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#### Question 1:

Choose the correct answer from the given four options
For some integer m, every even integer is of the form
(A) m
(B) m + 1
(C) 2m
(D) 2m + 1

An integer is said to be even if it is divisible by 2.

Let m be any integer. Then,
m = ..., −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, ...
so m can be even or not even.
Hence, 'm' cannot always be even.

Now, + 1 = ..., −4, −3, −2, −1, 0, 1, 2, 3, 4, ...
Here, 'm+1' cannot always be even.

Now consider 2m =..., −10, −8, −6, −4, −2, 0, 2, 4, 6, 8, 10, ...
Here, '2m' is always even.

Consider 2+ 1 = ..., −9, −7, −5, −3, −1, 1, 3, 5, 7, 9, 11, ....
Here, '2m+1' is always an odd integer.

Hence, the correct answer is option C.

#### Question 2:

Choose the correct answer from the given four options
For some integer q, every odd integer is of the form
(A) q
(B) q + 1
(C) 2q
(D) 2q + 1

An integer is said to be odd if it is not divisible by 2.

Let q be any integer i.e q = ..., −2, −1, 0, 1, 2, ...

Now multiplying both sides by 2,
2q = ..., −4, −2, 0, 2, 4, ...

2q + 1 = ..., −3, −1, 1, 3, 5, ...

Thus, for any integer q, (2q +1) is always odd.

Hence, the correct answer is option  D.

#### Question 3:

Choose the correct answer from the given four options
n2 – 1 is divisible by 8, if n is
(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer

Let a n2 − 1.
Here, n can be odd or even.

Case 1: When n = even. Let n = 4k, where k is any integer.
$⇒a={\left(4k\right)}^{2}-1$
$⇒a=16{k}^{2}-1$
which is not divisible by 8.

Case 2: When n = odd. Let n = 4k + 1, where k is any integer.

$⇒a=16{k}^{2}+1+8k-1\phantom{\rule{0ex}{0ex}}⇒a=16{k}^{2}+8k$
$⇒a=8\left(2{k}^{2}+k\right)$
which is always divisible by 8.

Thus, if n is odd, then ${n}^{2}-1$ is divisible by 8.

Hence, the correct answer  is option C.

#### Question 4:

Choose the correct answer from the given four options
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is
(A) 4
(B) 2
(C) 1
(D) 3

By Euclid's division algorithm,
b = aq + r, where

$⇒117=65×1+52\phantom{\rule{0ex}{0ex}}⇒65=52×1+13\phantom{\rule{0ex}{0ex}}⇒52=13×4+0$

∴ HCF(65,117) = 13         .....(1)

Also, given that HCF(65,117) = 65m − 117        .....(2)

From (1) and (2), we get:

$65m-117=13\phantom{\rule{0ex}{0ex}}⇒65m=130\phantom{\rule{0ex}{0ex}}⇒m=2$

Hence, the correct answer is option B.

#### Question 5:

Choose the correct answer from the given four options
The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively is
(A) 13
(B) 65
(C) 875
(D) 1750

Firstly, we subtract the remainders 5 and 8 respectively from corresponding numbers and then get the HCF of resulting numbers using Euclid's division algorithm to get the required number.

After subtracting these remainders from the numbers we have:
(70 − 5) = 65
(125 − 8) = 117

Now, required number is HCF(65,117).

Using Euclid's division algorithm,
$117=65×1+52\phantom{\rule{0ex}{0ex}}65=52×1+13\phantom{\rule{0ex}{0ex}}52=13×4+0$

∴  HCF = 13
Thus, 13 is the largest number which divides 70 and 125 leaving remainders 5 and 8.

Hence, the correct answer is option A.

#### Question 6:

Choose the correct answer from the given four options
If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is
(A) xy
(B) xy2
(C) x3y3
(D) x2y2

Given:
$a={x}^{3}{y}^{2}=x×x×x×y×y\phantom{\rule{0ex}{0ex}}b=x{y}^{3}=x×y×y×y$

The HCF is the largest common factor of two or more numbers.

