SEBA Class 10 Advanced Maths Chapter 3.3 Solutions: Arithmetic of Integers

In this article, we have solved all the questions from Chapter 3.3 of SEBA Class 10 Advanced Mathematics in a simple way. 100 % reliable solutions for all the questions of SEBA Class 10 Advanced Maths Chapter 3.3

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Chapter 3.2 of SEBA Class 10 Advanced Mathematics

Q1 : If $a \mid b$, then show that $-a \mid b$, $a \mid -b$, and $-a \mid -b$.

Solution:

1.To prove $-a \mid b$:

Given that $a \mid b$.

Then there exists an integer $c$ such that:

$b = ac$

$\implies b = (-a)(-c)$

So, $-a$ is one of the factor of $b$.

∴ $-a \mid b$.

2. To prove $a \mid -b$:

$b = ac$ [Given]

$\implies -b = -(ac)$ [Multiplying both sides by $-1$]

$\implies -b = a(-c)$

So, $a$ is one of the factor of $-b$.

∴ $a \mid -b$.

3. To prove $-a \mid -b$:

$b = ac$ [Given]

$\implies -b = -(ac)$ [Multiplying both sides by $-1$]

$\implies -b = (-a)(c)$

So, $-a$ is one of the factor of $-b$.

∴ $-a \mid -b$.

Q2: For given integers $a, b, c$ and $d$, show that: 
(i) $a \mid b \implies a \mid bc$ 
(ii) $a \mid b, a \mid c \implies a^2 \mid bc$

Solution (i):

Given $a \mid b$.

Then there exists an integer $k$ such that:

$b = ak$

$\implies bc = (ak)c$ [multiply both sides by $c$]

$\implies bc = a(kc)$

So, $a$ is one of the factor of $bc$

∴ $a \mid bc$ [Since $k$ and $c$ are integers, their product $kc$ is also an integer]

(ii) $a \mid b, a \mid c \implies a^2 \mid bc$

Solution:

Given $a \mid b$ and $a \mid c$.

Then there exist integers $k_1$ and $k_2$ such that:

So, $a^2$ is one of the factor of $bc$

[Since $k_1$ and $k_2$ are integers, their product $k_1 k_2$ is also an integer]

Q3: Examine if the following statements are true or false. (Supply proof when it is true and give counterexample when it is false) : 
(i) If $a \mid (b + c)$, then $a \mid b$ or $a \mid c$. 
(ii) $a \mid n, b \mid n \implies ab \mid n$.

Solution (i):

$a \mid (b + c)$
$\implies b + c = ak,\ \text{where } k \in \mathbb{Z}$
$\implies b = ak – c$ or $c = ak – b$

Since $a$ is neither a factor of  $b$ nor $c$

$\therefore a \nmid b$ or $a \nmid c$

Answer: False

Counterexample:

Let $a = 5$, $b = 3$, and $c = 2$.

Here, $b + c = 3 + 2 = 5$.

It is true that $5 \mid 5$ (so, $a \mid (b + c)$ is true).

However, $5 \nmid 3$ and $5 \nmid 2$ (so, $a \nmid b$ and $a \nmid c$).

Therefore, the statement is false.

(ii) $a \mid n, b \mid n \implies ab \mid n$.

Solution (ii):

$a \mid n$ and $b \mid n$

$\implies n = ap$ and $n = bq,\ \text{where } p, q \in \mathbb{Z}$

$\implies n^2 = (ap)(bq)$

$\implies n^2 = abpq$

$\implies n = \sqrt{abpq}$

∴ $ab$ is not a factor of $n$

∴ $ab \nmid n$

Answer: False

Counterexample:

Let $a = 4$, $b = 6$, and $n = 12$.

Here, $4 \mid 12$ and $6 \mid 12$ are both true.

However, $ab = 4 \times 6 = 24$, and $24 \nmid 12$.

Q4: For any natural number $n$, show that: 
(a) (i) $2^n + (-1)^{n+1}$ is divisible by 3. 
(ii) $3^{3n+1} + 2^{n+1}$ is divisible by 5. 
(b) $3^{2n-1} + 2^{n+1}$ is divisible by 7. 
(c) $3^{4n+2} + 5^{2n+1}$ is divisible by 14.

