NCERT Exemplar Class 10 Maths Solutions Chapter 1 Real Numbers

Question:1

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

Answer: [C]

If the value of m is an odd number, then option A gives an odd number. Therefore, option A is not the answer of a given question. When the value of m is even, then option B is always given an odd number.
For example let m = 2, then m + 1 = 2 + 1 = 3
Which is odd, hence B is not the answer.
When the value of m is odd/even, then option C always gives an even number.
For example let m = 2, then 2 m = 4;
let m = 3, then 2 m = 6
Hence, C is the correct option.
Option D always gives an odd integer for every value of m, hence, option D is not the answer.

Question:2

For some integer q, every odd integer is of the form
(A) q
(B) q + 1
(C) 2q
(D) 2q + 1

Answer. [D]

If the value of q is an even number, then option A gives an even number. Therefore, option A is not the answer to a given question.
When the value of q is odd, then option B always gives an even number.
For example let q = 3, then q + 1 = 3 + 1 = 4
Which is even and hence B is not the correct answer.
When the value of q is odd/even, then 2q always gives an even number.
For example let q = 2, then 2q = 4;
let q = 3, then 2q = 6
Hence, C is the incorrect option.
But 2q + 1 always gives an odd integer for every value of q.
For example, let q = 2 (even), then 2q + 1 = 2 × 2 + 1 $\Rightarrow$ 5,
and at q = 3 (odd), 2q + 1 = 7
Hence, option D is the answer.

Question:3

n² – 1 is divisible by 8 if n is
(A) an integer
(B) a natural number
(C) An odd integer
(D) An even integer

Answer. [C]

For any even natural number or even integer, n² is always even. So, n² - 1 will always be odd. An odd number can never be divisible by 8.
Therefore, options A, B, and D do not satisfy the given condition.
Now, n² - 1 = (n + 1)(n - 1)
If n is an odd integer,
(n + 1) will be even, and also (n - 1) will be even.
The multiplication of two consecutive even integers is always divisible by 8.
We can understand it through an example:
If n = 3, (n + 1) = 4, (n - 1) = 3 – 1 = 2
So, n²– 1 = 4 × 2 = 8, which is divisible by 8.
Hence, the correct answer is option (C).

Question: 4

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

Answer. [B]
Solution. First of all, let us find the HCF of 65 and 117

65 = 1 × 5 × 13
117 = 1 × 3 × 13 × 3
HCF = 13 ….(1)
According to question HCF = 65m – 117
Using (1)
13 = 65 m – 117
⇒ 13 + 117 = 65 m
⇒ $\frac{130}{65}=m$
⇒ m = 2
Therefore, option (b) is correct.

Question:5

The largest number that divides 70 and 125, leaving remainders 5 and 8, respectively, is
(A) 13 (B) 65 (C) 875 (D) 1750

Answer. [A]
Solution. First of all, subtract the remainder from the given number.
(70 – 5) = 65
(125 – 8) = 117
The required number is the HCF of 65 and 117

65 = 5 × 13 × 1
117 = 3 × 3 × 13 × 1
HCF = 13
Hence, 13 is the largest number that divides 70 and 120 & leaves the remainder 5 and 8.
Therefore, option (A) is correct.

Question:6

If two positive integers a and b are written as a = x³ y² and b = xy³; x, y are prime numbers, then HCF (a, b)
(A) xy (B) xy² (C) x³y³ (D) x²y²

Answer. [B]

$a=x^3{y}^2\Rightarrow x\times x\times x\times y\times y$
$b=x{y}^3\Rightarrow x\times y\times y\times y$
For HCF, we have to take all common prime numbers from both numbers.
Hence,
$\text{HCF = }x\times y\times y=x{y}^2$
$\text{HCF = }x{y}^2$
Hence, option B is correct.

Question:7

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

Answer. [C]
Solution.
$p=a{b}^2=a\times b\times b$
$q=a^{3}b=a\times a\times a\times b$
For LCM, we have to take all the primes from the prime factorization of the two numbers.
The common prime in both numbers has to be taken only once.
Therefore,
$\text{LCM =}a\times a\times a\times b\times b=a^3{b}^2$
Hence, option C is correct.

