The power of 2 is nothing but multiplying 2 by itself.

For example, if the power of 2 is 3, we have to multiply 2 by itself 3 times.

More clearly,

2^{3}** = **2 x 2 x 2

2^{3 }= 8

Powers of 2 get large very quickly. It’s true that 2^{1} is equal to
2 and 2^{2} equals just 4. But 2^{13} equals 8192, which is slightly
more than Earth’s diameter in miles !

Let us see, how the value of "powers of 2" gets larger and larger when we increase the powers from 0 to 10

2^{0} = 1

2^{1} = 2

2^{2} = 4

2^{3} = 8

2^{4} = 16

2^{5} = 32

2^{6} = 64

2^{7} = 128

2^{8} = 256

2^{9} = 512

2^{10} = 1024

**Freelance Computer Programmer :**

David is a freelance computer programmer contracted by a company that makes video games.

In his work, he often used the power of 2ⁿ to find the number of bits (or units of information) that can be arranged. The n stands for the number of bits.

In the video game of hidden treasures that he is programming, the main character wins if he collects 3 fewer hidden treasures than the highest number of bits that can be arranged in the system.

If David is working with a 16-bit system, is the main character a winner if he has collected 65,000 hidden treasures ? Explain your reasoning.

**Solution : **

The formula for winning number of hidden treasures is

**2 ^{n} - 3**

Since, David is working with 16-bit system, we have to substitute 16 for 'n' in the above formula.

Then, the winning number of treasures is

2^{16} - 3 = 65536 - 3

2^{16} - 3 = 65533

Hence, the main character is not a winner. Because he has only collected 65,000 hidden treasures which is less than the winning amount 65533.

**Problem 1 :**

A man has 5 friends. In how many ways, can he invite one or more of his friends to dinner ?

**Solution :**

We have '2' alternatives for each friend. That is, either he may invite or he may not invite.

Therefore, no. of all possible ways to invite 5 friends to dinner is

= 2 x 2 x 2 x 2 x 2

= 2^{5}

(But it includes the way of not inviting all the 5 friends)

So, no. of ways to invite one or more of his friends is

= 2^{5 }- 1

= 32 - 1

= 31

Hence, the no. of ways to invite one more of his friends is 31.

**Problem 2 :**

A examination paper with 10 questions consists of 6 questions in Algebra and 4 questions in geometry. At least one question is to be attempted from each section. In how many ways can this be done ?

**Solution :**

We have '2' alternatives for each question. That is, either we may attempt or we may not attempt.

Therefore,no. of ways to attempt six questions in Algebra is

= 2 x 2 x 2 x 2 x 2 x 2

= 2^{6}

(But it includes the way of not attempting all the questions)

So, no. of ways to attempt at least one question in Algebra is

= 2^{6 }- 1

Similarly, no.of ways to attempt atleast one question in Geometry is

= 2^{4 }- 1

Total no.ways for both the sections is

= (2^{6 }- 1)(2^{4 }- 1)

= (64 - 1)(16 - 1)

= 63 x 15

= 945

Hence, the no. of ways of attempting at least one question from each section is 945

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