that every positive integer is in the form of $4q + 2$.

2. “The product of two consecutive positive integers is divisible by 2”. is this statement true or false? Give reasons.

Ans: Two consecutive positive integers are always an even and an odd number.

The product of even and an odd number is always an even number.

Since, the even number is divisible by 2.

Therefore, the product of two consecutive positive integers is also divisible by 2.

Hence, the given statement “The product of two consecutive positive integers is divisible by 2”. is true.

3. “The product of three consecutive positive integers is divisible by $6$. Is this statement true or false”? Justify your answer

Ans: We can verify it by taking different random groups of three consecutive positive integers.

a) Let take 2, 3, 4 as 3 consecutive positive integers

So, $2 \times 3 \times 4 = 24$ which is divisible by 6

b) Let take 5, 6, 7 as 3 consecutive positive integers

So, $5 \times 6 \times 7 = 210;$ which is divisible by 6

Hence, The statement “The product of three consecutive positive integers is divisible by $6$.”

4. Write whether the square of any positive integer can be of the form $3m + 2$, where $m$ is a natural number. Justify your answer.

Ans: No

Consider any random positive integer say ‘a’.

According to Euclid’s division lemma; a can be written as

$a = bq + r{\text{ }};{\text{ }}0 \leqslant r < b$

Let b = 3 then,

$a = 3q + r{\text{ where }}0 \leqslant r < 3$

So, r can have the following values 0, 1, 2.

Since, any positive integer can have the following forms.

$r = 0{\text{ }} \Rightarrow {\text{ }}a = 3q$

$r = 1{\text{ }} \Rightarrow {\text{ }}a = 3q + 1$

$r = 2{\text{ }} \Rightarrow {\text{ }}a = 3q + 2$

When $r = 0,a = 3q$

$\Rightarrow {a^2} = {\left( {3q} \right)^2}$

$\Rightarrow 9{q^2} = 3\left( {3{q^2}} \right) = 3m$(where $m = 3{q^2}$)

When $r = 1,a = 3q + 1$

$\Rightarrow {a^2} = {\left( {3q + 1} \right)^2}$

$\Rightarrow 9{q^2} + 6q + 1 = 3\left( {3{q^2} + 2q} \right) + 1 = 3m + 1$ (where $m = 3{q^2} + 2q$)

When $r = 2,a = 3q + 2$

$\Rightarrow {a^2} = {\left( {3q + 2} \right)^2}$

$\Rightarrow 9{q^2} + 12q + 4 = 3\left( {3{q^2} + 4q + 3} \right) + 1 = 3m + 1$(where $m = 3{q^2} + 4q + 3$)

Hence, squares of positive integers cannot be expressed in the form $3m + 2$.

5. A positive integer is of the form $3q + 1,{\text{ }}q$ being a natural number. Can you write its square in any form other than$3m + 1$, i.e., $3m$ or $3m + 2$ for some integer m? Justify your answer.

Ans: According to Euclid’s division lemma,

Put $b = 3$

$a = bq + r$, where all of a, b, q, r are positive integers and $0 \leqslant r < b$ i.e., $0 \leqslant r < 3$.

$\Rightarrow a = 3q,3q + 1{\text{ or }}3q + 2$.

Now,

When $a = 3q$,

$\Rightarrow {a^2} = {\left( {3q} \right)^2}$

$= 9{q^2} = 3.3{q^2}$(here, $m = 3{q^2}$)

When $a = 3q + 1$,

$\Rightarrow {a^2} = {\left( {3q + 1} \right)^2}$

$= 9{q^2} + 6q + 1 = 3\left( {3{q^2} + 2q} \right) + 1$(here, $m = 3{q^2} + 2q$)$$

When $a = 3q + 2$,

$\Rightarrow {a^2} = {\left( {3q + 2} \right)^2}$

$= 9{q^2} + 12q + 4$

$= 3\left( {3{q^2} + 4q + 1} \right) + 1$(here, $m = 3{q^2} + 4q + 1$)

Therefore, the square of a positive integer can be expressed in the form of $3m + 1$.

6. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.

Ans: The common factors of $3000$ and $525$ is $75.$

By Euclid’s division lemma,

$3000 = 525 \times 5 + 375$

$525 = 375 \times 1 + 150$

375 = 150 \times 2 + 75$

150 = 75 \times 2 + 0$

By Euclid’s method, we get the HCF (525, 3000) as 75.

7. Explain why $3 \times 5 \times 7 + 7$ is a composite number.

Ans: A composite number can be expressed as product of primes

Considering the given number,

$ = \left( {3 \times 5 \times 7} \right) + 7$

$\Rightarrow 105 + 7 = 112$

On prime factorisation of 112,

$112 = 2 \times 2 \times 2 \times 2 \times 7$

$= {2^4} \times 7$

Here 2 and 7 both are prime numbers.

We know that:

The product of two prime numbers is always a composite number.

Therefore, $3 \times 5 \times 7 + 7$ is a composite number.

8. Can two numbers have $18$ as their HCF and $380$ as their LCM? Give reasons.

Ans: HCF of two numbers means highest common factor and LCM of two numbers means their lowest common multiple. So, as per definition LCM of two numbers is always divisible by the HCF of two numbers. HCF is the factor of LCM.

Since, 18 is not the factor of 380. Therefore, no numbers can have 18 and 380 as their HCF and LCM respectively.

9. Without actually performing the long division, find if $\frac{{987}}{{10500}}$ will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

Ans: Let $\frac{{987}}{{10500}}$

On simplifying the given number,

$\Rightarrow \frac{{987}}{{10500}} = \frac{{3 \times 7 \times 47}}{{3 \times 7 \times 500}}$

${\text{ }} = \frac{{47}}{{500}}$

${\text{ }} = \frac{{47}}{{{2^2} \times {5^3}}}$

Here, the denominator is in the form ${2^m} \times {5^n}$.

If the denominator is in the form of ${2^m} \times {5^n}$ then its terminating rational number.

Hence, the number$\frac{{987}}{{10500}}$ is a terminating rational number.

10. A rational number in its decimal expansion is $327.7081$. What can you say about the prime factors of $q$, when this number is expressed in the form$\frac{p}{q}$? Give reasons.