Question:8

The product of a non-zero rational and an irrational number is
(A) always irrational
(B) always rational
(C) Rational or irrational
(D) one

Answer. [A]

A rational number is a number that can be written:
In the form of p/q, where q is not equal to zero. (Here, P and Q have to be integers)
Irrational numbers cannot be written in the form of p/q.
Consider a rational number = 1/2
$\text{Irrational number =}$ $\sqrt{3}$
$\text{Product =}\frac{1}{2}\times \sqrt{3}=\frac{\sqrt{3}}{2}$ $\text{(irrational)}$
Therefore, option (A) is correct.

Question:9

The least number that is divisible by all the numbers from 1 to 10 is:
(A) 10 (B) 504 (C) 100 (D) 2520

Answer. [D]

Option A is not the answer to the question, because 10 is not divisible by 3, 4, 6, 7, 8, 9
Option B is also not the answer to the question, because 504 is not divisible by 5 and 10
Option (C) is also not the answer to the question because 100 is not divisible by 3, 6, 7, 8 and 9
Option (D) is the answer to the question because 2520 is divisible by all the numbers from 1 to 10.
Calculation:
Prime factorization of all the numbers:
1 = 1
2 = 2
3 =3
4 =2$\times$2
5 =5
6 = 3$\times$2
7 = 7
8 = 2$\times$2$\times$2
9 = 3$\times$3
10 = 2$\times$5
For LCM, we have to take all the primes from the prime factorization of the two numbers.
The common prime in both numbers has to be taken only once.
Therefore,
LCM = 2$\times$2$\times$2$\times$3$\times$3$\times$3$\times$5$\times$7 =2520
Therefore, option (D) is correct.

Question:10

The decimal expansion of the rational numbers $\frac{14587}{1250}$ will terminate after:
(A) one decimal place
(B) two decimal places
(C) there decimal places
(D) four decimal places

Answer. [D]
$\text{Rational number =}\frac{14587}{1250}$

$\Rightarrow$ $\frac{14587}{2^1\times {\left(5\right)}^4}$
$=\frac{14587}{2^1\times 5^1\times 5^3}$
$=\frac{14587}{10\times 5^3}\times \frac{2^3}{2^3}$ $\text{{Divide and multiply by 23}}$
$=\frac{116696}{10000}=11.6696$
Hence, the given rational number will terminate after four decimal places. (D) is the correct option.

Question:1

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

Answer. [false]

The above statement is not true in all conditions. For example, if we take the value of q as –1. Then 4q + 2 = – 4 + 2 = – 2, which is not a positive integer.
Moreover, if q is positive, then 4q + 2 will always be even. Therefore, we can never write an odd positive number in this form. Hence, the given statement is false.

Question:2

"The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Justify your answer.

Answer. [True]

The above statement is true in all conditions. Consider a number P = q (q + 1)
If q is odd, then q+1 will be even, hence P will be divisible by 2.
If q is even, then q+1 will be odd, but P will be divisible by 2.
For example:
Let us take numbers 2 and 3:
The product is 2 x 3 = 6, which is divisible by 2
Hence, the given statement is true.

Question:3

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

Answer. [True]

Let the 3 consecutive positive integers be n, n+1, n+2.

Any number divided by 3 leaves the remainder 0,1, or 2.

So, n = 3p, 3p+1 or 3p+2 where p is any integer.

If n = 3p, n is divisible by 3

If n = 3p + 1,

So, n + 2 = 3p + 1 + 2

= 3p + 3

= 3(p + 1) which is divisible by 3

If n = 3p + 2,

So, n + 1

= 3p + 2 + 1

= 3p + 3

= 3(p + 1) which is divisible by 3

Therefore , n(n+1)(n+2) is divisible by 3

When a number is divisible by 2, the remainder obtained is 0 or 1.

n = 2q or 2q + 1, where q is some integer.

If n = 2q

So, n + 2

= 2q + 2

= 2(q + 1) is divisible by 2.

If n = 2q + 1

So, n + 1

= 2q + 1 + 1

= 2q + 2

$\Rightarrow$ 2 (q + 1) is divisible by 2.

Therefore, n(n+1)(n+2) is divisible by 2.