Solution (a)(i):

Let $S(n) = 2^n + (-1)^{n+1}$ is divisible by $3$ for all $n \in \mathbb{N}$.

$S(1) = 2^1 + (-1)^{1+1} = 2 + 1 = 3$, which is divisible by $3$

Let $S(k) = 2^k + (-1)^{k+1} is divisible by $3$, for $k \in \mathbb{N}$

.i.e. $ 2^k + (-1)^{k+1} = 3p$ , where $p \in \mathbb{N}$

This implies $2^k = 3p – (-1)^{k+1}$.

We need to show that,

∴ $S (k+1)$ is divisible by $3$

Therefore, $S(k+1)$ is divisible by $3$ if $S(k)$ is divisible by $3$.
By the method of mathematical induction, $S(n)$ is divisible by $3$ for all $n \in \mathbb{N}$.

(a)(ii) $3^{3n+1} + 2^{n+1}$ is divisible by 5

Solution:

Let $S(n) = 3^{3n+1} + 2^{n+1}$ is divisible by $5$ for all $n \in \mathbb{N}$.

$S(1) = 3^{3(1)+1} + 2^{1+1} = 3^4 + 2^2 = 81 + 4 = 85$, which is divisible by $5$.

Let $S(k) = 3^{3k+1} + 2^{k+1}$ is divisible by $5$, for $k \in \mathbb{N}$.

.i.e. $3^{3k+1} + 2^{k+1} = 5p$, where $p \in \mathbb{N}$.

This implies $3^{3k+1} = 5p – 2^{k+1}$.  We need to show that,$S(k+1) = 3^{3(k+1)+1} + 2^{(k+1)+1}$ is divisible by $5$.

$\implies S(k+1) = 3^{3k+3+1} + 2^{k+2}$

$\implies S(k+1) = 3^3 \cdot 3^{3k+1} + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 27[5p – 2^{k+1}] + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 135p – 27 \cdot 2^{k+1} + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 135p – 25 \cdot 2^{k+1}$

$\implies S(k+1) = 5[27p – 5 \cdot 2^{k+1}]$ ∴ $S(k+1)$ is divisible by $5$.  Therefore, $S(k+1)$ is divisible by $5$ if $S(k)$ is divisible by $5$.

By the method of mathematical induction, $S(n)$ is divisible by $5$ for all $n \in \mathbb{N}$.

(b):$3^{2n-1} + 2^{n+1}$ is divisible by 7

Solution:

Let $S(n) = 3^{2n-1} + 2^{n+1}$ is divisible by $7$ for all $n \in \mathbb{N}$.

$S(1) = 3^{2(1)-1} + 2^{1+1} = 3^1 + 2^2 = 3 + 4 = 7$, which is divisible by $7$.

Let $S(k) = 3^{2k-1} + 2^{k+1}$ is divisible by $7$, for $k \in \mathbb{N}$.

.i.e. $3^{2k-1} + 2^{k+1} = 7p$, where $p \in \mathbb{N}$.

This implies $3^{2k-1} = 7p – 2^{k+1}$.

We need to show that,

$S(k+1) = 3^{2(k+1)-1} + 2^{(k+1)+1}$ is divisible by $7$.

$\implies S(k+1) = 3^{2k+2-1} + 2^{k+2}$

$\implies S(k+1) = 3^2 \cdot 3^{2k-1} + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 9[7p – 2^{k+1}] + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 63p – 9 \cdot 2^{k+1} + 2 \cdot 2^{k+1}$

$\implies S(k+1) = 63p – 7 \cdot 2^{k+1}$

$\implies S(k+1) = 7[9p – 2^{k+1}]$

∴ $S(k+1)$ is divisible by $7$.

Therefore, $S(k+1)$ is divisible by $7$ if $S(k)$ is divisible by $7$.

By the method of mathematical induction, $S(n)$ is divisible by $7$ for all $n \in \mathbb{N}$.

(c): $3^{4n+2} + 5^{2n+1}$ is divisible by 14.

Solution:

Let $S(n) = 3^{4n+2} + 5^{2n+1}$ is divisible by $14$ for all $n \in \mathbb{N}$.

