NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real number Exercise 1.4

Chapter 1 Real Number NCERT Exemplar Solutions Exercise 1.4 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 1 Real Numbers Exercise 1.4
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 1.1 Exercise 1.2 Exercise 1.3
Related Study	NCERT Solutions for Class 10 Maths

Exercise 1.4 Solutions

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.

Solution

We have,
6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5
So, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.
Cube of these integers will be:
Taking 6q,

Therefore, 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.

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. 

Solution

Using Euclid's Algorithm, 
Let the positive integer = n and b = 3. 
n = 3q + r, where q is the quotient and r is the remainder. 
Remainders can be 0, 1 and 2 
So, n can be 3q, 3q + 1, 3q + 2
For n = 3q 
n + 2 = 3q + 2 
n + 4 = 3q + 4
In this case n is only divisible by 3.
For n = 3q + 1 
n + 2 = 3q + 3
In this case n + 2 is divisible by  3.
For n = 3q + 2
n + 2 = 3q + 4
n + 4 = 3q + 2 + 4
= 3q + 6
In this case n + 4 is divisible by 3.
Therefore, we can say that one and only one out of n, n + 2 and n + 4 is divisible by 3.

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

Solution

Let the three consecutive number be n, n+1, n+2.
By Euclid’s algorithm, we have
n = 3q + r
Where q is quotient and r is remainder
The values of r can be = 0, 1, 2.
For r = 0,
n = 3q
n + 1 = 3q + 1
n+2 = 3q+2
So, here n is divisible by 3.
Now, For r = 1,
n = 3q+1
n+1=3q+2
n+2 = 3q+3
So, here n+2 is divisible by 3.
For r = 2,
n = 3q+2
n+1=3q+3
n+2 = 3q+4
So, here n+1 is divisible by 3.
Therefore one of any three consecutive positive integers must be divisible by 3.

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

Solution

Let x = n³ – n
x = n(n² – 1)
x = n(n – 1)(n + 1)
(n – 1), n, (n + 1) are consecutive integers so out of three consecutive numbers at least one will be even.
So, we can say that x is divisible by 2.
Sum of number = (n – 1) + n + (n + 1)
= n – 1 + n + n + 1
= 3n
Here, the sum of three consecutive numbers is divisible by 3, so one of them is divisible by 3.
So, out of n, (n – 1), (n + 1), one is divisible by 2 and one is divisible by 3 and
x = (n – 1) × n × (n + 1)
Hence, out of three factors of x, one is divisible by 2 and one is divisible by 3.
Therefore, x is divisible by 6 or n³ – n is divisible by 6.

5. Show that one and only one out of n, (n + 4), (n + 8), (n + 12), (n + 16) is divisible by 5, where n is any positive integer.
[Hint: Any positive integer can be written in the form 5q, (5q + 1), (5q + 2), (5q + 3), (5q +4)]

Solution

If a number n is divided by 5 then let the quotient is q and remainder is r. Then by Euclid’s algorithm,

n = 5q + r, where n, q, r are positive integers and 0 ≤ r < 5.

For, r = 0,

n = 5q + 0

= 5q

So, n is divisible by 5.

For, r = 1,

n = 5q + 1

n + 4 = (5q + 1) + 4

= 5q + 5

= 5(q + 1) divisible by 5.

We can say that (n + 4) is divisible by 5.

For r = 2,

n = 5q + 2

(n + 8) = (5q + 2) + 8

= 5q + 10 = 5(q + 2)

= 5m is divisible by 5.

We can say that, (n + 8) is divisible by 5.

For, r = 3,

n = 5q + 3

n + 12 = (5q + 3) + 12

= 5q + 15

= 5(q + 3) = 5m is divisible by 5.

We can say that, (n + 12) is divisible by 5.

For, r = 4,

n = 5q + 4

n + 16 = (5q + 4) + 16

= 5q + 20 = 5(q + 4)

(n + 16) = 5m is divisible by 5.

Therefore, n, (n + 4), (n + 8), (n + 12) and (n + 16) are divisible by 5.