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Class 10 Advanced Maths Chapter 3.2 Solutions: Arithmetic of Integers | SEBA

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SEBA Class 10 Advanced Maths Chapter 3.2 Solutions: Arithmetic of Integers 

In this article, we have solved all the questions from Chapter 3.2 of SEBA Class 10 Advanced Mathematics in a simple, uniform, step-by-step manner.

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Chapter 3.1 of SEBA Class 10 Advanced Mathematics

Chapter 3.2 of SEBA Class 10 Advanced Mathematics

Chapter 3.3 of SEBA Class 10 Advanced Mathematics

Chapter 3.4 of SEBA Class 10 Advanced Mathematics

Chapter 3.5 of SEBA Class 10 Advanced Mathematics


SEBA Class 10 Advanced Maths Chapter 3.2 Solutions

Q1. Show that the square of an odd integer is expressible in the form $3k$ or $3k+1$.

Solution:

Let us consider any integer $a$.

Applying the division algorithm to the integers $a$ and 3, we get

$$a = 3q + r, \quad \text{where } 0 \le r < 3 \text{ and } q, r \in \mathbb{Z}$$

Thus $a = 3q$, $3q+1$, or $3q+2$, since $r = 0,1,2$.

If $a = 3q$, then

$$a^2 = (3q)^2$$
$$a^2 = 9q^2$$
$$a^2 = 3(3q^2)$$
$$a^2 = 3k, \quad \text{where } k = 3q^2 \in \mathbb{Z}$$

If $a = 3q + 1$, then

$$a^2 = (3q + 1)^2$$
$$a^2 = 9q^2 + 6q + 1$$
$$a^2 = 3(3q^2 + 2q) + 1$$
$$a^2 = 3k + 1, \quad \text{where } k = 3q^2 + 2q \in \mathbb{Z}$$

If $a = 3q + 2$, then

$$a^2 = (3q + 2)^2$$
$$a^2 = 9q^2 + 12q + 4$$
$$a^2 = 9q^2 + 12q + 3 + 1$$
$$a^2 = 3(3q^2 + 4q + 1) + 1$$
$$a^2 = 3k + 1, \quad \text{where } k = 3q^2 + 4q + 1 \in \mathbb{Z}$$

Thus the square of an integer can be expressed in the form $3k$ or $3k+1$.

Q2. Show that the cube of an integer can be expressed in the form $9k$, $9k+1$, or $9k+8$.

Solution:

Let us consider any integer $a$.
Applying the division algorithm to the integers $a$ and 3, we get

$a = 3q + r, \quad \text{where } 0 \le r < 3 \text{ and } q, r \in \mathbb{Z}$

Thus $a= 3q$, $3q+1$, or $3q+2$, sice $ r= 0,1,2 $

If $a = 3q$, then

$a^3 = (3q)^3 $
$a^3 = 27q^3 $
$a^3 = 9(3q^3)$
$a^3 = 9k, \quad \text{where } k = 3q^3 \in \mathbb{Z}$

If $a = 3q + 1$, then

$a^3 = (3q + 1)^3$
$a^3 = 27q^3 + 27q^2 + 9q + 1$
$a^3 = 9(3q^3 + 3q^2 + q) + 1$
$a^3 = 9k + 1, \quad \text{where } k = 3q^3 + 3q^2 + q \in \mathbb{Z}$

If $a = 3q + 2$, then
$a^3 = (3q + 2)^3$
$a^3 = 27q^3 + 54q^2 + 36q + 8$
$a^3 = 9(3q^3 + 6q^2 + 4q) + 8$
$a^3 = 9k + 8, \quad \text{where } k = 3q^3 + 6q^2 + 4q \in \mathbb{Z}$

The cube of an integer can be expressed in the form $9k$, $9k+1$, or $9k+8$.

Q3. For any $n \in \mathbb{N}$, prove that $\frac{n(n+1)(2n+1)}{6}$ is an integer.