$S(1) = 3^{4(1)+2} + 5^{2(1)+1} = 3^6 + 5^3 = 729 + 125 = 854$, which is divisible by $14$ ($14 \times 61 = 854$).

Let $S(k) = 3^{4k+2} + 5^{2k+1}$ is divisible by $14$, for $k \in \mathbb{N}$.

.i.e. $3^{4k+2} + 5^{2k+1} = 14p$, where $p \in \mathbb{N}$.

This implies $3^{4k+2} = 14p – 5^{2k+1}$.

We need to show that,

$S(k+1) = 3^{4(k+1)+2} + 5^{2(k+1)+1}$ is divisible by $14$.

$\implies S(k+1) = 3^{4k+4+2} + 5^{2k+2+1}$

$\implies S(k+1) = 3^4 \cdot 3^{4k+2} + 5^2 \cdot 5^{2k+1}$

$\implies S(k+1) = 81[14p – 5^{2k+1}] + 25 \cdot 5^{2k+1}$

$\implies S(k+1) = 81(14p) – 81 \cdot 5^{2k+1} + 25 \cdot 5^{2k+1}$

$\implies S(k+1) = 81(14p) – 56 \cdot 5^{2k+1}$

$\implies S(k+1) = 14[81p – 4 \cdot 5^{2k+1}]$

∴ $S(k+1)$ is divisible by $14$.

Therefore, $S(k+1)$ is divisible by $14$ if $S(k)$ is divisible by $14$.

By the method of mathematical induction, $S(n)$ is divisible by $14$ for all $n \in \mathbb{N}$.

Question 5:Prove that:
(a) Square of every integer can be expressed as $3k$ or $3k + 1$. 
(b) Cube of any integer can be expressed in one of the forms $9k, 9k + 1, 9k + 8$.

Solution (a):

Applying the division algorithm on any 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

$\implies a^2 = (3q)^2 $
$\implies a^2 = 9q^2 $
$\implies a^2= 3(3q^2)$
∴$a^2= 3k$ (where $k = 3q^2$).

If ($a = 3q+1$), then

$\implies a^2 = (3q+1)^2$
$\implies a^2 = 9q^2 + 6q + 1 $
$\implies a^2 = 3(3q^2 + 2q) + 1 $
∴ $ a^2 = 3k + 1$. (where $k = 3q^2 + 2q$).

If ($a = 3q+2$), then

$\implies a^2 = (3q+2)^2$
$\implies a^2= 9q^2 + 12q + 4 $
$\implies a^2= 9q^2 + 12q + 3 + 1$
$\implies a^2= 3(3q^2 + 4q + 1) + 1 $
∴ $ a^2 = 3k + 1$.

Thus, the square of any integer is always of the form $3k$ or $3k+1$.

(b) Cube of any integer can be expressed in one of the forms $9k, 9k + 1, 9k + 8$.

Solution (b):

Applying the division algorithm on any 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

$\implies a^3 = (3q)^3$

$\implies a^3 = 27q^3$

$\implies a^3 = 9(3q^3)$

∴ $a^3 = 9k$ (where $k = 3q^3$).

If ($a = 3q+1$), then

$\implies a^3 = (3q+1)^3$

$\implies a^3 = 27q^3 + 27q^2 + 9q + 1$

$\implies a^3 = 9(3q^3 + 3q^2 + q) + 1$

∴ $a^3 = 9k + 1$ (where $k = 3q^3 + 3q^2 + q$).

If ($a = 3q+2$), then

$\implies a^3 = (3q+2)^3$

$\implies a^3 = 27q^3 + 54q^2 + 36q + 8$

$\implies a^3 = 9(3q^3 + 6q^2 + 4q) + 8$

∴ $a^3 = 9k + 8$ (where $k = 3q^3 + 6q^2 + 4q$).

Thus, the cube of any integer can be expressed in one of the forms $9k, 9k + 1, 9k + 8$.

Question 6 For any integer $a$, show that: 
(i) $2 \mid a(a+1)$ and $3 \mid a(2a^2+7)$. 
(ii) For any even integer, prove that $48 \mid a(a^2+20)$. 
(iii) If $a$ be any odd integer, prove that $24 \mid a(a^2-1)$.

