Number Theory Test 6

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Number Theory Test 6

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Weekly Quiz Competition

Question 1
1 / -0

If $$i^2 = - 1$$, then the value of $$\displaystyle \sum^{200}_{n = 1} i^n $$ is

A
50

B
-50

C
0

D
100

Solution

Given, $$i^2=-1$$
$$\displaystyle \sum_{n = 1}^{200} i^n = i + i^2 + i^3 + ......... + i^{200} = \displaystyle \frac{i (1 - i^{200})}{1 - i} $$ (since G. P.)
$$= \displaystyle \frac{\displaystyle i (1 - 1)}{\displaystyle 1 - i} = 0$$
Question 2
1 / -0

Which is the least prime number ?

A
$$1$$

B
$$0$$

C
$$2$$

D
$$3$$

Solution

The prime numbers are $$2,3,5.......$$
So least prime is $$2.$$
Question 3
1 / -0

How many prime numbers are there between 10 and 30?

A
$$6$$

B
$$5$$

C
$$7$$

D
$$8$$

Solution

$${\because}$$ The prime numbers between $$10$$ and $$30 $$ are $$11,13,17,19,23,29.$$
Question 4
1 / -0

By Goldbach's conjecture, every even number greater than 4 can be expressed as a sum of

A
Two even numbers

B
Two odd prime numbers

C
Two composite numbers

D
Two co-prime numbers

Solution

Goldbach's conjecture says that "Every integer greater than $$2$$ can be expressed as the sum of $$2$$ prime numbers."
Question 5
1 / -0

If $$z = \displaystyle \frac{(3 + 4i)(5 - 7 i)}{(7 + 5i)(4 - 3i)}$$ then $$|z| = ?$$

A
5

B
4

C
1

D
None of these

Solution

Given, $$z = \displaystyle \frac{(3 + 4i)(5 - 7 i)}{(7 + 5i)(4 - 3i)}$$
$$\therefore |z| = \displaystyle \frac{|(3 + 4i) ||(5 - 7i)|}{|(7 + 5i)||(4 - 3i)|}$$

$$= \displaystyle \frac{\sqrt{9 + 16} \sqrt{25 + 49}}{\sqrt{49 + 25} \sqrt{16 + 9}} = 1$$
Question 6
1 / -0

Which of the following is not a composite number ?

A
$$4$$

B
$$6$$

C
$$7$$

D
$$10$$

Solution

A composite number is a positive integer that has at least one positive divisor other than one and number itself.
Here 7 is not composite number as it is cannot be divided evenly orwe can say 7 has divisor 1 and 7 itself that means it is a prime number.
Question 7
1 / -0

If 2009 = $$\displaystyle p^{a}.q^{b}$$ where p and q are prime numbers then find the value of p + q

A
3

B
48

C
51

D
2009

Solution

Given $$2009 = p^a q^b$$ ...(i)
Where p and q are prime numbers
We know that $$2009 = 7\times 7\times 41= 7^2\times 41^1$$ ...(ii)
Comparing equation (1) and (2) we get
$$p=7, q=41, a=2 and b= 1$$
$$\therefore p+q = 7+41=48$$
Question 8
1 / -0

1+$$i^2 + i^4 + i^6 + ........+ i^{2n}$$ is

A
Positive

B
Negative

C
Zero

D
Cannot be determined

Solution

$$1+i^2+i^4+........+i^{2n}$$

$$=1-1+1-1+1-1......$$

Hence it can b positive or negative depending on the value of $$2n$$

Hence It cannot be determined from the given data.
Question 9
1 / -0

The number of prime numbers between $$0$$ and $$20$$ is

A
$$7$$

B
$$8$$

C
$$6$$

D
$$9$$

Solution

We know that, a prime number is a number having exactly two factors, $$1$$ and the number itself.
Hence, the prime numbers between $$0$$ and $$20$$ are $$2, 3, 5, 7, 11, 13, 17$$ and $$19$$

$$\therefore \ $$ The number of prime numbers between $$0$$ and $$20$$ is $$8.$$
Question 10
1 / -0

Which of the following is not a composite number?

A
3

B
6

C
7

D
8

Solution

A composite number is a positive integer that has at least one positive divisor other than one and number itself.
Here 3 & 7 are not composite number as both cannot be divided evenly so 3 & 7 are not composite number.

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Re-Attempt Weekly Quiz Competition

Number Theory Test 6

Result

Number Theory Test 6

Score

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Rank

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SHARING IS CARING

Question 1

50

-50

0

100

Solution

Question 2

$$1$$

$$0$$

$$2$$

$$3$$

Solution

Question 3

$$6$$

$$5$$

$$7$$

$$8$$

Solution

Question 4

Two even numbers

Two odd prime numbers

Two composite numbers

Two co-prime numbers

Solution

Question 5

5

4

1

None of these

Solution

Question 6

$$4$$

$$6$$

$$7$$

$$10$$

Solution

Question 7

3

48

51

2009

Solution

Question 8

Positive

Negative

Zero

Cannot be determined

Solution

Question 9

$$7$$

$$8$$

$$6$$

$$9$$

Solution

Question 10

3

6

7

8

Solution

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