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
$$a^2 = 3k, \quad \text{where } k = 3q^2 \in \mathbb{Z}$$
If $a = 3q + 1$, then
$$a^2 = 3(3q^2 + 2q) + 1$$
$$a^2 = 3k + 1, \quad \text{where } k = 3q^2 + 2q \in \mathbb{Z}$$
If $a = 3q + 2$, then
$$a^2 = 9q^2 + 12q + 3 + 1$$
$$a^2 = 3(3q^2 + 4q + 1) + 1$$
$$a^2 = 3k + 1, \quad \text{where } k = 3q^2 + 4q + 1 \in \mathbb{Z}$$
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
$$\frac{n(n+1)(2n+1)}{6} = \frac{6k(6k+1)(12k+1)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = k(6k+1)(12k+1) \in \mathbb{Z}$$
If $n = 6k + 1$, then
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+1)(6k+2)(12k+3)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+1) \cdot 2(3k+1) \cdot 3(4k+1)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{6(6k+1)(3k+1)(4k+1)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = (6k+1)(3k+1)(4k+1) \in \mathbb{Z}$$
If $n = 6k + 2$, then
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+2)(6k+3)(12k+5)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{2(3k+1) \cdot 3(2k+1) \cdot (12k+5)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{6(3k+1)(2k+1)(12k+5)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = (3k+1)(2k+1)(12k+5) \in \mathbb{Z}$$
If $n = 6k + 3$, then
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+3)(6k+4)(12k+7)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{3(2k+1) \cdot 2(3k+2) \cdot (12k+7)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{6(2k+1)(3k+2)(12k+7)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = (2k+1)(3k+2)(12k+7) \in \mathbb{Z}$$
If $n = 6k + 4$, then
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+4)(6k+5)(12k+9)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{2(3k+2) \cdot (6k+5) \cdot 3(4k+3)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{6(3k+2)(6k+5)(4k+3)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = (3k+2)(6k+5)(4k+3) \in \mathbb{Z}$$
If $n = 6k + 5$, then
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+5)(6k+6)(12k+11)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = \frac{(6k+5) \cdot 6(k+1) \cdot (12k+11)}{6}$$
$$\frac{n(n+1)(2n+1)}{6} = (6k+5)(k+1)(12k+11) \in \mathbb{Z}$$
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
$$n(7n^2 + 5) = 6q[7(6q)^2 + 5]$$
$$n(7n^2 + 5) = 6q(252q^2 + 5)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = q(252q^2 + 5) \in \mathbb{N} \cup \{0\}$$
If $n = 6q + 1$, then
$$n(7n^2 + 5) = (6q+1)[7(6q+1)^2 + 5]$$
$$n(7n^2 + 5) = (6q+1)[7(36q^2 + 12q + 1) + 5]$$
$$n(7n^2 + 5) = (6q+1)[252q^2 + 84q + 7 + 5]$$
$$n(7n^2 + 5) = (6q+1)[252q^2 + 84q + 12]$$
$$n(7n^2 + 5) = 6(6q+1)(42q^2 + 14q + 2)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = (6q+1)(42q^2 + 14q + 2) \in \mathbb{N}$$
If $n = 6q + 2$, then
$$n(7n^2 + 5) = (6q+2)[7(6q+2)^2 + 5]$$
$$n(7n^2 + 5) = (6q+2)[7(36q^2 + 24q + 4) + 5]$$
$$n(7n^2 + 5) = (6q+2)[252q^2 + 168q + 28 + 5]$$
$$n(7n^2 + 5) = (6q+2)[252q^2 + 168q + 33]$$
$$n(7n^2 + 5) = 2(3q+1) \cdot 3(84q^2 + 56q + 11)$$
$$n(7n^2 + 5) = 6(3q+1)(84q^2 + 56q + 11)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = (3q+1)(84q^2 + 56q + 11) \in \mathbb{N}$$
If $n = 6q + 3$, then
$$n(7n^2 + 5) = (6q+3)[7(6q+3)^2 + 5]$$
$$n(7n^2 + 5) = (6q+3)[7(36q^2 + 36q + 9) + 5]$$