Since n(n+1)(n+2) is divisible by both 2 and 3. It is divisible by 6.

The product of three consecutive positive integers is divisible by 6.

Hence proved.

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

Answer. [false]

Any positive integer can be written in the form of 3q or 3q+1, or 3q + 2.
Since square of 3q = 9q² = 3(3q²) = 3m (where 3q² = m)
Square of (3q+1) = (3q+1) ² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1
(where, m = 3q² + 2q)
Square of (3q+2) = (3q+2) ² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (where, m = 3q²+2q)

Thus, there is not any square of a positive integer that can be written in the form of 3m + 2.
Hence, the given statement is false.

Question:5

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

Answer. [false]

Let us take the square of (3q + 1)
(3q + 1)² = (3q)² + (1)² + 2 × 3q × 1
$=$ 9q² + 1 + 6q
$=$ 9q² + 6q + 1
$=$ 3(3q² + 2q) + 1
$=$ 3m + 1
{where m = 3q² + 2q}
Hence, the square of 3q + 1 (where q is a positive integer) cannot be
written in any form other than 3m + 1
Hence, the given statement is false.

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.

Answer:

The highest common divisor of any two numbers is called HCF.
Here, 75 is the highest common divisor.
We can also find HCF:
Applying Euclid’s division algorithm on 525 and 3000, we get

Hence, 75 is the HCF of the numbers 525 and 3000.

Question:7

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

Answer:

Composite number – a number that is divisible by a number other than one and itself. Here, 3 x 5 x 7 + 7 = 112 and 112 is divisible by 2, 4, 7, 8, 14, 16, 28 and 56. Hence, it is a composite number.

Question:8

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

Answer:

As we know, LCM is a multiple of HCF
$\frac{LCM}{HCF}$has to be an integer.
If the given statement is true:
380 = 18n, where n is any integer.
But there is no such type of n that exists, as we can say that 380 is not divisible by 18.
Hence, two numbers, with 18 as their HCF and 380 as their LCM, do not exist.

Question:9

Without actually performing the long division, find if 987/10500 will have a terminating or non-terminating decimal expansion. Give a reason for your answer.

Answer:

We can also write this number as
:$\frac{987}{10500}=\frac{3\times 7\times 47}{3\times 7\times 2^2\times 5^3}=\frac{47}{2^2\times 5^3}$

= $\frac{47}{2^2\times 5^3}\times \frac{2}{2}=\frac{47\times 2}{5^3\times 2^3}$
= $\frac{94}{{\left(10\right)}^3}=\frac{94}{1000}=0\cdot 094$
Here, we have a finite number of digits after the decimal point.
Hence, it is a 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 p/q? Give a reason.

Answer:

The given decimal number is 327.7881.
327.7081 $=\frac{3277081}{10000}=\frac{p}{q}$
The q is 10000, therefore its prime factorization can be written as:
q = 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
= 2⁴ × 5⁴ = (2 × 5)⁴

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.

Answer:

Any positive integer can be written in the form of 4m or 4m + 1 or 4m + 2, or 4m + 3.
Case 1:
A=4m
$\Rightarrow$(4m)²=16 m²
$=$4(4m²)
$=$4q(Here q=4m²)
Case 2:
A=4m+1
$=$(4m+1)²=(4m)² + 1 + 8m
$=$16m²+8m+1
$=$4(4m²+2m) + 1
$=$4q+1 (here q=4m²+2m)
Case 3:
A = 4m + 2
$=$(4m + 2) ²
$=$16 m² + 4 + 16 m
$=$4 × (4m² + 1 + 4 m)

$=$4q
Here q = (4m² + 1 + 4m)
Case 4:
A = 4m + 3
(4m + 3)² = 16 m² + 9 + 24 m
$=$ 4(4m² + 6m + 2) + 1
$=$ 4q + 1
here q = (4m² + 6m +2)
Hence square of any positive integer is either of the form 4q or 4q + 1.