Ans:

$327.7081$ can be written as,

$\frac{{3277081}}{{10000}} = \frac{{32777081}}{{104}} = \frac{p}{q}$

$q = 10000$

$= {10^4}$

$= {\left( {2 \times 5} \right)^4}$

Hence, the denominator can be expressed in the form of ${2^m} \times {5^n}$.

### Sample Questions

1. Using Euclid’s division algorithm, find which of the following pairs of numbers are co-prime:

(i) $231,396$

Ans: Let's calculate the HCF of each number pair.

Here, Co-primes are two numbers that have only $1$ as a common factor.

Then, by Euclid's lemma $a = bq + r,0 \leqslant r < b$

(i) $231,396$

So, HCF of $231,396$is

$\Rightarrow 396 = 1 \times 231 + 165$

$\Rightarrow 231 = 1 \times 165 + 66$

$\Rightarrow 165 = 2 \times 66 + 33$

$\Rightarrow 66 = 2 \times 33 + 0$

The HCF of $231,396$ is $33$ which is not equal to$1$,

Therefore, $231,396$are not co-prime.

(ii) $847,2160$

Let's calculate the HCF of each number pair.

Here, Co-primes are two numbers that have only $1$ as a common factor.

Then, by Euclid's lemma $a = bq + r,0 \leqslant r < b$

So, HCF of $847$ and $2160$ is

$2160 = 847 \times 2 + 466$

$\Rightarrow 847 = 466 \times 1 + 381$

$\Rightarrow 466 = 381 \times 1 + 85$

$\Rightarrow 85 = 41 \times 2 + 3$

$\Rightarrow 41 = 3 \times 13 + 2$

$\Rightarrow 3 = 2 \times 1 + 1$

$\Rightarrow 2 = 1 \times 2 + 0$

Since, The HCF of $847$ and $2160$is $1$.

Therefore, $847$ and $2160$ are co-primes.

2. Show that the square of an odd positive integer is of the form $8m + 1$, for some whole number$m$.

Ans: Let us take $2p + 1$ is any odd positive integer for which$p \in I$

$\Rightarrow {(2p + 1)^2} = 4{p^2} + 1 + 4p$

$\Rightarrow {(2p + 1)^2} = 4{p^2} + 4p + 1$

$\Rightarrow {(2p + 1)^2} = 4p(p + 1) + 1$

$p(p + 1)$is even and also a product of two consecutive integers .

Let $p(p + 1) = 2m$ for some whole number $m \in {\text{W}}$

$\Rightarrow {(2p + 1)^2}$

$\Rightarrow {\left( {2p} \right)^2} + 1 + 4p$

$\Rightarrow 4{p^2} + 4p + 1$

$\Rightarrow 4p\left( {p + 1} \right) + 1$

$\Rightarrow 4(2m) + 1$

$\Rightarrow 8m + 1$

Therefore, the square of an odd integer is of the form $8{\text{m}} + 1$ for some whole number $m \in {\text{W}}$.

3. Prove that $\sqrt 2 + \sqrt 3 $ is irrational.

Ans: Let us assume $\sqrt 2 + \sqrt 3 $is a rational number.

So it can be represented in the form of $a/b$

$\Rightarrow \sqrt 2 + \sqrt 3 = \frac{a}{b}$

$\Rightarrow \sqrt 2 = \frac{a}{b} - \sqrt 3 $

Now, squaring on both sides, we get

$\Rightarrow {\left( {\sqrt 2 } \right)^2} = {\left( {\frac{a}{b} - \sqrt 3 } \right)^2}$

Expand ${\left( {\frac{a}{b} - \sqrt 3 } \right)^2} = {\frac{a}{{{b^2}}}^2} + 3 - 2\left( {\frac{a}{b}} \right)\sqrt 3 $

$\Rightarrow 2 = \frac{{{a^2}}}{{{b^2}}} + 3 - 2\sqrt 3 \left( {\frac{a}{b}} \right)$

$\Rightarrow \frac{{{a^2}}}{{{b^2}}} + 3 - 2 = 2\sqrt 3 \frac{a}{b}$

$\Rightarrow \frac{{{a^2}}}{{{b^2}}} + 1 = 2 \times \sqrt 3 \left( {\frac{a}{b}} \right)$

$\Rightarrow \frac{{\left( {{a^2} + {b^2}} \right)}}{{{b^2}}} \times \frac{b}{{2a}} = \sqrt 3 $

$\Rightarrow \frac{{\left( {{a^2} + {b^2}} \right)}}{{2ab}} = \sqrt 3 $

$\frac{{\left( {{a^2} + {b^2}} \right)}}{{2ab}}$ which is a contradiction.

Since, $a,b$ are integers. Therefore,$\frac{{\left( {{a^2} + {b^2}} \right)}}{{2ab}}$ is a rational number but $\sqrt 3 $ is an irrational number. Therefore, it contradicts the fact that $\sqrt 2 + \sqrt 3 $ is rational.

Hence, it is an irrational number.

### EXERCISE 1.3:

1. Show that the square of any positive integer is either of the form $4q$ or $4q + 1$ for some integer$q$.

Ans: Let us consider $a$ be any positive integer

Using Euclid's division lemma $a = bm + r$

$a = 4q + r$,$0 \leqslant r < 4$when$b = 4$

So, $r = 0,1,2,3$

Now, squaring on both sides the equation $a = 4q + r$ in every case

Case. i) when $r = 0$

$a = 4m$

${a^2} = {(4m)^2}$

${a^2} = 4\left( {4{m^2}} \right)$

${a^2} = 4q$

where $q = 4{m^2}$

When $r = 1$

$a = 4m + 1$

${a^2} = {(4m + 1)^2}$

${a^2} = 16{m^2} + 1 + 8m$

${a^2} = 4\left( {4{m^2} + 2m} \right) + 1$

${a^2} = 4q + 1$, where $q = 4{m^2} + 2m$

Case. ii) when $r = 2$

$a = 4m + 2$

${a^2} = {(4m + 2)^2}$

${a^2} = 16{m^2} + 4 + 16m$

${a^2} = 4\left( {4{m^2} + 4m + 1} \right)$

${a^2} = 4q$, Where $q = 4{m^2} + 4m + 1$

Case. iii) when $r = 3$

$a = 4m + 3$

${a^2} = {(4m + 3)^2}$

${a^2} = 16{m^2} + 9 + 24m$

${a^2} = 16{m^2} + 24m + 8 + 1$

${a^2} = 4\left( {4{m^2} + 6m + 2} \right) + 1$

${a^2} = 4q + 1$, where $q = 4{m^2} + 6m + 2$

Therefore, Square of any positive integer is in the form of $4q$ or $4q + 1$, where $q$ is any integer.