Here,
HCF(a, b) = HCF(x3y2, xy3)
$=x×y×y\phantom{\rule{0ex}{0ex}}=x{y}^{2}$

Hence, the correct answer is option B.

#### Question 7:

Choose the correct answer from the given four options
If two positive integers p and q can be expressed as p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is
(A) ab
(B) a2b2
(C) a3b2
(D) a3b3

Given that,
$p=a{b}^{2}=a×b×b\phantom{\rule{0ex}{0ex}}q={a}^{3}b=a×a×a×b$

The LCM is the smallest positive integer that is divisible by both the given numbers.

Here,
LCM(pq) = LCM(ab2, a3b)
$=a×b×b×a×a\phantom{\rule{0ex}{0ex}}={a}^{3}{b}^{2}$

Hence, the correct answer is option C.

#### Question 8:

Choose the correct answer from the given four options
The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) rational or irrational
(D) one

Consider the rational number as $\frac{7}{2}$ and the irrational number as $\frac{\sqrt{5}}{2}$.
Their product is given as:

$\frac{7}{2}×\frac{\sqrt{5}}{2}=\frac{7\sqrt{5}}{4}$,which is irrational.

Hence, the correct answer is option A.

#### Question 9:

Choose the correct answer from the given four options
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(A) 10
(B) 100
(C) 504
(D) 2520

The LCM is the smallest positive integer that is divisible by the given numbers.

Here, we need to find LCM of the numbers from 1 to 10 (both inclusive).
Factors of 1 to 10 numbers:
$\begin{array}{rcl}1& =& 1\\ 2& =& 1×2\\ 3& =& 1×3\\ 4& =& 1×2×2\\ 5& =& 1×5\\ 6& =& 1×2×3\\ 7& =& 1×7\\ 8& =& 1×2×2×2\\ 9& =& 1×3×3\\ 10& =& 1×2×5\end{array}$

∴ LCM of numbers from 1 to 10 = LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
⇒ LCM = $1×2×2×2×3×3×5×7=2520$

Hence, the correct answer is option D.

#### Question 10:

Choose the correct answer from the given four options
The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after:
(A) one decimal place
(B) two decimal places
(C) three decimal places
(D) four decimal places

For terminating decimal expansion of a rational number, the denominator is of the form ${2}^{m}×{5}^{n}$.
Here,
$\frac{14587}{1250}=\frac{14587}{{2}^{1}×{5}^{4}}$

$⇒\frac{14587}{10×{5}^{3}}×\frac{{2}^{3}}{{2}^{3}}=\frac{14587×8}{10×1000}=\frac{116696}{10000}=11.6696\phantom{\rule{0ex}{0ex}}$

Thus, the decimal expansion of the given rational number terminates after four decimal places.

Hence, the correct answer is option D.

#### Question 1:

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

By Euclid's division lemma.
b = aq + r , where b is any positive integer and 0 ≤ a.

Now,  if we divide b by 4, we get
b = 4q + r for 0 ≤ < 4.
∴ r = 0, 1, 2, 3

So, b can be in the form of 4q, 4+ 1, 4+ 2, 4+ 3.

Hence, every positive integer cannot be of the form 4+ 2.

#### Question 2:

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

Let the two consecutive positive integers be n and + 1.

Now, the product of two consecutive positive integers will be n(+ 1).
Thus, the following two cases are possible:

Case 1:
If n is even, then + 1 will be odd.
i.e. if n = 2, then + 1 = 3.
This implies that n(+ 1) will always be even, since one of these two numbers is always divisible by 2.

Case 2:
If n is odd, then+ 1 will be even.
i.e. if = 1, then+ 1 = 2.
This implies that n(+ 1) will always be even, since one of these two numbers is always divisible by 2.