Question:2

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

Answer:

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is the remainder when we divide A by 4, therefore:
$0\le r$ $<4$ $r=0,1,2,3$
A = 4q + r …(1)
Case 1:
A = 4q
$\Rightarrow$ A³ = (4q)³ = 64q³ = 4(16q³)
$\Rightarrow$ A³ = 4m
(m = 16 q³)
Case 2:
A = 4q + 1
$\Rightarrow$ (4q + 1)³ = (4q)³ + (1)³ + 3(4q)² (1) + 3(4q) (1)²
(Using (a + b)³ = a³+ b³ + 3a²b + 3ab²)
$\Rightarrow$ 64q³ + 1 + 48q² + 12q
$\Rightarrow$ 4(16q³ + 12q² + 3q) + 1
$\Rightarrow$ A³ = 4m + 1
(m = 16q³ + 12q² + 3q)
Case 3:
A = 4q + 2
$\Rightarrow$ (4q + 2)³ = (4q)³ + (2)³ + 3(4q)² (2) + 3(4q) (2)²
(using (a + b)³ = a³ + b³ + 3a²b + 3ab²)
$\Rightarrow$ 64q³ + 8 + 96q² + 48q
$\Rightarrow$ 4(16q³ + 2 + 24q²+ 12q)
$\Rightarrow$ 4m
(m = 16q³ + 2 + 24q² + 12 q)
Case 4:
A = 4q + 3
$\Rightarrow$ A³ = (4q + 3)² = (4q)³ + (3)³ + 3(4q)² (3) + 3(4q) (3)²
(Using (a + b)³ = a³ + b³ + 3a²b + 3ab²)
$\Rightarrow$ 64q³ + 27 + 144q² + 108q
$\Rightarrow$ 64q³ + 24 + 144 q² + 108q + 3
$\Rightarrow$ 4(16q³ + 6 + 36q² + 27q) + 3
$\Rightarrow$ 4m + 3
(m = 16q³ + 6 + 36q² + 27q)
Hence, any positive integer’s cube can be written in the form of
4m, 4m + 1, 4m + 3.

Question:3

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

Answer:

By Euclid’s division –
Any positive integer can be written as:
A = bm + r
Here b = 5
r is remainder when we divide A by 5, therefore:
0 $\le$ r $<$ 5, r = 0, 1, 2, 3.4
A = 5m + r …(1)
Case 1:
A = 5m
$\Rightarrow$ (5m)² = 25 m²
$\Rightarrow$ 5(5m²)
$\Rightarrow$ 5q
(Here q = 5m²)
Case 2:
A = 5m + 1
$\Rightarrow$ (5m + 1)² = 25 m² + 1 + 10 m
$\Rightarrow$ 25m² + 10 m + 1
$\Rightarrow$ 5(5m² + 2m) + 1
= 5q + 1
(Here q = 5m² + 2m)
Similarly, we can verify it for 5m + 2, 5m + 3, 5m + 4
Here, the square of any positive integer is in the form of 5q, 5q + 1
Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Question:4

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

Answer:

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is the remainder when we divide A by 5, therefore:
0 $\le$ r $<$ 6, r = 0, 1, 2, 3,4,5
A = 6q + r …(1)
Case 1:
A = 6q
$\Rightarrow$ A² = (6q)² = 36q² = 6(6q²)
$\Rightarrow$ A² = 6m
(m = 6q²)
Case 2:
A = 6q + 1
$\Rightarrow$ A² = (6q + 1)²
$\Rightarrow$ A² = 36q² + 1 + 2(6q)
(Using (a + b)² = a² + b² + 2ab)
$\Rightarrow$ A² = 36q² + 1 + 12q
$\Rightarrow$ A² = 6(6q² + 2q) + 1
$\Rightarrow$ A² = 6m + 1
(m = 6q² + 2q)
Similarly, we can verify it for 6q + 2, 6q + 3, 6q + 4, 6q + 5.
Now, it is clear that the square of any positive integer cannot be of the form 6m + 2, 6m + 5.

