Number Theory Test 2

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Number Theory Test 2

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Weekly Quiz Competition

Question 1
1 / -0

$$\sqrt{-3}\sqrt{-75}=$$

A
$$15$$

B
$$15i$$

C
$$-15$$

D
$$-15i$$

Solution

$$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$$
$$=\sqrt{3}i\times\sqrt{75}i$$
$$=\sqrt{225}i^{2}$$
$$= - 15$$
Question 2
1 / -0

The sum of two complex numbers $$a + ib$$ and $$c +id$$ is a real number if

A
$$a + c = 0$$

B
$$b + d = 0$$

C
$$a + b= 0$$

D
$$b + c = 0$$

Solution

It is given that

$$z_{1}=a+ib$$ and

$$z_{2}=c+id$$

Then
$$z_{1}+z_{2}=(a+c)+i(b+d)$$

Now
$$(z_{1}+z_{2})$$ is purely real.

Then the imaginary part has to be $$0$$.

Hence
$$b+d=0$$.
Question 3
1 / -0

The locus of complex number z such that z is purely real and real part is equal to - 2 is

A
Negative y-axis

B
Negative x-axis

C
The point (-2, 0)

D
The point (2, 0)

Solution

$$z=x+iy$$
$$(x,y)$$
$$z$$ is purely real and the real part equals $$-2$$
$$\therefore y=0$$ & $$x=-2$$
$$z=-2$$
Hence, this would be represented by the point (-2,0) on the Argand Plane.
Question 4
1 / -0

$$\dfrac{1}{i-1}+\dfrac{1}{i+1}$$ is

A
positive rational number

B
purely imaginary

C
positive Integer

D
negative integer

Solution

Let $$Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$$

$$=\dfrac{i+1+i-1}{(i-1)(i+1)}$$

$$=\dfrac{i+i}{(i^2-1^2)}$$

$$=\dfrac{2i}{-2}$$

$$ \therefore Z=-i$$
Question 5
1 / -0

The sum of two complex numbers $$a + ib$$ and $$c+ id$$ is purely imaginary if

A
$$a + c = 0$$

B
$$a + d = 0$$

C
$$b + d = 0$$

D
$$b + c = 0$$

Solution

It is given that
$$z_{1}=a+ib$$ and
$$z_{2}=c+id$$
$$z_{1}+z_{2}=(a+c)+i(b+d)$$
$$z_{1}+z_{2}$$ is purely imaginary. (Given)
Then the real part has to be $$0$$.
Hence
$$a+c=0$$.
Question 6
1 / -0

If $$(x+iy)(2-3i)=4+i$$ then (x, y) =

A
$$\left ( 1,\dfrac{1}{13} \right )$$

B
$$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$$

C
$$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$$

D
$$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$$

Solution

$$(x+iy)(2-3i)=4+i$$

$$2x-(3x)i+(2y)i-3yi^{2}=4+i$$

$$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$$

Comparing the real & imaginary parts,
$$2x+3y=4$$--------------------------(1)
$$2y-3x=1$$----------------------------(2)
Solving eq(1) & eq(2),
$$4x+6y=8$$
$$-9x+6y=3$$
$$13x=5\Rightarrow x=\dfrac{5}{13}$$
$$y=\dfrac{14}{13}$$
$$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$$
Question 7
1 / -0

Which of the following is not a prime number ?

A
5

B
7

C
8

D
11

Solution

The factors of 8 are 1, 2, 4 and 8.
As it has factors other than itself and 1, so it is not a prime number.
Question 8
1 / -0

Which of the following is not a composite number?

A
$$4$$

B
$$6$$

C
$$7$$

D
$$8$$
Question 9
1 / -0

The number of prime numbers between 0 and 20 is

A
7

B
8

C
6

D
9

Solution

The prime numbers between $$ 0 $$ and $$ 20 $$ are $$ 2, 3, 5, 7, 11, 13, 17 $$ and $$ 19 $$.
Hence, the number of prime numbers between $$ 0 $$ and $$ 20 $$ is $$ 8 $$.
Question 10
1 / -0

The units digit of every prime number (other than $$2$$ and $$5$$) must be necessarily

A
$$1, 3$$ or $$5$$

B
$$1, 3, 7$$ or $$9$$

C
$$7$$ or $$9$$

D
$$1$$ or $$7$$

Solution

All even numbers (other than $$2$$) are composite, so prime numbers cannot end with any of the digits $$0,2,4,6,8$$.

Also, a number with unit digit $$5$$ will be divisible by $$5.$$

Therefore, the unit's digit of every prime number(other than $$5$$) must be necessarily $$1, 3, 7$$ or $$9$$.

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Re-Attempt Weekly Quiz Competition

Number Theory Test 2

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Number Theory Test 2

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SHARING IS CARING

Question 1

$$15$$

$$15i$$

$$-15$$

$$-15i$$

Solution

Question 2

$$a + c = 0$$

$$b + d = 0$$

$$a + b= 0$$

$$b + c = 0$$

Solution

Question 3

Negative y-axis

Negative x-axis

The point (-2, 0)

The point (2, 0)

Solution

Question 4

positive rational number

purely imaginary

positive Integer

negative integer

Solution

Question 5

$$a + c = 0$$

$$a + d = 0$$

$$b + d = 0$$

$$b + c = 0$$

Solution

Question 6

$$\left ( 1,\dfrac{1}{13} \right )$$

$$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$$

$$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$$

$$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$$

Solution

Question 7

5

7

8

11

Solution

Question 8

$$4$$

$$6$$

$$7$$

$$8$$

Question 9

7

8

6

9

Solution

Question 10

$$1, 3$$ or $$5$$

$$1, 3, 7$$ or $$9$$

$$7$$ or $$9$$

$$1$$ or $$7$$

Solution

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