2. Show that cube of any positive integer is of the form $4m,4m + 1$ or $4m + 3$, for some integer

Ans: Let us consider $a$ be any positive integer

Using Euclid's division lemma $a = bq + r$

$a = 4q + r$ ,$0 \leqslant r < 4$when$b = 4$

So, $r = 0,1,2,3$

Now, cubing on both sides of $a = 4q + r$in every case

Case 1: When $r = 0$

$a = 4q$

${(a)^3} = {(4q)^3}$

${(a)^3} = 64{q^3} = 4\left( {16{q^3}} \right)$

${a^3} = 4m$ , where $m = 16{q^3}$

Case. 2) when $r = 1$

$a = 4q + 1$

${a^3} = {(4q + 1)^3}$

Since, ${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$

${a^3} = {(4q)^3} + {(1)^3} + 3{(4q)^2} \times 1 + 3(4q) \times {(1)^2}$

${a^3} = 64{q^3} + 1 + 3 \times 16{q^2} + 12q$

${a^3} = 64{q^3} + 1 + 48{q^2} + 12q$

${a^3} = 4\left( {16{q^3} + 12{q^3} + 3q} \right) + 1$

${a^3} = 4m + 1$, where $m = 16{q^3} + 12{q^2} + 3q$

Case. 3) when $r = 2$

$a = 4q + 2$

${a^3} = {(4q + 2)^3}$

${a^3} = {(4q)^3} + {(2)^3} + 3 \times {(4q)^2} \times 2 + 3 \times 4q{\left( 2 \right)^2}$

${a^3} = 64{q^3} + 8 + 3 \times 16{q^2} \times 2 + 3 \times 4q \times 4$

${a^3} = 64{q^3} + 8 + 96{q^2} + 48q$

${a^3} = 4\left( {16{q^3} + 2 + 24{q^3} + 12q} \right)$

${a^3} = 4m$, Where $m = 16{q^3} + 2 + 24{q^3} + 12q$

Case. 4) when $r = 3$

$a = 4q + 3$

${a^3} = {(4q + 3)^3}$

${a^3} = \left( {64{q^3} + 24 + 144{q^2} + 108q} \right) + 3$

${a^3} = 4\left( {16{q^3} + 6 + 36{q^2} + 27q} \right) + 3$

${a^3} = 4m + 3$, where $m = 16{q^3} + 6 + 36{q^2} + 27q$

Therefore, ${a^3}$ is of the form of $4m,4m + 1,4m + 3$.

Hence, the cube of any positive integer is of the form $4m,4m + 1$ or $4m + 3$, for some integer $m$.

3. Show that the square of any positive integer cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

Ans: Let us consider $a$ be any positive integer

Using Euclid's division lemma $a = bm + r$

$a = 5m + r$, $0 \leqslant r < 5$ when $b = 5$

So, $r = 0,1,2,3,4$

Now, Squaring both sides of an equation $a = 6m + r$in every case

Case 1: when$r = 0$

$a = 5m$

${a^2} = 25{m^2}$

${a^2} = 5\left( {5{m^2}} \right)$

${a^2} = 5q,{\text{ where q = }}5{m^2}$

Case 2: when $r = 1$

$a = 5m + 1$

${a^2} = {(5m + 1)^2}$

${a^2} = 25{m^2} + 10m + 1$

${a^2} = 5\left( {5{m^2} + 2m} \right) + 1$

$= 5q + 1$ , where $q = 5{m^2} + 2m$

Case 3: when $r = 2$

$a = 5m + 2$

${a^2} = 25{m^2} + 20m + 4$

${a^2} = 5\left( {5{m^2} + 4m} \right) + 4$

${a^2} = 5q + 4$, where $q = 5{m^2} + 4m$

Case 4: when $r = 3$

$a = 5m + 3$

${a^2} = 25{m^2} + 30m + 9$

$= 25{m^2} + 30m + 5 + 4$

$= 5\left( {5{m^2} + 6m + 1} \right) + 4$

$= 5q + 4$, where $q = 5{m^2} + 6m + 1$

Case 5: when $r = 4$

$a = 5m + 4$

${a^2} = 25{m^2} + 40m + 16$

${a^2} = 25{m^2} + 40m + 15 + 1$

$= 5\left( {5{m^2} + 8m + 3} \right) + 1$

$= 5q + 1$ , Where $q = 5{m^2} + 8m + 3$

From the above cases, $(5q + 2)$ and, $(5q + 3)$ are not perfect squares for some value of $q$.

Hence, the square of any positive integer cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

4. Show that the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer $m$.

Ans: Let us consider $a$ be any positive integer.

Using Euclid's division lemma $a = bm + r$

$a = 6q + r$, $0 \leqslant r < 6$ when $b = 6$

So that $r = 0,1,2,3,4,5$

Now, squaring both sides of an equation $a = 6q + r$ in every case.

