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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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Advanced Mathematics Chapter 3.1 Solutions

SEBA Class 10 Advanced Maths Chapter 3.2 Solutions

Question 1

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 $3k+1$.

Question 2

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 $9k+8$.

Question 3

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.

Question 4

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 $6k$, $k \in \mathbb{N}$.

Question 5

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 $k \in \mathbb{Z}$.

Question 6

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.

Question 7

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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