Solution: Let us consider any natural number $n$. Applying the division algorithm to the integers $n$ and 6, we get

$$n = 6k + r, \quad \text{where } 0 \le r < 6 \text{ and } k, r \in \mathbb{Z} \text{ with } k \ge 0$$

Thus $n = 6k$, $6k+1$, $6k+2$, $6k+3$, $6k+4$, or $6k+5$, since $r = 0,1,2,3,4,5$.

If $n = 6k$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{6k(6k+1)(12k+1)}{6}$$
$$= k(6k+1)(12k+1) \in \mathbb{Z}$$

If $n = 6k + 1$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+1)(6k+2)(12k+3)}{6}$$
$$= \frac{(6k+1) \cdot 2(3k+1) \cdot 3(4k+1)}{6}$$
$$= \frac{6(6k+1)(3k+1)(4k+1)}{6}$$
$$= (6k+1)(3k+1)(4k+1) \in \mathbb{Z}$$

If $n = 6k + 2$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+2)(6k+3)(12k+5)}{6}$$
$$= \frac{2(3k+1) \cdot 3(2k+1) \cdot (12k+5)}{6}$$
$$= \frac{6(3k+1)(2k+1)(12k+5)}{6}$$
$$= (3k+1)(2k+1)(12k+5) \in \mathbb{Z}$$

If $n = 6k + 3$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+3)(6k+4)(12k+7)}{6}$$
$$= \frac{3(2k+1) \cdot 2(3k+2) \cdot (12k+7)}{6}$$
$$ = \frac{6(2k+1)(3k+2)(12k+7)}{6}$$
$$= (2k+1)(3k+2)(12k+7) \in \mathbb{Z}$$

If $n = 6k + 4$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+4)(6k+5)(12k+9)}{6}$$
$$= \frac{2(3k+2) \cdot (6k+5) \cdot 3(4k+3)}{6}$$
$$= \frac{6(3k+2)(6k+5)(4k+3)}{6}$$
$$= (3k+2)(6k+5)(4k+3) \in \mathbb{Z}$$

If $n = 6k + 5$, then

$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+5)(6k+6)(12k+11)}{6}$$
$$= \frac{(6k+5) \cdot 6(k+1) \cdot (12k+11)}{6}$$
$$ = (6k+5)(k+1)(12k+11) \in \mathbb{Z}$$

Thus for any $n \in \mathbb{N}$, $\frac{n(n+1)(2n+1)}{6}$ is an integer.

Q4. For any $n \in \mathbb{N}$, show that $n(7n^2+5)$ is expressible in the form $6k$, $k \in \mathbb{N}$.

Solution: Let us consider any natural number $n$. Applying the division algorithm to the integers $n$ and 6, we get

$$n = 6q + r, \quad \text{where } 0 \le r < 6 \text{ and } q, r \in \mathbb{Z} \text{ with } q \ge 0$$

Thus $n = 6q$, $6q+1$, $6q+2$, $6q+3$, $6q+4$, or $6q+5$, since $r = 0,1,2,3,4,5$.

If $n = 6q$, then

$$n(7n^2 + 5) = 6q[7(6q)^2 + 5]$$
$$= 6q(252q^2 + 5)$$
$$= 6k, \quad \text{where } k = q(252q^2 + 5) \in \mathbb{N}$$

If $n = 6q + 1$, then

$$n(7n^2 + 5) = (6q+1)[7(6q+1)^2 + 5]$$
$$ = (6q+1)[7(36q^2 + 12q + 1) + 5]$$
$$ = (6q+1)[252q^2 + 84q + 7 + 5]$$
$$ = (6q+1)[252q^2 + 84q + 12]$$
$$= 6(6q+1)(42q^2 + 14q + 2)$$
$$ = 6k, \quad \text{where } k = (6q+1)(42q^2 + 14q + 2) \in \mathbb{N}$$