(i) $2 \mid a(a+1)$ and $3 \mid a(2a^2+7)

Solution

Part 1: Show $2 \mid a(a+1)$

The expression $a(a+1)$ is the product of two consecutive integers. One of them must be even, so it is always divisible by 2.

Part 2: Show $3 \mid a(2a^2+7)$

Now, $(a – 1)a(a + 1)$ is a product of 3 consecutive integers, so $3 \mid (a – 1)a(a + 1)$.

Also, $9a$ is divisible by 3.

$\therefore 3 \mid 2(a – 1)a(a + 1) + 9a = a(2a^2+7)$

(ii) For any even integer, prove that $48 \mid a(a^2+20)$

Solution:

Since $a$ is given as an even integer, so

So it is divisible by 16.

If $m$ is an odd integer ($m = 2k+1$)

Then $m^2$ is odd, which means $m^2 + 5$ must be an even integer (since an odd number plus an odd number is always even).

Let $m^2 + 5 = 2y$.

So it is divisible by 16.

Therefore, in all cases:

Thus, $3 \mid a(a – 1)(a + 1)$.

Also, $21a$ is divisible by 3.

The greatest common divisor of 16 and 3 is 1:

Solution:

Since $a$ is given as an odd integer, so

Now, $(a – 1)a(a + 1)$ is a product of 3 consecutive integers.

Thus, $3 \mid a(a^2 – 1) \quad \text{— (Equation 1)}$

Again, substituting $a = 2m + 1$ into our expression:

Now, $m(m + 1)$ is the product of two consecutive integers. One of them must be even, so it is always divisible by 2.

Let $m(m + 1) = 2k$.

So it is divisible by 8.

Question 7: Compute by using Euclidean Algorithm:
(i) $\text{GCD}(72,120)$
(ii) $\text{GCD}(56,-36)$
(iii) $\text{GCD}(13,80)$

(i) $\text{GCD}(72,120)$

Solution:

The given numbers are $72$ and $120$

Now,

$120 = 72 \times 1 + 48$

$72 = 48 \times 1 + 24$

$48 = 24 \times 2 + 0$

The last non-zero remainder is $24$.

So, $\text{GCD}(72,120) = 24$.

(ii) $\text{GCD}(56,-36)$

Solution:

Note: $\text{GCD}(56,-36) = \text{GCD}(56,36)$.

$56 = 36 \times 1 + 20$

$36 = 20 \times 1 + 16$

$20 = 16 \times 1 + 4$

$16 = 4 \times 4 + 0$

The last non-zero remainder is $4$.

So, $\text{GCD}(56,-36) = 4$.

(iii) $\text{GCD}(13,80)$

Solution:

$80 = 13 \times 6 + 2$

$13 = 2 \times 6 + 1$

$2 = 1 \times 2 + 0$

The last non-zero remainder is $1$.

So, $\text{GCD}(13,80) = 1$.

Question 8: Find two integers $u$ and $v$ such that:
(i) $20u+63v=1$
(ii) $30u+72v=12$
(iii) $52u-91v=78$
(iv) $24u+138v=6$

(i) $20u+63v=1$

Solution:

$63 = 20 \times 3 + 3 \quad \text{— (i)}$

$20 = 3 \times 6 + 2 \quad \text{— (ii)}$

$3 = 1 \times 2 + 1\quad \text{— (iii)}$

$2 = 1 \times 2 + 0$

So, $\text{GCD}(20,63) = 1$.

$\implies 1 = 3 – 1 \times 2$ [From eq(iii)]$\implies 1 = 3 – 1 \times (20 – 3 \times 6)$ [From eq(ii)]$\implies 1 = 3 – 1 \times 20 + 3 \times 6$
$\implies 1 = 3 ( 1 + 6 ) – 1 \times 20 $
$\implies 1 = 3 \times 7 – 20 \times 1$
$\implies 1 = (63 – 20 \times 3) \times 7 – 20 \times 1$ [ From equation (i)]$\implies 1 = 63 \times 7 – 20 \times3\times 7- 20 \times 1$
$\implies 1 = 63 \times 7 – 21 \times20- 1 \times 20$
$\implies 1 = 63 \times 7 – 20 ( 21 + 1 ) $
$\implies 1 = 63 \times 7 – 20 \times 22$
$\implies 1 = 20(-22) + 63(7) $

So, $u = -22, v = 7$.