Question:5

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

Answer:

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 4
r is remainder when we divide A by 4, therefore:
0 $\le$ r $<$ 4, r = 0, 1, 2, 3
Even: A = 4q and 4q + 2
Odd: A = 4q + 1 and 4q + 3
Case 1:
Now, squaring the odd terms, we get
(4q + 1)² = (4q) ² + (1)² + 8q
$\Rightarrow$ 16q² + 8q + 1
$\Rightarrow$ 4(4q² + 2q) + 1
$\Rightarrow$ 4m + 1
Here m = 4q² + 2q.
Case 2:
(4q + 3)² = (4q)² + (3)² +24q
$\Rightarrow$ 16q² + 8 + 1 + 24q

$\Rightarrow$ 16q² + 24q + 8 + 1
$\Rightarrow$ 4(4q² + 6q + 2) + 1
$\Rightarrow$ 4m + 1
Here, m = 4q² + 6q + 2.
Hence, we can say that the square of any odd integer is of the form 4q + 1 for some integer q.

Question:6

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

Answer:

We can represent any odd positive integer in the form of (4q + 1) or (4q + 3) for some integer q.
Case 1:
If n = 4q + 1
$\Rightarrow$ n² – 1 = (4q + 1)² –1
$\Rightarrow$ n²– 1 = (4q)² + (1)² + 2(4q) (1) – 1
Using (a + b)² = a² + b² + 2ab
$\Rightarrow$ n² – 1 = 16q² + 1 + 8q – 1
$\Rightarrow$ n² – 1 = 16q² + 8q = 8q (2q + 1)
Which is divisible by 8.
Case 2:
If n = 4q + 3
$\Rightarrow$ n² – 1 = (4q + 3)² – 1
$\Rightarrow$ n² – 1 = (4q)² + (3)² + 2(4q) (3) – 1
(Using (a + b)² = a² + b² + 2ab)
$\Rightarrow$ n² – 1 = 16q² + 9 + 24q –1
$\Rightarrow$ n² – 1 = 16q² + 8 + 24q = 8(2q² + 1 + 3q)
Which is divisible by 8.
Hence, any odd positive integer n, when written in n² – 1 form, is divisible by 8.

Question:7

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

Answer:

We can represent any positive integer in the form of (2q +1) for any integer q.
Let x = 2q + 1; y = 2p + 1
$\Rightarrow$ x² + y² = (2q + 1)² + (2p + 1)²
$\Rightarrow$ x² + y² = (2q)² + (1)² + 2(2q) (1) + (2p)² + (1)² + 2(2p) (1)
(Using (a + b) ² = a² + b + 2ab)
$\Rightarrow$ x² + y² = 4q² + 1 + 4q + 4p² + 1 + 4p
$\Rightarrow$ x² + y² = 2(2q² + 2q + 2p² + 2p + 1)
(2q² + 2q + 2p² + 2p + 1) is odd
Hence, x² + y² is divisible by 2 but not divisible by 4.

Question:8

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

Answer:

The Euclidean Algorithm to determine the HCF (A, B) is:

If A = 0, then HCF (A, B) = B,

As HCF (0, B) = B, we can stop.

If B = 0, then HCF (A, B) = A,

As HCF (A, 0) = A, we can stop.

Now let us write A in quotient remainder form, i.e A = BQ + R

By using the Euclidean Algorithm as HCF (A, B) = HCF(B, R), we can determine the HCF (B, R)

We know that, HCF of 441 and 567 is

567 = 441 × 1 + 126

441 = 126 × 3 + 63

126 = 63 × 2 + 0

Remainder is 0,

Therefore, H.C.F of (441, 567) is = 63.

H.C.F of 63 and 693 is 693 = 63 × 11 + 0

Therefore, H.C.F of (441, 567, 693) = 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.

Answer:

As we know that 1, 2 and 3 are the remainders of the given numbers 1251, 9377 and 15628
After subtracting the remainder, we get
1251 – 1 = 1250
9377 – 2 = 9375
15628 – 3 = 15625
Now HCF of 1250, 9375, 15625 can be calculated as:
First, we will calculate the HCF of 9375 and 15625;

HCF (15625, 9375) = 3125. Now we will calculate the HCF of 3125 and 1250.

HCF of 3125 and 1250 is 625. Here HCF of (1250, 9375, 15625) = 625.
Hence, the largest number is 625.