Hence, we can conclude that product of two consecutive positive integers is divisible by 2.

#### Question 3:

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

Let the three consecutive positive integers be n, + 1 and + 2.

Now, the product of three consecutive positive integers will be n(+ 1)(+ 2).

Case 1:
If n is even, then + 1 will be odd and + 2 will be even.
i.e. if n = 2, then + 1 = 3 and n + 2 = 4.
This implies that n(+ 1)(+ 2) will always be divisible by 2 and 3.

Case 2:
If n is odd, then n + 1 will be even and + 2 will be odd.
i.e. if = 1, then n + 1 = 2 and + 2 = 3.
This implies that n(+ 1)(+ 2) will always be divisible by 2 and 3.

Now, any number divisible by 2 and 3 is also divisible by 6.
Hence, we can conclude that product of three consecutive positive integers is divisible by 6.

#### Question 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.

By Euclid's division lemma,
b = aq + r and 0 ≤ a, where b can be any positive integer.

Now, if we divide b by 3, we get
b = 3q + r for 0 ≤ < 3

Now, b can be written in the form of 3q, 3q + 1, 3q + 2.

Squaring each possible form of b, there are three possible cases:

Case I:
If b = 3q, then

Case II:
If b = 3q + 1, then

Case III:
If b = 3q + 2, then

So, the square of any positive integer can be of the form 3m and 3+ 1.
Hence, square of any positive number cannot be of the form 3+ 2.

#### Question 5:

A positive integer is of the form 3q + 1, 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.

By Euclid's division lemma,
b = aq + r, 0 ≤ a, where b can be any positive integer.

Now, dividing b by 3, we get
b = 3q + r and 0 ≤ < 3

Thus, b can be written in the form of 3q, 3q + 1, 3q + 2.

Now, squaring 3+ 1,

Hence, the square of the form 3+ 1 can be written as 3+ 1 always. Thus, it cannot be written in the form of 3m or 3+ 2.

#### Question 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.

By Euclid's division algorithm, we can find the HCF(525, 3000).

$\begin{array}{rcl}& ⇒& 3000=525×5+375\\ & ⇒& 525=375×1+150\\ & ⇒& 375=150×2+75\\ & ⇒& 150=75×2+0\end{array}$

Now, the numbers 3, 5, 15, 25 and 75 are the common factors of 525 and 3000, but the highest common factor is 75.
So, we can say that HCF(525, 3000) is 75.

#### Question 7:

Explain why 3 × 5 × 7 + 7 is a composite number.

A number which has more than two factors is known as composite number.

Here,
$3×5×7+7=105+7=112$
$⇒112=2×2×2×2×7={2}^{4}×7$

So, it is the product of prime factors 2 and 7. i.e. it has more than two factors.

Hence, it is a composite number.

#### Question 8:

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

We know,
LCM × HCF = Product of two numbers

The HCF will always be the factor of LCM.

Since, 18 is not factor of 380.

Therefore, we can say that two numbers cannot have 18 as their HCF and 380 as their LCM.

#### Question 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.

In terminating decimal expansion of a rational number, the denominator is always of the form 2m × 5n.

$\begin{array}{rcl}& ⇒& \frac{987}{10500}=\frac{47}{500}\\ & =& \frac{47}{{5}^{3}×{2}^{2}}\end{array}$

Since, the denominator is of the form 2m × 5n, where m = 2 and n = 3.
Hence, the rational number has terminating decimal expansion.

#### Question 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.

Given that the number 327.7081 is a terminating decimal number. So, it represents a rational number and also its denominator must be of the form 2m × 5n.

Thus,
$327.7081=\frac{3277081}{10000}=\frac{p}{q}$
$\begin{array}{rcl}& ⇒& q={10}^{4}\\ & =& {2}^{4}×{5}^{4}={\left(2×5\right)}^{4}\end{array}$

Hence, the prime factors of q are 2 and 5.