Case 1: when $r = 0$

${a^2} = 6\left[ {6{q^2} + 2q \times 0} \right] + {0^2}$

$\Rightarrow {a^2} = 36{q^2}$

$\Rightarrow {a^2} = 6m$, where $m = 6{q^2}$

Case 2: when $r = 1$

${a^2} = 6\left[ {6{q^2} + 2q \times 1} \right] + {1^2}$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 2q} \right] + 1$

$\Rightarrow {a^2} = 6m + 1$, where $m = 6{q^2} + 2q$

Case 3: When $r = 2$

${a^2} = 6\left[ {6{q^2} + 2q \cdot 2} \right] + {2^2}$

$\Rightarrow {a^2} = 6m + 4$, where $m = \left( {6{q^2} + 4q} \right)$

Case 4: When $r = 3$

${a^2} = 6\left[ {6{q^2} + 2q \cdot 3} \right] + {3^2}$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 6q} \right] + 6 + 3$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 6q + 1} \right] + 3$

$\Rightarrow {a^2} = 6m + 3$, where $m = 6{q^2} + 6q + 1$

Case 5: when $r = 4$

${a^2} = 6\left[ {6{q^2} + 2q \cdot 4} \right] + {4^2}$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 8q} \right] + 12 + 4$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 8q + 2} \right] + 4$

$\Rightarrow {a^2} = 6m + 4$, where $m = 6{q^4} + 8q + 2$

Case 6: When $r = 5$

${a^2} = 6\left[ {6{q^2} + 2q \cdot 5} \right] + {5^2}$

$\Rightarrow {a^2} = 6\left[ {6{q^2} + 10q} \right] + 24 + 1$

$\Rightarrow 6\left[ {6{q^2} + 10q + 4} \right] + 1$

$\Rightarrow {a^2} = 6m + 1$, where $m = 6{q^2} + 10q + 4$

Therefore, the form of $6m,(6m + 1),(6m + 3)$ and $(6m + 4)$ are squares and $(6m + 2)$ and $(6m + 5)$ are not perfect square for some value of $m$.

Hence, the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer$m$.

5. Show that the square of any odd integer is of the form $4q + 1$, for some integer $q$.

Ans: Using Euclid’s division lemma method, i.e., $a = bq + r,0 \leqslant r < b$.

Let $b = 4$ then ${\text{a}} = 4{\text{q}} + {\text{r}}$, where $0 \leqslant r < 4$ i.e., ${\text{r}} = 0,1,2,3 \ldots $ (i)

Case 1: If $r = 0 \Rightarrow a = 4q$

$4q$is divisible by $2$.

$4q$is even

Case 2: If $r = 1 \Rightarrow a = 4q + 1$

$(4q + 1)$is not divisible by $2$.

Case 3: If $r = 2 \Rightarrow a = 4q + 2$

$2(2q + 1)$is divisible by $2$

$2(2q + 1)$is even.

Case 4: If $r = 3 \Rightarrow a = 4q + 3$

$(4q + 3)$ is not divisible by $2$.

$(4q + 1)$ and $(4q + 3)$ are odd integers, which are not divisible by $2$.

Now, squaring the odd number.

${a^2} = {(4q + 1)^2}$

Since, $\left[ {{{(a + b)}^2} = {a^2} + 2ab + {b^2}} \right]$

${a^2} = 16{q^2} + 1 + 8q$

${a^2} = 4(4{q^2} + 2q) + 1$

${a^2} = 4(4{q^2} + 2q) + 1$ is a square.

Consider $m = \left( {4{q^2} + 2q} \right)$

Now, it is in the form of $4m + 1$

${a^2} = {(4q + 3)^2}$

Since, $\left[ {{{(a + b)}^2} = {a^2} + 2ab + {b^2}} \right]$

${a^2} = 16{q^2} + 9 + 24q$

${a^2} = 4(4{q^2} + 6q) + 9$

${a^2} = 4(4{q^2} + 6q + 2) + 1$ is a square.

Consider $m = \left( {4{q^2} + 6q + 2} \right)$

Now, it is of the form $4m + 1$.

Therefore, the square of any odd integer is of the form $4m + 1$ for some integer $m$.

6. If $n$ is an odd integer, then show that ${n^2} - 1$ is divisible by $8$.

Ans: As we know, odd numbers can be expressed in form $4q + 1$ or $4q + 3$, where $q$ is an integer.

So, consider $n = 4q + 1$

Then

${n^2} - 1 = {(4q + 1)^2} - 1$

${n^2} - 1 = 16{q^2} + 8q + 1 - 1$

${n^2} - 1 = 8q(2q + 1)$

$8q(2q + 1)$ is divisible by $8$.

Now, consider $n = 4q + 3$

Then

${n^2} - 1 = {(4q + 3)^2} - 1$

${n^2} - 1 = 16{q^2} + 24q + 9 - 1$

${n^2} - 1 = 8\left( {2{q^2} + 3q + 1} \right)$

$8\left( {2{q^2} + 3q + 1} \right)$ is divisible by $8$.

From the above equations,

$n$ is an odd positive integer ${n^2} - 1$ is divisible by $8$.

7. Prove that if $x$ and $y$ are both odd positive integers, then ${x^2} + {y^2}$ is even but not divisible by $4$.

Ans: $x$ and $y$ are both odd positive integers

If$x$ is odd positive integers, then the form of x is $2q + 1$ where $q$ is an integer.

$x = 2q + 1$

Let us consider that $x$ and $y$ are both odd positive integers of the form $2q + 1$, where $q$ is an integer.

So, $x = 2m + 1$ and $y = 2n + 1$ for some other integers of $m$ and $n$.

Now, consider, ${x^2} + {y^2}$ is even but not divisible by $4$.

Substitute the values of $x = 2m + 1$ and $y = 2n + 1$in ${x^2} + {y^2}$.

${x^2} + {y^2} = {(2m + 1)^2} + {(2n + 1)^2}$

${x^2} + {y^2} = 4{m^2} + 1 + 4m + 4{n^2} + 1 + 4n$

${x^2} + {y^2} = 4{m^2} + 4{n^2} + 4m + 4n + 2$

${x^2} + {y^2} = 4\left( {{m^2} + {n^2}} \right) + 4(m + n) + 2$

${x^2} + {y^2} = 4\left\{ {\left( {{m^2} + {n^2}} \right) + (m + n)} \right\} + 2$

${x^2} + {y^2} = 4q + 2$, where $q = \left( {{m^2} + {n^2}} \right) + (m + n)$

By Euclid’s division lemma,

$2$ is the remainder of ${x^2} + {y^2}$which is even when divided by $4$.

Therefore, ${x^2} + {y^2}$ is even but not divisible by $4$

8. Use Euclid's division algorithm to find the HCF of $441,567,693$.

Ans: The Euclidean Algorithm for finding HCF (A, B) is as follows:

If A=0 then HCF (A, B) = B, since the HCF (0, B) = B, and we can stop.

If B = 0 then HCF (A, B) = A, since the HCF (A, 0) = A, and we can stop.