If $n = 6q + 2$, then

$$n(7n^2 + 5) = (6q+2)[7(6q+2)^2 + 5]$$
$$ = (6q+2)[7(36q^2 + 24q + 4) + 5]$$
$$ = (6q+2)[252q^2 + 168q + 28 + 5]$$
$$ = (6q+2)[252q^2 + 168q + 33]$$
$$ = 2(3q+1) \cdot 3(84q^2 + 56q + 11)$$
$$ = 6(3q+1)(84q^2 + 56q + 11)$$
$$= 6k, \quad \text{where } k = (3q+1)(84q^2 + 56q + 11) \in \mathbb{N}$$

If $n = 6q + 3$, then

$$n(7n^2 + 5) = (6q+3)[7(6q+3)^2 + 5]$$
$$ = (6q+3)[7(36q^2 + 36q + 9) + 5]$$
$$ = (6q+3)[252q^2 + 252q + 63 + 5]$$
$$ = (6q+3)[252q^2 + 252q + 68]$$
$$ = 3(2q+1) \cdot 2(126q^2 + 126q + 34)$$
$$ = 6(2q+1)(126q^2 + 126q + 34)$$
$$= 6k, \quad \text{where } k = (2q+1)(126q^2 + 126q + 34) \in \mathbb{N}$$

If $n = 6q + 4$, then

$$n(7n^2 + 5) = (6q+4)[7(6q+4)^2 + 5]$$
$$ = (6q+4)[7(36q^2 + 48q + 16) + 5]$$
$$ = (6q+4)[252q^2 + 336q + 112 + 5]$$
$$= (6q+4)[252q^2 + 336q + 117]$$
$$= 2(3q+2) \cdot 3(84q^2 + 112q + 39)$$
$$= 6(3q+2)(84q^2 + 112q + 39)$$
$$= 6k, \quad \text{where } k = (3q+2)(84q^2 + 112q + 39) \in \mathbb{N}$$

If $n = 6q + 5$, then

$$n(7n^2 + 5) = (6q+5)[7(6q+5)^2 + 5]$$
$$= (6q+5)[7(36q^2 + 60q + 25) + 5]$$
$$ = (6q+5)[252q^2 + 420q + 175 + 5]$$
$$= (6q+5)[252q^2 + 420q + 180]$$
$$ = 6(6q+5)(42q^2 + 70q + 30)$$
$$ = 6k, \quad \text{where } k = (6q+5)(42q^2 + 70q + 30) \in \mathbb{N}$$

Thus for any $n \in \mathbb{N}$, $n(7n^2+5)$ is expressible in the form $6k$, $k \in \mathbb{N}$.

Q5. If $n$ is any odd integer then show that $n^4+4n^2+11$ is expressible in the form $16k$ where $k \in \mathbb{Z}$.

Solution: Let us consider any odd integer $n$. Applying the division algorithm to the integers $n$ and 4, we get

$$n = 4q + r, \quad \text{where } 0 \le r < 4 \text{ and } q, r \in \mathbb{Z}$$

Since $n$ is an odd integer, $r$ can only take the odd values 1 or 3. Thus $n = 4q + 1$ or $4q + 3$.

If $n = 4q + 1$, then

$n^4 + 4n^2 + 11 = (4q+1)^4 + 4(4q+1)^2 + 11$
$ = (256q^4 + 256q^3 + 96q^2 + 16q + 1) + 4(16q^2 + 8q + 1) + 11$
$ = 256q^4 + 256q^3 + 96q^2 + 16q + 1 + 64q^2 + 32q + 4 + 11$
$ = 256q^4 + 256q^3 + 160q^2 + 48q + 16$
$ = 16(16q^4 + 16q^3 + 10q^2 + 3q + 1)$
$ = 16k, \quad \text{where } k = 16q^4 + 16q^3 + 10q^2 + 3q + 1 \in \mathbb{Z}$