Question:10

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

Answer:

We will do it by the method of contradiction: We will assume $\sqrt{3}+\sqrt{5}$ is a rational number. If it leads to some absurd outcome, then it is a wrong assumption.
Let $\sqrt{3}+\sqrt{5}$ is a rational number
$\therefore \sqrt{3}+\sqrt{5}=\frac{a}{b},b\ne 0,a,bϵz$
$\sqrt{5}=\frac{a}{b}-\sqrt{3}$
Squaring both sides.
$\Rightarrow$ ${\left(\sqrt{5}\right)}^2={(\frac{a}{b}-\sqrt{3})}^2$
$\Rightarrow$ $5={\left(\frac{a}{b}\right)}^2+{\left(\sqrt{3}\right)}^2-2\left(\frac{a}{b}\right)\left(\sqrt{3}\right)$
(Using (a – b)² = a² + b² = 2ab)
$\Rightarrow$ ${\left(\frac{a}{b}\right)}^2+3-2\left(\frac{a}{b}\right)\left(\sqrt{3}\right)=5$
$\Rightarrow$ $\frac{a^2}{b^2}+3-5=2\sqrt{3}\left(\frac{a}{b}\right)$
$\Rightarrow$ $\frac{a^2-2b^2}{b^2}=2\sqrt{3}\left(\frac{a}{b}\right)$
$\Rightarrow$ $\frac{b(a^2-2b^2)}{2a{b}^2}=\sqrt{3}$
$\Rightarrow$ $\frac{a^2-2b^2}{2ab}=\sqrt{3}$
LHS is a rational number, but RHS is an irrational number.
This is not possible, hence, our assumption was wrong.
Hence, $\sqrt{3}+\sqrt{5}$ is a irrational number.

Question:11

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

Answer:

We know that if any number ends up with digit 0 or 5, then it must be divisible by 5.
Here 12ⁿ = (2 × 2 × 3)ⁿ
$=$12ⁿ = (2² × 3)ⁿ
$=$12ⁿ = (2²ⁿ × 3ⁿ)
is not there, in the prime factorization form.
Hence, 12ⁿ can’t end with the digit 0 or 5.

Question:12

On a morning walk, three people 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?

Answer:

Step length of first person = 40 cm
Step length of second person = 42 cm
Step length of third person = 45 cm
For finding the minimum distance which will be multiple of all step lengths. We have to find the LCM of (40, 42, 45)

Prime factorization forms: 40 = 2 × 2 × 2 × 5 × 1
42 = 2 × 3 × 7 × 1
45 = 5 × 3 × 3 × 1
LCM = 2 × 3 × 5 × 2 × 2 × 7 × 3 = 2520
Hence, the minimum distance is 2520 cm.

Question:13

Write the denominator of the rational number 257/5000 in the form 2^m × 5ⁿ, where m and n are non-negative integers. Hence, write its decimal expansion, without actual division.

Answer:

Here, denominator = 5000
It can be written as:
5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5
Hence the given fraction $\frac{257}{2^3\times 5^4}$ can be written as:
$\Rightarrow \frac{257}{(2^3\times 5^4)5^1}$
$\Rightarrow \frac{257}{1000\times 5^1}\times \frac{2^1}{2^1}$ (On dividing & multiply by 21)
$\Rightarrow \frac{514}{10000}$
$\Rightarrow 0\cdot 0514$
Hence, it is a terminating decimal.

Question:14

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

Answer:

We will do it by the method of contradiction: We will assume $\sqrt{p}+\sqrt{q}$ is a rational number. If it leads to some absurd outcome, then it is a wrong assumption.
Let $\sqrt{p}+\sqrt{q}$ is a rational number
$\sqrt{p}+\sqrt{q}=\frac{a}{b},b\ne a,a,bϵz$
$=$$\sqrt{q}=\frac{a}{b}-\sqrt{p}$
Squaring both sides
$=$${\left(\sqrt{q}\right)}^2={(\frac{a}{b}-\sqrt{p})}^2$
$=$$q={\left(\frac{a}{b}\right)}^2+{\left(\sqrt{p}\right)}^2-2\left(\frac{a}{b}\right)\left(\sqrt{p}\right)$
(Using (a – b)² = a² + b² – 2ab)
$=$$\frac{a^2}{b^2}+p-q=2\sqrt{p}\frac{a}{b}$
$=$$2\sqrt{p}\frac{a}{b}=\frac{a^2}{b^2}+p-q$
$=$$\sqrt{p}=\frac{b(a^2+b^2(p-q))}{2a{b}^2}$
$=$$\sqrt{p}=\frac{a^2+b^2(p-q)}{2ab}$
RHS is a rational number, but LHS is an irrational number.
This is not possible, hence, our assumption was wrong.
Here $\sqrt{p}+\sqrt{q}$ has to be irrational.