Write A in quotient remainder form (A=BQ+R)

Find HCF (B, R) using the Euclidean Algorithm since HCF (A, B) = HCF(B, R)

Given $441,567,693$

Using Euclid’s division lemma method, we have:

$a = bq + r,0 \leqslant r < b$.

Calculate HCF for $\left( {693,567} \right)$

$693 = 567 \times 1 + 126$

$567 = 126 \times 4 + 63$

$126 = 63 \times 2 + 0$

Since, remainder is $0$

The HCF of $\left( {693,567} \right)$is $63$ .

Now, calculate HCF for $\left( {441,63} \right)$

$441 = 63 \times 7 + 0$

Since, remainder is $0$

Therefore, HCF of $\left( {693,567,441} \right)$ is $63$.

9. Using Euclid’s division algorithm, find the largest number that divides $1251,9377$and $15628$leaving remainders $1, 2$ and $3$respectively.

Ans: The largest number that divides $1251,9377$and $15628$leaving remainders $1,{\text{ }}2$ and $3$respectively.

Subtract the remainders from the given numbers then

1251 - 1 = 1250,

9377 - 2 = 9375

and 15628 - 3 = 15625

We know that:

The largest number that divides $1251,9377$and $15628$leaving remainders $1,{\text{ }}2$ and $3$ respectively is the HCF of $\left( {1251,9377,15628} \right)$.

Using Euclid’s division lemma method, i.e., $a = bq + r,0 \leqslant r < b$

Calculate HCF for $\left( {15625,9375} \right)$

$\Rightarrow 15625 = 9375 \times 1 + 6250$

$\Rightarrow 9375 = 6250 \times 1 + 3125$

$\Rightarrow 6250 = 3125 \times 2 + 0$

Since, remainder is $0$

The HCF of $\left( {15625,9375} \right)$ is $3125$

Now calculate HCF for $\left( {3125,1250} \right)$

$\Rightarrow 3125 = 1250 \times 2 + 625$

$\Rightarrow 1250 = 625 \times 2 + 0$

Since, remainder is $0$

Therefore, HCF of $\left( {1251,9377,15628} \right)$is $625$

$625$ is the required largest number that divides $1251,9377$and $15628$leaving remainders $1,{\text{ }}2$ and $3$respectively.

10. Prove that $\sqrt 3 + \sqrt 5 $ is irrational.

Ans: Let us assume $\sqrt 3 + \sqrt 5 $is a rational number.

So it can be represented in the form of $\frac{a}{b}$

$\Rightarrow \sqrt 3 + \sqrt 5 = \frac{a}{b}$

$\Rightarrow \sqrt 3 = \frac{a}{b} - \sqrt 5 $

Now, Squaring on both sides

$\Rightarrow {\left( {\sqrt 3 } \right)^2} = {\left( {\frac{a}{b} - \sqrt 5 } \right)^2}$

Expand ${\left( {\frac{a}{b} - \sqrt 5 } \right)^2} = {\frac{a}{{{b^2}}}^2} + 5 - 2\left( {\frac{a}{b}} \right)\sqrt 5$

$\Rightarrow 3 = \frac{{{a^2}}}{{{b^2}}} + 5 - 2\sqrt 5 \left( {\frac{a}{b}} \right)$

$\Rightarrow \frac{{{a^2}}}{{{b^2}}} + 5 - 3 = 2\sqrt 5 \frac{a}{b}$

$\Rightarrow \frac{{{a^2}}}{{{b^2}}} + 2 = 2 \times \sqrt 5 \left( {\frac{a}{b}} \right)$

$\Rightarrow \frac{{\left( {{a^2} + 2{b^2}} \right)}}{{{b^2}}} \times \frac{b}{{2a}} = \sqrt 5 $

$\Rightarrow \frac{{\left( {{a^2} + 2{b^2}} \right)}}{{2ab}} = \sqrt{5}$

Here,$\frac{{\left( {{a^2} + 2{b^2}} \right)}}{{2ab}}$ is a rational number but $\sqrt 5 $ is an irrational number. Therefore, it contradicts the fact that $\sqrt 3 + \sqrt 5 $ is rational.

Hence, $\sqrt 3 + \sqrt 5 $ is an irrational number.

11. Show that $1{2^n}$ cannot end with the digit 0 or 5 for any natural number n.

Ans: ${12^n} = {(2 \times 2 \times 3)^n}$

${\text{ }} = {2^n} \times {2^n} \times {3^n}$

Hence, ${12^n}$ cannot end with the digit 0 or 5.

12. On a morning walk, three persons step off together and their step measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

Ans: The minimum distance to walk = LCM (Least Common Multiple) of 40, 42 and 45.

$40 = 2 \times 2 \times 2 \times 5$

$42 = 2 \times 3 \times 7$

$45 = 3 \times 3 \times 5$

LCM = (product of the highest power of each prime factor involved)

$= {2^3} \times {3^2} \times 5 \times 7$

$= 2520$

13. Write the denominator of the rational number$\frac{{257}}{{5000}}$ in the form${2^m} \times {5^n}$ , where m, n are non-negative integers. Hence, write its decimal expansion, without actual division.

Ans: The denominator of the rational number$\frac{{257}}{{5000}}$ is 5000.

$5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5$

${\text{ = }}{2^3} \times {5^4}$

$m = 3,n = 4$

In order to find the decimal expansion for $\frac{{257}}{{5000}}$ we must multiply the numerator and denominator by 2,

$\frac{{257}}{{5000}} = \frac{{257 \times 2}}{{5000 \times 2}}$

${\text{ }} = \frac{{514}}{{{2^3} \times {5^4} \times 2}}$

${\text{ }} = \frac{{514}}{{{{(2 \times 5)}^4}}}$

${\text{ = }}\frac{{514}}{{{{10}^4}}}$

${\text{ }} = 0.0514$

14. Prove that $\sqrt p + \sqrt q $ is irrational, where $p,{\text{ }}q$ are primes.

Ans: Let’s assume that $p$ and $q$ are rational , where $p$ and $q$ are primes.

$p{\text{ }} + {\text{ }}q{\text{ }} = x$ , where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we will get rational numbers on both sides.