#### Question 1:

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

By Euclid's division lemma,
aq r and 0 ≤ a, where b is any positive integer.

Now, if we divide b by 4, then b can written in the form of 4m, 4+ 1,4+ 2, 4+ 3.
Thus, four cases are possible:

Case I:
If b = 4m, then squaring both sides,
$\begin{array}{rcl}{b}^{2}& =& {\left(4m\right)}^{2}\\ & =& 16{m}^{2}\\ & =& 4\left(4{m}^{2}\right)\end{array}$
$⇒{b}^{2}=4q$, where q is any positive integer.

Case II:
If b = 4+ 1, then squaring both sides,

Case III:
If b = 4+ 2, then squaring both sides,

Case IV:
If b = 4+ 3, then squaring both sides,

Hence, the square of any positive integer is either of the form 4q or 4+ 1 for some integer.

#### Question 2:

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

By Euclid's division lemma,
aq and 0 ≤ a, where b is any positive integer.

Now, if we divide b by 4, then b can written in the form of 4q, 4+ 1, 4+ 2, 4+ 3.
Thus, four cases are possible:

Case I:
If b = 4q, then taking cube both sides,

Case II:
If b = 4+ 1, then taking cube both sides

Case III:
If b = 4q + 2, then taking cube both sides,

Case IV:
If b = 4q + 3, then taking cube both sides,

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

#### Question 3:

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

By Euclid's division lemma,
aq r and 0 ≤ a, where b is any positive integer.

Now, if b is divided by 5, then b can written in the form of 5m, 5+ 1, 5+ 2, 5+ 3, 5+ 4.
Thus, five cases are possible:

Case I:
If b = 5m, then squaring both sides,

Case II:
If b = 5+ 1, the squaring both sides,

Case III:
If b = 5+ 2, then squaring both sides,

Case IV:
If b = 5+ 3, then squaring both sides,

Case V:
If b = 5+ 4, then squaring both sides,

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

#### Question 4:

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

By Euclid's division lemma,
aq r and 0 ≤ a, where b is any positive integer.

Now, if b is divided by 6, then b can written in the form of 6q, 6+ 1, 6+ 2, 6+ 3, 6+ 4, 6+ 5.
Thus, six cases are possible:

Case I:
If b = 6q, then squaring both sides,

Case II:
If b = 6+ 1, then squaring both sides,

Case III:
If b = 6+ 2, then squaring both sides,

Case IV:
If b = 6+ 3, then squaring both sides,

Case V:
If b = 6+ 4, then squaring both sides,

Case VI:
If b = 6+ 5, then squaring both sides,

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

#### Question 5:

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

By Euclid's division lemma,
aq r and 0 ≤ a, where b is any positive integer.

Now, if b is divided by 4, then b can written in the form of 4q, 4+ 1, 4+ 2, 4+ 3.
Now, 4q and 4+ 2 are even integers, since they are divisible by 2.
Hence, the forms 4q and 4+ 2 are ignored.

Thus, two cases are possible:

Case I:
If b = 4+ 1, the squaring both sides,

Case II:
If b = 4+ 3, then squaring both sides,

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

#### Question 6:

If n is an odd integer, then show that n2 – 1 is divisible by 8.

By Euclid's division lemma,
aq r and 0 ≤ a, where n is any positive integer.

If n is divided by 4, then n can written in the form of 4q, 4+ 1, 4+ 2, 4+ 3.
Now, 4q and 4+ 2 are even integers, since they are divisible by 2.
Hence, the forms 4q and 4+ 2 are ignored.

Thus, two cases are possible:

Case I:
If n = 4+ 1, then
$\begin{array}{rcl}{n}^{2}-1& =& {\left(4q+1\right)}^{2}-1\\ & & =16{q}^{2}+1+8q-1\phantom{\rule{0ex}{0ex}}=8q\left(2q+1\right)\end{array}$
which is clearly divisible by 8.