Question:1

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

Answer:

By Euclid’s division –
Any positive integer can be written as:
A = bq + r
Here b = 6
r is remainder when we divide A by 5, therefore:
0 $\le$ r $<$ 6, r = 0, 1, 2, 3,4,5
A = 6q + r
Here A = 6q + r
A³ = (6q + r)³ = 216q³ + r³ + 3.6q.r (6q + r)
[$\because$ (a + b)³ = a³ + b³ + 3ab (a + b)]
A³ = (216q³ + 108q²r + 18qr²) + r³ …(1)
Put r = 0, 1, 2, 3, 4, 5
Case 1:
A = 6q
$\Rightarrow$ A³ = 216q³
$\Rightarrow$ A³ = 6 (36q³)
$\Rightarrow$ A³ = 6m

{m = 36q³}
Case 2:
A = 6q + 1
$\Rightarrow$ A³ = (216q³ + 108q² + 18q) + 1

$\Rightarrow$ A³ = 6(36q³ + 18q² + 3q) + 1
$\Rightarrow$ A³ = 6m + 1
{m = 36q³ + 18q² + 3q}
Case 3:
A = 6q + 2
$\Rightarrow$ A³ = (216q³ + 216q³ + 72q) + 8
$\Rightarrow$ A³ = (216q³ + 216q² + 72q + 6) + 2
$\Rightarrow$ A³ = 6m + 2

{m = 216q³ + 216q² + 72q + 6}
Case 4:
A = 6q + 3
$\Rightarrow$ A³= 216q³ + 324q² + 162q + 24 + 3
$\Rightarrow$ A³ = 6(36q³ + 54q² + 27q + 4) + 3
$\Rightarrow$ A³ = 6m + 3
{m = 36 q³ + 54q² + 27q + 4}
Case 5:
A = 6q + 4
$\Rightarrow$ A³ = (216 q³ + 432 q² + 288q) + 64
$\Rightarrow$ A³ = 6(36q³ + 72q² + 48 q) + 60 + 4
$\Rightarrow$ A³ = 6(36q³ + 72q² + 48q + 10) + 4
$\Rightarrow$ A³ = 6m + 4
{m = 36q³ + 72q² + 48q + 10}
Case 6:
A = 6q + 5
$\Rightarrow$ A³ = (216 q³ + 540 q² + 450 q) + 125
$\Rightarrow$ A³ = 216 q³ + 540 q² + 450 q + 120 + 5
$\Rightarrow$ A³ = 6 (36q³ + 90q² + 75 q + 20) + 5
$\Rightarrow$ A³ = 6m + 5
(where m = 36 q³ + 90q² + 75q + 20)
Hence, the cube of a positive integer of the form 6q + r, where q is an integer, can be expressed in the form 6m + r, where r = 0, 1, 2, 3, 4 and 5.

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.

Answer:

Any positive integer can be written in the form of 3m, 3m + 1 or 3m + 2
Case-1:
n = 3m
3m is divisible by 3
Adding two on both sides
n + 2 = 3m + 2
Here, when n + 2 is divided by 3, it leaves the remainder of 2
Adding four on both sides
n + 4 = 3m + 4
$\Rightarrow$ 3m + 3 + 1
$\Rightarrow$ 3(m + 1) + 1
Here on dividing by 3 leaves the remainder of 1
In this case n is divisible by 3 but (n + 2) and (n + 4) are not divisible by 3
Case-2 :
n = 3m + 1
On dividing by 3, it leaves the remainder of 1
Adding two on both sides
$\Rightarrow$ n + 2 = 3m + 1 + 2
$\Rightarrow$ n + 2 = 3m + 3
$\Rightarrow$ 3(m + 1)
On dividing by 3, it leaves the remainder of 0
Hence, it is divisible by 3.
adding four on both sides
$\Rightarrow$ x + 4 = 3m + 1 + 4
$\Rightarrow$ x + 4 = 3m + 3 + 2
$\Rightarrow$ 3(m + 1) + 2
On dividing by 3, it leaves the remainder of 2.
In this, (n + 2) is divisible by 3, but n and (n + 4) are not divisible by 3.
Hence, through both cases, we conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.
Similarly, we can show for n = 3m + 2 form.