$$n(7n^2 + 5) = (6q+3)[252q^2 + 252q + 63 + 5]$$
$$n(7n^2 + 5) = (6q+3)[252q^2 + 252q + 68]$$
$$n(7n^2 + 5) = 3(2q+1) \cdot 2(126q^2 + 126q + 34)$$
$$n(7n^2 + 5) = 6(2q+1)(126q^2 + 126q + 34)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = (2q+1)(126q^2 + 126q + 34) \in \mathbb{N}$$
If $n = 6q + 4$, then
$$n(7n^2 + 5) = (6q+4)[7(6q+4)^2 + 5]$$
$$n(7n^2 + 5) = (6q+4)[7(36q^2 + 48q + 16) + 5]$$
$$n(7n^2 + 5) = (6q+4)[252q^2 + 336q + 112 + 5]$$
$$n(7n^2 + 5) = (6q+4)[252q^2 + 336q + 117]$$
$$n(7n^2 + 5) = 2(3q+2) \cdot 3(84q^2 + 112q + 39)$$
$$n(7n^2 + 5) = 6(3q+2)(84q^2 + 112q + 39)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = (3q+2)(84q^2 + 112q + 39) \in \mathbb{N}$$
If $n = 6q + 5$, then
$$n(7n^2 + 5) = (6q+5)[7(6q+5)^2 + 5]$$
$$n(7n^2 + 5) = (6q+5)[7(36q^2 + 60q + 25) + 5]$$
$$n(7n^2 + 5) = (6q+5)[252q^2 + 420q + 175 + 5]$$
$$n(7n^2 + 5) = (6q+5)[252q^2 + 420q + 180]$$
$$n(7n^2 + 5) = 6(6q+5)(42q^2 + 70q + 30)$$
$$n(7n^2 + 5) = 6k, \quad \text{where } k = (6q+5)(42q^2 + 70q + 30) \in \mathbb{N}$$
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
$$n^4 + 4n^2 + 11 = (4q+1)^4 + 4(4q+1)^2 + 11$$
$$n^4 + 4n^2 + 11 = (256q^4 + 256q^3 + 96q^2 + 16q + 1) + 4(16q^2 + 8q + 1) + 11$$
$$n^4 + 4n^2 + 11 = 256q^4 + 256q^3 + 96q^2 + 16q + 1 + 64q^2 + 32q + 4 + 11$$
$$n^4 + 4n^2 + 11 = 256q^4 + 256q^3 + 160q^2 + 48q + 16$$
$$n^4 + 4n^2 + 11 = 16(16q^4 + 16q^3 + 10q^2 + 3q + 1)$$
$$n^4 + 4n^2 + 11 = 16k, \quad \text{where } k = 16q^4 + 16q^3 + 10q^2 + 3q + 1 \in \mathbb{Z}$$
If $n = 4q + 3$, then
$$n^4 + 4n^2 + 11 = (4q+3)^4 + 4(4q+3)^2 + 11$$
$$n^4 + 4n^2 + 11 = (256q^4 + 768q^3 + 864q^2 + 432q + 81) + 4(16q^2 + 24q + 9) + 11$$
$$n^4 + 4n^2 + 11 = 256q^4 + 768q^3 + 864q^2 + 432q + 81 + 64q^2 + 96q + 36 + 11$$
$$n^4 + 4n^2 + 11 = 256q^4 + 768q^3 + 928q^2 + 528q + 128$$
$$n^4 + 4n^2 + 11 = 16(16q^4 + 48q^3 + 58q^2 + 33q + 8)$$
$$n^4 + 4n^2 + 11 = 16k, \quad \text{where } k = 16q^4 + 48q^3 + 58q^2 + 33q + 8 \in \mathbb{Z}$$
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 $a = 6q + r$, since $r = 0,1,2,3,4,5$.
Now, the cube of the integer is $a^3$. Let us find $a^3$:
$$a^3 = (6q)^3 + 3(6q)^2r + 3(6q)r^2 + r^3$$
$$a^3 = 216q^3 + 108q^2r + 18qr^2 + r^3$$
$$a^3 = 6(36q^3 + 18q^2r + 3qr^2) + r^3$$
$$a^3 = 6m + r^3, \quad \text{where } m = 36q^3 + 18q^2r + 3qr^2 \in \mathbb{Z}$$
Let us analyze the value of $r^3$ for each possible remainder $r$:
If $r = 0$, then $r^3 = 0^3 = 0 = 6(0) + 0$
If $r = 1$, then $r^3 = 1^3 = 1 = 6(0) + 1$
If $r = 2$, then $r^3 = 2^3 = 8 = 6(1) + 2$
If $r = 3$, then $r^3 = 3^3 = 27 = 6(4) + 3$
If $r = 4$, then $r^3 = 4^3 = 64 = 6(10) + 4$
If $r = 5$, then $r^3 = 5^3 = 125 = 6(20) + 5$
In each case, $r^3$ can be written as $6q’ + r$, which yields the same remainder $r$.
Substituting this back into $a^3$:
$$a^3 = 6k + r, \quad \text{where } k = m + q’ \in \mathbb{Z}$$
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
$$a^2 – a = (2q)^2 – 2q$$
$$a^2 – a = 2(2q^2 – q)$$
$$a^2 – a = 2m, \quad \text{where } m = 2q^2 – q \in \mathbb{Z}$$
If $a = 2q + 1$, then
$$a^2 – a = (2q + 1)^2 – (2q + 1)$$
$$a^2 – a = (4q^2 + 4q + 1) – 2q – 1$$
$$a^2 – a = 2(2q^2 + q)$$
$$a^2 – a = 2m, \quad \text{where } m = 2q^2 + q \in \mathbb{Z}$$
Thus, the difference of a number and its square is always an even integer.