$\Rightarrow {(\sqrt p + \sqrt q )^2} = {x^2}$

$\Rightarrow p + 2\sqrt {pq} + q = {x^2}$

$\Rightarrow 2\sqrt {pq} = {x^2} - p - q$

$\Rightarrow \sqrt {pq} = \frac{{({x^2} - p - q)}}{2}$

Here, $\frac{{({x^2} - p - q)}}{2}$ is rational.

But as $p{\text{ and }}q$ are both primes, then $pq$ will not be a perfect square. Therefore, $pq$ is not rational. But this is a contradiction, so our original assumption must be wrong.

So, $p{\text{ and }}q$ are irrational, where$p{\text{ and }}q$ are primes.

### Sample Question

1. Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.

Ans: We know that, any positive integer can be expressed in the form of $6m,6m + 1,6m + 2,6m + 3,6m + 4{\text{ }}or{\text{ }}6m + 5,$ for any integer m.

If we take an odd positive integer it will be of the form$6m + 1,6m + 3or6m + 5$.

Therefore,

${\left( {6m + 1} \right)^2} = 36{m^2} + 12m + 1 = 6(6{m^2} + 2m) + 1 = 6q + 1$

${\left( {6m + 3} \right)^2} = 36{m^2} + 36m + 9 = 6(6{m^2} + 6m + 1) + 3 = 6q + 3$

${\left( {6m + 5} \right)^2} = 36{m^2} + 60m + 25 = 6(6{m^2} + 10m + 4) + 1 = 6q + 1$

Where q is an integer.

Thus, the square of an odd positive integer can be of the form $6q + 1or6q + 3$ .

### EXERCISE 1.4

1. Show that the cube of a positive integer of the form $6q + {\text{ }}r,{\text{ }}q$ is an integer and $r = 0, 1, 2, 3, 4, 5$ is also of the form$6m + r$ .

Ans: Let us assume $6q{\text{ }} + {\text{ }}r$ is a positive integer, where q is an integer and $r = 0,1,2,3,4,5$.

Then, the positive integers are of the form $6q,{\text{ }}6q + 1,{\text{ }}6q + 2,{\text{ }}6q + 3,{\text{ }}6q + 4{\text{ and }}6q + 5$ .

Taking cube of each term, we have,

${\left( {6q} \right)^3} = 216{q^3} = 6 \times 36{q^3} + 0 = 6m,{\text{ }}$ where $m$= $36{q^3}$is an integer

${\left( {6q + 1} \right)^3} = 216{q^3} + 108{q^2} + 18q + 1{\text{ }}$

$= 6\left( {36{q^3} + 18{q^2} + 3q} \right) + 1{\text{ }}$

$= {\text{ }}6m + 1,$ where $m$= $36{q^3} + 18{q^2} + 3q$ is an integer.

## ${\left( {6q + 2} \right)^3} = 216{q^3} + 216{q^2} + 72q + 8{\text{ }}$

## $= {\text{ }}6\left( {36{q^3} + 36{q^2} + 12q + 1} \right) + 2{\text{ }}$

## $= {\text{ }}6m + 2,{\text{ }}$ where $m$= $36{q^3} + 36{q^2} + 12q + 1$is an integer.

## ${\left( {6q + 3} \right)^3} = 216{q^3}{\text{ + }}324{q^2}{\text{ + }}162q + 27$

## $= 6\left( {36{q^3} + 54{q^2} + 27q + 4} \right) + 3{\text{ }}$

## $= {\text{ }}6m{\text{ + }}3,{\text{ }}$ where $m$= $36{q^3} + 54{q^2} + 27q + 4$is an integer.

${\left( {6q + 4} \right)^3} = 216{q^3} + 432{q^2} + 288q + 64{\text{ }}$

$= 6\left( {36{q^3} + 72{q^2} + 48q + 10} \right) + 4$

$= {\text{ }}6m{\text{ }} + {\text{ }}4,{\text{ }}$ where $m$= $36{q^3} + 72{q^2} + 48q + 10$is an integer.

${\left( {6q + 5} \right)^3} = 216{q^3} + 540{q^2} + 450q + 125{\text{ }}$

$= 6\left( {36{q^3} + 90{q^2} + 75q + 20} \right){\text{ + }}5{\text{ }}$

$= {\text{ }}6m{\text{ }} + {\text{ }}5,{\text{ }}$where $m$= $36{q^3} + 90{q^2} + 75q + 20$is an integer

Hence, the cube of a positive integer of the form $6q{\text{ }} + {\text{ }}r,{\text{ }}q$ is an integer and $r = 0,1,2,3,4,5$ is also of the form $6m + r$.

2. Prove that one and only one out of $n, n + 2 and n + 4$ is divisible by 3, where n is any positive integer.

Ans: Using Euclid’s division lemma,

we have $a{\text{ }} = {\text{ }}3q{\text{ }} + {\text{ }}r;{\text{ }}0{\text{ }} \leqslant r{\text{ }} < {\text{ }}b$

For$a{\text{ }} = {\text{ }}n$ , we have

$n{\text{ }} = {\text{ }}3q{\text{ }} + {\text{ }}r$ , ………$\left( i \right)$

where q is an integer and$0{\text{ }} \leqslant {\text{ }}r{\text{ }} < {\text{ }}3$ , i.e.$r{\text{ }} = {\text{ }}0,{\text{ }}1,{\text{ }}2$ .

Putting $r = 0$ in $\left( i \right)$$$ , we get $n = 3q$

So, $n$ is divisible by 3.

$n + 2 = 3q + 2$ so, $n + 2$ is not divisible by 3.

$n + 4 = 3q + 4$ so,$n + 4$ is not divisible by 3.

Putting $r = 1$ in$\left( i \right)$ , we get $n = 3q + 1$ so, n is not divisible by 3$n + 2 = 3q + 3{\text{ }} = 3\left( {q + 1} \right)$ so, $n + 2$ is divisible by 3

$n + 4 = 3q + 5$ so, $n + 4$ is not divisible by 3.

Putting r = 2 in$\left( i \right)$ , we get $n{\text{ }} = {\text{ }}3q{\text{ }} + {\text{ }}2$ so, n is not divisible by 3.

$n{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}3q{\text{ }} + {\text{ }}4$ so,$n{\text{ }} + {\text{ }}2$ is not divisible by 3.

