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.
2³ = 2 x 2 x 2 = 8
Powers of 2 get large very quickly. It’s true that 2¹ is equal to 2 and 2² equals just 4. But 2¹³ 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
2⁰ = 1
2¹ = 2
2² = 4
2³ = 8
2⁴ = 16
2⁵ = 32
2⁶ = 64
2⁷ = 128
2⁸ = 256
2⁹ = 512
2¹⁰ = 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.
The formula for winning number of hidden treasures is
2ⁿ - 3
Since, David is working with 16-bit system, we have to plug n = 16 in the above formula.
Then, the winning number of treasures is
2¹⁶ - 3 = 65536 - 3
2¹⁶ - 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 ?
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
(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
= 2⁵- 1
= 32 - 1
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 ?
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
(But it includes the way of not attempting all the questions)
So, no. of ways to attempt atleast one question in Algebra is
= 2⁶- 1
Similarly, no.of ways to attempt atleast one question in Geometry is
= 2⁴- 1
Total no.ways for both the sections is
= (2⁶- 1)(2⁴- 1)
= 63 x 15
Hence, the no. of ways of attempting at least one question from each section is 945
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