CBSE Class 10 Maths: Real Numbers NCERT Solutions

The correct option is (C) 2m.

An even integer is defined as any integer that is exactly divisible by 2. This means that an even integer must be a multiple of 2. Therefore, for any integer 'm', an even integer can always be expressed in the form $2m$. For example, if m=1, the even integer is 2; if m=2, the even integer is 4, and so on.

The correct option is (D) $2q\; +\; 1$.

An odd integer is an integer that is not exactly divisible by 2. When any integer is multiplied by 2, the result is an even integer. If we add 1 to any even integer, the result is an odd integer. Therefore, for any integer 'q', every odd integer can be expressed in the form $2q\; +\; 1$.

The correct option is (C) an odd integer.

Let's test the divisibility of n^2 - 1 by 8 for both even and odd values of n.

Case 1: n is an even integer.

If n is even, we can write n = 2x, where x is an integer.

Substituting this into the expression: n^2 - 1 = (2x)^2 - 1 = 4x^2 - 1.

If x = 1, 4(1)^2 - 1 = 3, which is not divisible by 8.

If x = 2, 4(2)^2 - 1 = 15, which is not divisible by 8.

Thus, when n is even, n^2 - 1 is not always divisible by 8.

Case 2: n is an odd integer.

If n is odd, we can write n = 2x + 1, where x is an integer.

Substituting this into the expression: n^2 - 1 = (2x + 1)^2 - 1.

Expanding the square: (2x + 1)^2 - 1 = (4x^2 + 4x + 1) - 1 = 4x^2 + 4x.

We can factor out 4x: 4x(x + 1).

Now, consider the term x(x+1). This represents the product of two consecutive integers. One of these consecutive integers must be even. Therefore, their product x(x+1) is always divisible by 2.

So, n^2 - 1 = 4 \times (\text{an even number}) = 4 \times (2k) = 8k, where k is an integer.

This shows that n^2 - 1 is always divisible by 8 when n is an odd integer.

Therefore, n^2 - 1 is divisible by 8 if n is an odd integer.

First, we need to find the Highest Common Factor (HCF) of 65 and 117 using the Euclidean Algorithm.

Step 1: Divide 117 by 65.

117 = 1 \times 65 + 52

Step 2: Divide 65 by the remainder 52.

65 = 1 \times 52 + 13

Step 3: Divide 52 by the remainder 13.

52 = 4 \times 13 + 0

The last non-zero remainder is 13. So, HCF(65, 117) = 13.

Now, we are given that the HCF is expressible in the form 65m - 117. We set the HCF equal to this expression:

13 = 65m - 117

To find the value of m, we rearrange the equation:

13 + 117 = 65m

130 = 65m

Now, divide both sides by 65:

m = \frac{130}{65}

m = 2

Thus, the value of m is 2.