If $n = 4q + 3$, then

$n^4 + 4n^2 + 11 = (4q+3)^4 + 4(4q+3)^2 + 11$
$ = (256q^4 + 768q^3 + 864q^2 + 432q + 81) + 4(16q^2 + 24q + 9) + 11$
$ = 256q^4 + 768q^3 + 864q^2 + 432q + 81 + 64q^2 + 96q + 36 + 11$
$ = 256q^4 + 768q^3 + 928q^2 + 528q + 128$
$ = 16(16q^4 + 48q^3 + 58q^2 + 33q + 8)$
$ = 16k, \quad \text{where } k = 16q^4 + 48q^3 + 58q^2 + 33q + 8 \in \mathbb{Z}$

Thus, if $n$ is any odd integer, then $n^4+4n^2+11$ is expressible in the form $16k$ where $k \in \mathbb{Z}$.

Q6. Show that an integer and its cube give the same remainder when divided by 6.

Solution:

Let us consider any integer $a$.

Applying the division algorithm to the integers $a$ and 6, we get

$$a = 6q + r, \quad \text{where } 0 \le r < 6 \text{ and } q, r \in \mathbb{Z}$$

Thus $a = 6q$, $6q+1$, $6q+2$, $6q+3$, $6q+4$, or $6q+5$, since $r = 0,1,2,3,4,5$.

The difference between the cube of the integer and the integer itself is given by $a^3 – a$.

If $a = 6q$, then

$$a^3 – a = (6q)^3 – 6q$$

$$ = 216q^3 – 6q$$

$$ = 6(36q^3 – q)$$

$$ = 6m, \quad \text{where } m = 36q^3 – q \in \mathbb{Z}$$, which is divisible by 6

If $a = 6q + 1$, then

$$a^3 – a = (6q + 1)^3 – (6q + 1)$$

$$ = (216q^3 + 108q^2 + 18q + 1) – 6q – 1$$

$$ = 216q^3 + 108q^2 + 12q$$

$$ = 6(36q^3 + 18q^2 + 2q)$$

$$ = 6m, \quad \text{where } m = 36q^3 + 18q^2 + 2q \in \mathbb{Z}$$, which is divisible by 6

If $a = 6q + 2$, then

$$a^3 – a = (6q + 2)^3 – (6q + 2)$$

$$ = (216q^3 + 216q^2 + 72q + 8) – 6q – 2$$

$$ = 216q^3 + 216q^2 + 66q + 6$$

$$ = 6(36q^3 + 36q^2 + 11q + 1)$$

$$ = 6m, \quad \text{where } m = 36q^3 + 36q^2 + 11q + 1 \in \mathbb{Z}$$, which is divisible by 6.

Similarly, we can show that $a^3 – a$ is divisible by $6q+3$, $6q+4$,  $6q+5$

Since $a^3 – a = 6m$, it means $a^3 – a$ is  divisible by 6. Therefore, $a^3$ and $a$ must leave the exact same remainder when divided by 6.

Thus, an integer and its cube give the same remainder when divided by 6.

Q7. Show that the difference of a number and its square is always an even integer.

Solution:

Let us consider any integer $a$.

Applying the division algorithm to the integers $a$ and 2, we get

$$a = 2q + r, \quad \text{where } 0 \le r < 2 \text{ and } q, r \in \mathbb{Z}$$

Thus $a = 2q$, or $2q+1$, since $r = 0,1$.

The difference of the number and its square is given by $a^2 – a$.

If $a = 2q$, then

$$a^2 – a = (2q)^2 – 2q$$
$$ = 4q^2 – 2q$$
$$ = 2(2q^2 – q)$$
$$ = 2m, \quad \text{where } m = 2q^2 – q \in \mathbb{Z}$$

If $a = 2q + 1$, then

$$a^2 – a = (2q + 1)^2 – (2q + 1)$$
$$ = (4q^2 + 4q + 1) – 2q – 1$$
$$ = 4q^2 + 2q$$
$$ = 2(2q^2 + q)$$
$$ = 2m, \quad \text{where } m = 2q^2 + q \in \mathbb{Z}$$

Thus, the difference of a number and its square is always an even integer.

 

 

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