Number Theory Test 14

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Number Theory Test 14

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Weekly Quiz Competition

Question 1
1 / -0

The sum of prime numbers, out of the numbers $$17, 8, 21, 13, 41, 2, 27, 31, 51$$ is:

A
$$125$$

B
$$102$$

C
$$104$$

D
$$155$$

Solution

Prime numbers out of $$17,8,21,13,41,2,27,31,51$$ are $$17,13,41,2,31$$.
Sum of prime numbers $$= 17+13+41+2+31=104$$.
Question 2
1 / -0

If $$z_1=4+i,z_2=4-i $$ find $$z_1z_2$$

A
$$17$$

B
$$16$$

C
$$17-i$$

D
$$16i$$

Solution

$$z_1=4+i\\z_2=4-i\\z_1z_2=(4+i)(4-i)\\16-i^2=16+1=17$$
Question 3
1 / -0

$$z_1=9+8i\ \ \ |z|=$$

A
$$\sqrt {145}$$

B
$$\sqrt {163}$$

C
$$\sqrt {117}$$

D
$$\sqrt {137}$$

Solution

$$z=9+8i\\|z|=\sqrt {a^2+b^2}\\\\|z|=\sqrt {9^2+8^2} =\sqrt {81+64}=\sqrt {145}$$
Question 4
1 / -0

Mark against the correct answer in each of the following .
$$i^{-38}=$$?

A
$$i$$

B
$$-i$$

C
$$1$$

D
$$-1$$

Solution

$$i^{-38}\\=\dfrac {1}{i^{-38}}\times \dfrac {i^2}{i^2}\\=\dfrac {-1}{i^{40}}\\=\dfrac {-1}{(i^4)^{10}}\\=\dfrac {-1}{(1)^{10}}\\=\dfrac {-1}{1}\\=-1$$
Question 5
1 / -0

Which of the following is a prime number?

A
$$117$$

B
$$171$$

C
$$179$$

D
none of these

Solution

$$179$$
Since $$179$$ have no factors other than $$1$$ and itself
Option (c) is the correct answer
Question 6
1 / -0

Which of the following is a prime number?

A
$$323$$

B
$$361$$

C
$$263$$

D
none of these

Solution

a) $$323$$ can be written as $$17 \times 19$$
Hence $$323$$ is not a prime number
b) $$361$$ can be written as $$19 \times 19$$
Hence $$361$$ is not a prime number
c) $$263$$ is a prime number
Option (c) is the correct answer
Question 7
1 / -0

Mark against the correct answer in each of the following .
$$i^{-75}=$$?

A
$$1$$

B
$$-1$$

C
$$i$$

D
$$-i$$

Solution

$$i^{-75}\\=\dfrac {1}{i^{75}}\\=\dfrac {1}{i^{75}\times i}\times i\\=\dfrac {1}{i^{76}}\times i\\=\dfrac {1}{(i^{4})^{19}}\times i\\=1\times i\\=i$$
Question 8
1 / -0

Mark against the correct answer in each of the following .
$$i^{124}=$$?

A
$$-1$$

B
$$1$$

C
$$i$$

D
$$-i$$

Solution

$$i^{124}\\=(i^4)^{31}\\=(1)^{31}\\=1$$
Question 9
1 / -0

The real part of $$\left ( \dfrac{1+i}{3-i} \right )^2=$$

A
$$1$$

B
$$16$$

C
$$16\omega ^2$$

D
$$\displaystyle \frac{-3}{25}$$

Solution

$$\displaystyle \frac{1+i}{3-i}=\frac{(1+i)(3+i)}{(3-i)(3+i)}=\frac{3-1+4i}{3^{2}+(1)^{2}}=\frac{2+4i}{10}=\frac{1+2i}{5}$$

$$\displaystyle \therefore \left ( \frac{1+2i}{5} \right )^{2}$$

$$\displaystyle =\frac{1^{2}-2^{2}+4i}{25}$$

$$\displaystyle =\frac{-3+4i}{25}$$

$$\displaystyle \therefore $$ Real part$$=\dfrac{-3}{25}$$
Question 10
1 / -0

If $$z_1$$ and $$z_2$$ are two complex numbers, then $$Re(z_1z_2)$$ is:

A
$$Re(z_1)Re(z_2)$$

B
$$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$$

C
$$Im(z_1).Re(z_2)$$

D
$$Re(z_1).Im(z_2)$$

Solution

$$z_{1}=a_{1}+ib_{1}; z_{2}=a_{2}+ib_{2}$$
$$z_{1}z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})$$
$$=a_{1}a_{2}+(a_{1}b_{2})i+(a_{2}b_{1})i-b_{1}b_{2}$$
$$=a_{1}a_{2}-b_{1}b_{2}+(a_{1}b_{2}+a_{2}b_{1})i$$
$$\therefore Re(z_{1}z_{2})=a_{1}a_{2}-b_{1}b_{2}$$
$$=Re(z_{1})\cdot Re(z_{2})-Im(z_{1})\cdot Im(z_{2})$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 14

Result

Number Theory Test 14

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Rank

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SHARING IS CARING

Question 1

$$125$$

$$102$$

$$104$$

$$155$$

Solution

Question 2

$$17$$

$$16$$

$$17-i$$

$$16i$$

Solution

Question 3

$$\sqrt {145}$$

$$\sqrt {163}$$

$$\sqrt {117}$$

$$\sqrt {137}$$

Solution

Question 4

$$i$$

$$-i$$

$$1$$

$$-1$$

Solution

Question 5

$$117$$

$$171$$

$$179$$

none of these

Solution

Question 6

$$323$$

$$361$$

$$263$$

none of these

Solution

Question 7

$$1$$

$$-1$$

$$i$$

$$-i$$

Solution

Question 8

$$-1$$

$$1$$

$$i$$

$$-i$$

Solution

Question 9

$$1$$

$$16$$

$$16\omega ^2$$

$$\displaystyle \frac{-3}{25}$$

Solution

Question 10

$$Re(z_1)Re(z_2)$$

$$Re(z_1).Re(z_2)-Im(z_1).Im(z_2)$$

$$Im(z_1).Re(z_2)$$

$$Re(z_1).Im(z_2)$$

Solution

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