$n + 4 = 3q + 6 = 3\left( {q + 2} \right)$ so, $n + 4$ is divisible by 3.

For every value of r so that$0{\text{ }} \leqslant {\text{ }}r{\text{ }} < {\text{ }}3$ only one out of$n,{\text{ }}n + 2{\text{ and }}n + 4$ is divisible by 3.

3. Prove that one of any three consecutive positive integers must be divisible by 3.

Ans: Let the three consecutive positive integers be$x,x + 1{\text{ and }}x + 2$ , where $x$ is any integer.

By Euclid’s division lemma, we have $c = dq{\text{ }} + {\text{ }}r;{\text{ }}0 \leqslant {\text{ }}r{\text{ }} < d$ For $c = x{\text{ and }}d{\text{ }} = {\text{ }}3$ , we have $x = 3q + r......\left( i \right)$ , Where $q$ is an integer and$0{\text{ }} \leqslant {\text{ }}r{\text{ }} < {\text{ }}3$ ,$i.e.{\text{ }}r{\text{ }} = {\text{ }}0,{\text{ }}1,{\text{ }}2$.

Putting $r{\text{ }} = {\text{ }}0{\text{ in }}\left( i \right)$ , we get

$x = 3q$ So, $x$ is divisible by 3.

$x + 1 = 3q + 1$ so, $x + 1$ is not divisible by 3.

$x + 2 = 3q + 2$ so, $x + 2$ is not divisible by 3.

Putting$r{\text{ }} = {\text{ }}1{\text{ in }}\left( i \right)$ , we get

$x = 3q + 1$ so, $x$ is not divisible by 3.

$x + 1 = 3q + 2{\text{ so}},{\text{ }}x + 1$ is not divisible by 3.

$x + 2 = 3q + 3 = 3\left( {q + 1} \right){\text{ so}},{\text{ }}x + 2$ is divisible by 3.

Putting $r{\text{ }} = {\text{ }}2{\text{ in }}\left( i \right)$ , we get

$x = $$3q{\text{ }} + {\text{ }}2$ so, $x$ is not divisible by 3.

$x + 1 = 3q + 3 = 3\left( {q + 1} \right)$ so, $x + 1$ is divisible by 3.

$x + 2 = 3q + 4$ so, $x + 2$is not divisible by 3.

Thus for each value of r such that $0{\text{ }} \leqslant {\text{ }}r{\text{ }} < {\text{ }}3$ only one out of $x,x + 1{\text{ and }}x + 2$ is divisible by 3.

4. For any positive integer n, prove that ${n^3} - n$ is divisible by 6.

Ans: factorizing

${n^3} - n$ is divisible by 6, the possible reminders are 0, 1 and 2.

[$\because if{\text{ }}P{\text{ }} = {\text{ }}ab{\text{ }} + {\text{ }}r$, then $0{\text{ }} \leqslant {\text{ }}r{\text{ }} < {\text{ }}a$ by Euclid lemma ]

$\therefore {\text{Let }}n{\text{ }} = {\text{ }}3r{\text{ }},{\text{ }}3r{\text{ }} + 1{\text{ }},{\text{ }}3r{\text{ }} + {\text{ }}2$ , where r is an integer

Case 1 :- when $n{\text{ }} = {\text{ }}3r$

Then, is divisible by 3

[ , it is divisible by 3 ]

Case 2 :- when $n{\text{ }} = {\text{ }}3r{\text{ }} + {\text{ }}1$

$n - 1 = 3r + 1 - 1 = 3r$

Then , it is divisible by 3.

Case 3 :- when $n = 3r - 1$

$n + 1 = 3r - 1 + 1 = 3r$

Then , it is divisible by 3.

Therefore is divisible by 3 , where $n$ is any positive integer.

5: Show that one and only one out of $n,{\text{ }}n + 4,{\text{ }}n + 8,{\text{ }}n + 12{\text{ and }}n + 16$ is divisible by 5, where n is any positive integer.

Ans:

$n,{\text{ }}n + 4,{\text{ }}n + 8,{\text{ }}n + 12{\text{ and }}n + 16$ is divisible by 5, where n is any positive integer.

Therefore, Any positive integer is of the form $5q,{\text{ }}5q + 1,{\text{ }}5q + 2,{\text{ }}5q + 3,{\text{ }}5q + 4$

$b{\text{ }} = {\text{ }}5$

$r{\text{ }} = {\text{ }}0{\text{ }},{\text{ }}1{\text{ }},{\text{ }}2{\text{ }},{\text{ }}3{\text{ }},{\text{ }}4$

When $r = 1$,

$n{\text{ }} = {\text{ }}5q{\text{ }} + {\text{ }}n{\text{ }} = {\text{ }}5q{\text{ }} + {\text{ }}1$ [ not divisible by 5]

$n + 4 = 5q + 5{\text{ }} = {\text{ 5}}\left[ {q + 1} \right]$ [ divisible by 5]

$n + 8 = 5q + 9$ [ not divisible by 5]

$n + 6 = 5q + 7$ [ not divisible by 5]

$n{\text{ }} + {\text{ }}12{\text{ }} = {\text{ }}5q{\text{ }} + {\text{ }}13$ [ not divisible by 5]

When $r = 2$

$n = 5q + 2n = 5q + 2$ [ not divisible by 5]

$n + 4 = 5q + 6$ [ not divisible by 5]

$n + 8 = 5q + 10 = 5\left[ {{\text{q + 2 }}} \right]$ [ divisible by 5]

$n + 6 = 5q + 8$ [ not divisible by 5]

$n + 12 = 5q + 14$ [ not divisible by 5]

When $r = 3$

$n = 5q + 3n = 5q + 3$ [ not divisible by 5]

$n + 4 = 5q + 7$ [ not divisible by 5]

$n + 8 = 5q + 11$ [ not divisible by 5]

$n + 6 = 5q + 9$ [ not divisible by 5]

$n + 12 = {\text{5q + 15}} = {\text{5}}\left[ {{\text{q + 3 }}} \right]$ [ divisible by 5]

When $r = 4$,

$n = 5q + 4n = 5q + 4$ [not divisible by 5]

$n{\text{ }} + {\text{ }}4{\text{ }} = {\text{ }}5q{\text{ }} + {\text{ }}8{\text{ }}$ [not divisible by 5]

$n + 8 = 5q + 12$ [not divisible by 5]

$n + 6 = 5q + 10 = {\text{5}}\left[ {{\text{ q + 2 }}} \right]$ [divisible by 5]

$n + 12 = 5q + 16$ [not divisible by 5]

From equations it is clear that one and only one out of n, n+4, n+8, n+12 and n+6 is divisible by 5.

