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
23 = 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 substitute 16 for 'n' 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
= 2 x 2 x 2 x 2 x 2
(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
= 2 x 2 x 2 x 2 x 2 x 2
(But it includes the way of not attempting all the questions)
So, no. of ways to attempt at least 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)
= (64 - 1)(16 - 1)
= 63 x 15
Hence, the no. of ways of attempting at least one question from each section is 945
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