Question:3

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

Answer:

Let three consecutive integers are n, n + 1, n + 2 when n is a natural number
i.e., n = 1, 2, 3, 4, ….
for n = 1, chosen numbers : (1, 2, 3), and 3 is divisible by 3
for n = 2, chosen numbers : (2, 3,4), and 3 is divisible by 3
for n = 3, chosen numbers: (3, 4, 5), and 3 is divisible by 3
for n = 4, chosen numbers : (4, 5, 6), and 6 is divisible by 3
for n = 5, (chosen numbers : (5, 6, 7), and 6 is divisible by 3
Proof:
Case 1: Let n be divisible by 3; it means n can be written as :
n = 3m,
Now, n +1 = 3m + 1 ; it gives remainder 1, when divided by 3
Now, n +2 = 3m + 2; it gives remainder 2, when divided by 3
Case 2: Let n + 1 be divisible by 3, it means n +1 can be written as :
n +1 = 3m,
Now, n = 3m – 1= 3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Now, n +2 = 3m + 1; it gives remainder 1, when divided by 3
Case 3: Let n + 2 be divisible by 3, it means n +2 can be written as :
n +2 = 3m,
Now, n = 3m – 2= 3(m – 1) + 1 ; it gives remainder 1, when divided by 3
Now, n +1 = 3m – 1=3(m – 1) + 2 ; it gives remainder 2, when divided by 3

Question:4

For any positive integer n, prove that n³–n is divisible by 6.

Answer:

Let three consecutive positive integers be x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
.$\therefore$n = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
$\Rightarrow$ x (x + 1) (x + 2) is divisible by 3.
Divisibility by 2: Whenever a number is divided by 2, the remainder obtained is 0 or 1.
$\therefore$ n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.$\Rightarrow$
x (x + 1) (x + 2) is divisible by 2.
Since x (x + 1) (x + 2) is divisible by 2 and 3.
$\therefore$x (x + 1) (x + 2) is divisible by 6.
Now the given number is:
$\Rightarrow$ P = n³ – n
$\Rightarrow$ P = n (n² – 1)
$\Rightarrow$ P = n (n – 1) (n + 1)
$\Rightarrow$ P = (n – 1) n (n + 1)
Therefore, P is the product of three consecutive numbers.
Now, P can be written as:
P = x (x + 1) (x + 2), which is divisible by 6.
Hence, P is divisible by 6.
Hence proved.

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.

Answer:

By Euclid’s division –
Any positive integer can be written as:
n = bm + r
Here b = 5
r is remainder when we divide n by 5, therefore:
0 $\le$ r $<$ 5, r = 0, 1, 2, 3.4
n = 5m + r, therefore n can have values:
n = 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4.
Here, m is a natural number
Case 1: Let n be divisible by 5; it means n can be written as :
n = 5m,
Now, n +4 = 5m + 4;it gives remainder 4, when divided by 5
Now, n + 8 = 5m + 8= 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 12 = 5m + 12= 5(m +2) +2; it gives remainder 2, when divided by 5
Now, n + 16 = 5m + 16= 5(m +3) +1; it gives remainder 1, when divided by 5
Case 2: Let n + 4 is divisible by 5, it means n + 4 can be written as :
n + 4 = 5m,
Now, n = 5m – 4 = 5(m – 1) + 1 ; it gives remainder 1, when divided by 5
Now, n + 8 = 5m + 4; it gives remainder 4, when divided by 5
Now, n + 12 = 5m + 8 = 5(m +1) +3; it gives remainder 3, when divided by 5
Now, n + 16 = 5m + 12 = 5(m +2) +2; it gives remainder 2, when divided by 5
Similarly, we can show for other cases.