### What's the Difference Between NCERT Solutions and NCERT Exemplar solutions?

The NCERT textbooks essentially set the groundwork for a subject. The Exemplar goes above and beyond the basics, including twists in questions and posing questions at a higher level than the standard NCERT. The National Council for Educational Research and Training (Ncert) publishes the Ncert exemplar. These books are designed for students studying in grades 6 to 12. If you have completed your entire syllabus, practiced all the questions towards the end of every chapter thoroughly, then NCERT Exemplar is the next step to pin the last nail in the coffin.

### Why Choose Vedantu for NCERT Exemplar for Class 10 Maths - Real Numbers?

Vedantu is a knowledge network in which any student can immediately contact a teacher and learn in a tailored, anytime-anywhere manner. With a wide variety of quality teachers, we have curated a step-by-step solution guide for NCERT Exemplar for Class 10 Maths - Real Numbers. The solutions you’ll find in this PDF are solved by our academic professionals for a better understanding of students. By referring to these solutions, one can easily understand the working of the problem and try to assess where he/she is lacking.

### Concepts Included in NCERT Class 10 Maths - Real Numbers

Definition of what Real Numbers mean?

Euclid’s Division Lemma ( Working)

Euclid’s Division Lemma (Proof)

Meaning of the HCF (Highest Common Factor)

What do Prime Numbers mean?

What are Composite Numbers

The explanation for Fundamental Theorem of Arithmetic

Using prime factorization method using HCF & LCM

Meaning and examples of irrational numbers

Proving rational and irrational numbers

### Number of Exercises under NCERT Exemplar for Class 10 Maths - Real Numbers

Exercise 1.1 (10 Questions)

Exercise 1.2 (10 Questions)

Exercise 1.3 (14 Questions)

Exercise 1.4 (5 Questions)

### Tips to Keep in Mind While solving Questions for Class 10 Maths - Real Numbers

For some students, maths can be a comparatively difficult subject to ace. But in reality, it’s the most scoring subject of all time. If you have solved all the questions correctly and have shown proper step-by-step working for the same, then there is no chance that the examiner can deduct your marks. But for those who still struggle in navigating through, here are some tips to look at while solving the problems:

Always have a look at the question twice before beginning to solve it. Sometimes, a job done in a hassle doesn’t reap fruitful results. So, take your time with it and try to understand the problem with an open and calm mind.

Sometimes you might find yourself stuck while solving a problem. Don’t panic. Try going through your solution step-by-step and try to assess what you missed. If you still can’t figure it out, move on to the next question.

After you’re done solving the entire paper, have a look at the solutions once again. A thorough revision of all the work will only make you more confident in your performance.

Always answer the problem with a complete sentence at the end, this way you’ll be emphasizing the solution and make it easier for the examiner to give you full marks.

Start with easy problems. By doing this you’ll finish up the first two sections of your paper very quickly and also you’ll save up time for the comparatively difficult questions.

Maths is not a difficult subject. It just requires a lot of practice. So, if you’re someone who is looking for exercises to solve, visit vedantu and find the resources you require to ace your exam.

## FAQs on NCERT Exemplar for Class 10 Maths Chapter 1 - Real Numbers (Book Solutions)

**1. How many sums are there in the NCERT Class 10 Maths Chapter 1- Real Numbers?**

There are 4 exercises in the NCERT Class 10 Maths Chapter 1- Real Numbers. There are 10 sums in the first exercise, Ex.-1.1, and 10 sums in the second exercise, Ex.-1.2. There are 14 sums in the third exercise, Ex.-1.3, and 5 sums in the fourth exercise, Ex.- 1.4. There are sub-questions for most of the sums in this chapter. All the exercise sums are solved and explained in the NCERT Exemplar for Class 10 Maths Chapter 1- Real Numbers.

**2. What is meant by real numbers?**

Any positive or negative number can be referred to as real numbers. All the natural numbers, integers, and rational numbers are included in the umbrella term of real numbers. The floating-point numbers are the real numbers having decimal points.

Unlike real numbers, imaginary numbers include a factor of √-1, termed as j or i.

**3. Are the NCERT Exemplar for Class 10 Maths Chapter 1- Real Numbers reliable?**

Yes, the NCERT Exemplar for Class 10 Maths Chapter 1- Real Numbers are very reliable. All the sums of these chapters are solved and explained in a detailed stepwise manner by the subject matter experts at Vedantu. These NCERT Solutions are prepared in accordance with the CBSE guidelines for Class 10 Mathematics. So you can rely on these solutions for your exam preparation. Also, you can address all your doubts more effectively when you compare the sums that you worked-out to these NCERT Solutions. These stepwise NCERT Solutions make a comprehensive, self-explanatory study material for revision purposes, before the examination.

**4. Can I download the NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers for free?**

Yes, you can download the NCERT Exemplar for Class 10 Maths Chapter 1- Real Numbers for free from Vedantu. These NCERT book solutions are available in PDF format, so you can download them and study offline.

Our free downloadable NCERT Solutions are accessible for all students with internet connectivity and a digital screen. The main objective of Vedantu is to provide the best study resources to all students across the globe. These free downloadable PDF files of NCERT Solutions guarantee access for all students.

**5. Why should I refer to NCERT Exemplar for Class 10 Maths - Real Numbers?**

The Exemplar focuses on your conceptual clarity. You will notice that the questions in the exemplar differ from the questions asked in the NCERT textual exercises and illustrations in both areas. This certain distinction aids you in adapting to varied asking methods and familiarizing yourself with the kind of twists to expect from your board paper. CBSE frequently asks the same questions as the exemplar. So, if you’re someone who is aiming to score 100/100 in your maths exam, then, the NCERT exemplar is a must.