Case II:
If n = 4+ 3, then
$\begin{array}{rcl}{n}^{2}-1& =& {\left(4q+3\right)}^{2}-1\\ & =& 16{q}^{2}+9+24q-1\\ & =& 16{q}^{2}+8+24q\\ & =& 8\left(2{q}^{2}+3q+1\right)\end{array}$
which is clearly divisible by 8.

Hence, ${n}^{2}-1$ is divisible by 8 if n is odd integer.

#### Question 7:

Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

Let m be any positive integer.
Now, 2m will always be even integer.
This implies that 2+ 1 and 2+ 3 will always be odd integer.

Let x = 2+ 1 and y = 2+ 3 are odd positive integers, for every positive integer m.

Thus, x2y2 can be written as:
$\begin{array}{rcl}{x}^{2}+{y}^{2}& =& {\left(2m+1\right)}^{2}+{\left(2m+3\right)}^{2}\\ & =& 4{m}^{2}+1+4m+4{m}^{2}+9+12m\\ & =& 8{m}^{2}+16m+10\\ & =& 2\left(4{m}^{2}+8m+5\right)\end{array}$
which is always even, but not always divisible by 4.

Hence, x2y2 is even for every positive integer m but not divisible by 4.

#### Question 8:

Use Euclid’s division algorithm to find the HCF of 441, 567, 693.

Let = 693, = 567 and= 441.

By Euclid's division algorithm,
= br and 0 ≤ b, where a is any positive integer.

First, consider a and b and find their HCF.
$⇒693=567×1+126\phantom{\rule{0ex}{0ex}}⇒567=126×4+63\phantom{\rule{0ex}{0ex}}⇒126=63×2+0\phantom{\rule{0ex}{0ex}}$

∴ HCF(693, 567) = 63.

Now, we take c and HCF(693, 567), then find their HCF.
This implies that using Euclid's division algorithm,
$⇒441=63×7+0$

∴ HCF(693, 567, 441) = 63.

#### Question 9:

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

Given that, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively.
After subtracting these remainders from the numbers,
$1251-1=1250\phantom{\rule{0ex}{0ex}}9377-2=9375\phantom{\rule{0ex}{0ex}}15628-3=15625$
which will be divisible by required number.

Thus, the required number will be HCF(1250, 9375, 15625).

First, we take the largest numbers, 15625 and 9375 and find their HCF.
Using Euclid's division algorithm,
$⇒15625=9375×1+6250\phantom{\rule{0ex}{0ex}}⇒9375=6250×1+3125\phantom{\rule{0ex}{0ex}}⇒6250=3125×2+0\phantom{\rule{0ex}{0ex}}$

∴ HCF(15625, 9375) = 3125.

Now, we take 1250 and HCF(15625, 9375) and find their HCF, we get
Using Euclid's division algorithm,
$⇒3125=1250×2+625\phantom{\rule{0ex}{0ex}}⇒1250=625×2+0$

∴ HCF(1250, 9375, 15625) = 625.

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3 respectively.

#### Question 10:

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

Let us suppose that $\sqrt{3}+\sqrt{5}$ is rational.

Thus, $\sqrt{3}+\sqrt{5}=a$, where a is rational number.
$⇒\sqrt{3}=a-\sqrt{5}$

Squaring both sides,
${\left(\sqrt{3}\right)}^{2}={\left(a-\sqrt{5}\right)}^{2}$
$⇒3={a}^{2}+5-2\sqrt{5}a\phantom{\rule{0ex}{0ex}}⇒2\sqrt{5}a={a}^{2}+2$

∴ $\sqrt{5}=\frac{{a}^{2}+2}{2a}$, which is of the form $\frac{p}{q}$.

Therefore, it is a contradiction since the right hand side is rational number while $\sqrt{5}$ is irrational.
Hence, our assumption is wrong and $\sqrt{3}+\sqrt{5}$ is irrational number.

#### Question 11:

Show that 12n cannot end with the digit 0 or 5 for any natural number n.