(ii) $30u+72v=12$

Solution:

$72 = 30 \times 2 + 12 \quad \text{— (i)}$

$30 = 12 \times 2 + 6 \quad \text{— (ii)}$

$12 = 6 \times 2 + 0$

So, $\text{GCD}(30,72) = 6$.

Now, $\implies 12 = 72 – 30 \times 2$ [From eq(i)]

$\implies 12 = 30(-2) + 72(1)$

So, $u = -2, v = 1$.

(iii) $52u-91v=78$

Solution:

$91 = 52 \times 1 + 39 \quad \text{— (i)}$

$52 = 39 \times 1 + 13 \quad \text{— (ii)}$

$39 = 13 \times 3 + 0$

So, $\text{GCD}(52,91) = 13$.

Now, $\implies 13 = 52 – 39 \times 1$ [From eq(ii)]

$\implies 13 = 52 – 1 \times (91 – 52 \times 1)$ [From eq(i)]

$\implies 13 = 52 – 1 \times 91 + 52 \times 1$

$\implies 13 = 52(1 + 1) – 91 \times 1$

$\implies 13 = 52 \times 2 – 91 \times 1$

Multiplying both sides by $6$ to get $78$ on the left side ($13 \times 6 = 78$):

So, $u = 12, v = 6$.

(iv) $24u+138v=6$

Solution:

$138 = 24 \times 5 + 18 \quad \text{— (i)}$

$24 = 18 \times 1 + 6 \quad \text{— (ii)}$

$18 = 6 \times 3 + 0$

So, $\text{GCD}(24,138) = 6$.

Now, $\implies 6 = 24 – 18 \times 1$ [From eq(ii)]

$\implies 6 = 24 – 1 \times (138 – 24 \times 5)$ [From eq(i)]

$\implies 6 = 24 – 1 \times 138 + 24 \times 5$

$\implies 6 = 24(1 + 5) – 138 \times 1$

$\implies 6 = 24 \times 6 – 138 \times 1$

$\implies 6 = 24(6) + 138(-1)$

So, $u = 6, v = -1$.

(ii) $30u+72v=12$

Solution:

Find $\text{GCD}(72,30)$:

1. $72 = 30 \times 2 + 12 \implies 12 = 72 – 30 \times 2$

We can directly see that:

So, $u = -2, v = 1$.

Question 9: Show that:
(i) For any integer $a$, $\text{GCD}(2a+1,9a+4)=1$.
(ii) For any odd integer $a$, $\text{GCD}(3a,3a+2)=1$.

Solution(i):

Let, $\text{GCD}(2a+1, 9a+4) = d$

Solution (ii):

Let, $\text{GCD}(3a, 3a+2) = d$

∴ $d \mid 3a \quad \text{and} \quad d \mid (3a+2)$

$\implies d \mid \left[(3a)x + (3a+2)y\right]$ [Take any integer $x = -1$ and $y = 1$]

$\implies d \mid \left[(3a)(-1) + (3a+2)(1)\right]$

$\implies d \mid (-3a + 3a + 2)$

$\implies d \mid 2$

Thus $d$ can be $1$ or $2$.

Since $a$ is an odd integer, $3a$ is also an odd integer.

An odd integer cannot be divided by $2$, so $d \ne 2$.

$\implies \therefore d = 1$

∴ $\text{GCD}(3a, 3a+2) = 1$

Question 10: Find $\text{LCM}(306,657)$ and $\text{LCM}(111,273)$.

$\therefore \text{GCD}(306, 657) = 9$

We know, $\text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b$

$\implies \text{LCM}(a, b) = \frac{a \times b}{\text{GCD}(a, b)}$

$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$

$\text{LCM}(306, 657) = 34 \times 657 = \mathbf{22,338}$

Again,

$\therefore \text{GCD}(111, 273) = 3$

Again, we know,

$\implies \text{LCM}(a, b) = \frac{a \times b}{\text{GCD}(a, b)}$

$\text{LCM}(111, 273) = \frac{111 \times 273}{3}$