Any number ending with 0 or 5 should always be divisible by 5.
This implies that 12n should end with the digit zero, if it must be divisible by 5.
Now, this is possible only if prime factorisation of 12n contains the prime number 5.

$⇒12=2×2×3={2}^{2}×3\phantom{\rule{0ex}{0ex}}⇒{12}^{n}={\left({2}^{2}×3\right)}^{n}={2}^{2n}×{3}^{n}$
Now, there is no term that contains 5.

Hence, we conclude there is no value of n for which 12n ends with digit 0 or 5.

#### Question 12:

On a morning walk, three persons step off together and their steps 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?

Here, we require the minimum distance each should walk.
This implies that we need to find least possible distance, so that each can cover the same distance in each steps.
Hence, we have to find LCM of 40 cm, 42 cm and 45 cm.

Now,
40 = $2×2×2×5$
42 = $2×3×7$
45 = $5×3×3$

Thus, LCM(40, 42, 45) is given as:
$=2×3×5×2×2×3×7\phantom{\rule{0ex}{0ex}}=30×12×7\phantom{\rule{0ex}{0ex}}=210×12\phantom{\rule{0ex}{0ex}}=2520$

Hence, minimum distance each should walk is 2520 cm so that each can cover the same distance in compete steps.

#### Question 13:

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

The denominator of the rational number $\frac{257}{5000}$ is 5000.
Factors of 5000 = $2×2×2×5×5×5×5={2}^{3}×{5}^{4}$, which is of the type 2m ⨯ 5n, where = 3 and = 4 are non-negative integers.

$\begin{array}{rcl}\therefore \frac{257}{5000}& =& \frac{257}{{2}^{3}×{5}^{4}}×\frac{2}{2}\\ & =& \frac{514}{{2}^{4}×{5}^{4}}\\ & =& \frac{514}{{10}^{4}}\\ & =& \frac{514}{10000}\\ & =& 0.0514\end{array}$

Hence, the decimal expansion of the rational number $\frac{257}{5000}$ is 0.0514.

#### Question 14:

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

Let us suppose $\sqrt{p}+\sqrt{q}$ is rational.

So, $\sqrt{p}+\sqrt{q}=a$, where is rational number.
$⇒\sqrt{q}=a-\sqrt{p}$

Squaring both sides,
${\left(\sqrt{q}\right)}^{2}={\left(a-\sqrt{p}\right)}^{2}$
$⇒$$q={a}^{2}+p-2a\sqrt{p}$

∴ $\sqrt{p}=\frac{{a}^{2}+p-q}{2a}$ which is of the form .
Thus, it is a contradiction as the right hand side is rational while $\sqrt{p}$ is irrational, where p and q are prime numbers.
Hence, $\sqrt{p}+\sqrt{q}$ is irrational number.

#### Question 1:

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

By Euclid's division lemma,
aq and 0 ≤ a, where b is any positive integer.

If b is divided by 6, then b can written in the form of 6q, 6+ 1, 6+ 2, 6+ 3, 6+ 4, 6+ 5.
Thus, six cases are possible:

Case I:
If b = 6q, then taking cube both sides,

Case II:
If b = 6+ 1, then taking cube both sides,

Case III:
If b = 6+ 2, then taking cube both sides,

Case IV:
If b = 6+ 3, then taking cube both sides,

Case V:
If b = 6+ 4, then taking cube both sides,

Case VI:
If b = 6+ 5, then taking cube both sides,

Hence, the cube of a positive integer of the form 6r, where q is an integer and = 0, 1, 2, 3, 4, 5 is also of the form 6m, 6+ 1, 6+ 2, 6+ 3, 6+ 4, 6+ 5 i.e. 6r.

#### Question 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.

By Euclid's division lemma,
aq r and 0 ≤ a, where n is any positive integer.

If n is divided by 3, then n can written in the form of 3q, 3+ 1, 3+ 2.
Thus, there are three possible cases:

Case I:
If n = 3q, then n is divisible by 3.
However, n + 2 and n + 4 are not divisible by 3.

Case II:
If n = 3q + 1, then
n + 2 = 3q + 3
= 3(q + 1), which is divisible by 3.
However, n and n + 4 are not divisible by 3.

Case III:
If n = 3q + 2, then
n + 4 = 3q + 6
= 3(q + 2), which is divisible by 3.
However, n and n + 2 are not divisible by 3.

Hence, one and only one out of n, n + 2 and n + 4 is divisible by 3.

#### Question 3:

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

Let n, n + 1, n + 2 be three consecutive positive integers.
By Euclid's division lemma,
n = aq and 0 ≤ a, where n is any positive integer.

If n is divided by 3, then n can be written in the form of 3q, 3+ 1, 3+ 2.
Thus, there are three possible cases:

Case I:
If = 3q, then n is only divisible by 3.
However, + 1 and n + 2 are not divisible by 3.

Case II:
If n = 3q + 1, then
n + 2 = 3q + 3
= 3(q + 1), which is only divisible by 3.
However, n and n + 1 are not divisible by 3.

Case III:
If n = 3q + 2, then
n + 1 = 3q + 3
= 3(q + 1), which is only divisible by 3.
However, n and n + 2 are not divisible by 3.

Hence, one and only one out of nn + 1 and n + 2 is divisible by 3.

#### Question 4:

For any positive integer n, prove that n3n is divisible by 6.

Let $a={n}^{3}-n$, where is any positive integer.
$⇒a=n\left({n}^{2}-1\right)\phantom{\rule{0ex}{0ex}}⇒a=n\left(n-1\right)\left(n+1\right)$

We know that,
I. If a number is completely divisible by 2 and 3, then it is also divisible by 6.
II. If the sum of digits of any number is divisible by 3, then it is divisible by 3.
III. If a number is an even number, then it is divisible by 2.

Now, sum of digits $=n-1+n+n+1=3n$, which is a multiple of 3.

Also, product of the numbers = $n\left(n-1\right)\left(n+1\right)$.
This is always be even as they are consecutive positive integers and at least one of them is even.

Therefore, conditions II and III are completely satisfied.
Hence, using condition I, the number ${n}^{3}-n$ is divisible by 6.

#### Question 5:

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Given numbers are n, (n + 4), (n + 8), (n + 12) and n + 16.
By Euclid's division lemma,
aq r and 0 ≤ a, where n is any positive integer.

If n is divided by 5, then n can written in the form of 5q, 5+ 1, 5+ 2, 5+ 3, 5+ 4.
Thus, there are five possible cases:

Case I:
If n = 5q, then n is only divisible by 5.
However (n + 4), (n + 8), (+ 12) and (n + 16) are not divisible by 5.

Case II:
If n = 5+ 1, then
n + 4 = 5+ 1 + 4
= 5q + 5
= 5(q + 1), which is divisible by 5.
However n, (n + 8), (+ 12) and (n + 16) are not divisible by 5.

Case III:
If  n = 5+ 2, then
n + 8 = 5+ 2 + 8
= 5+ 10
= 5(+ 2), which is divisible by 5.
However n, (n + 4), (+ 12) and (n + 16) are not divisible by 5.

Case IV:
If  n = 5+ 3, then
n + 12 = 5+ 3 + 12
= 5+ 15
= 5(+ 3), which is divisible by 5.
However n, (n + 4), (+ 8) and (n + 16) are not divisible by 5.

Case V:
​If  n = 5+ 4, then
n + 16 = 5+ 4 + 16
= 5+ 20
= 5(+ 4), which is divisible by 5.
However n, (n + 4), (+ 8) and (n + 12) are not divisible by 5.

Hence, one and only one out of n, (n + 4), (n + 8), (+ 12) and (n + 16) is divisible by 5

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