|
Question 443434: Who wants the Money, a game show, distributes prizes that are powers of 2. What is the minimum number of prizes that could be distributed for $900?
Found 3 solutions by swincher4391, Alan3354, richard1234:
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Hahaha, I like this problem.
Alright, so you need to know your powers of 2. If you think about it long enough, 2^9 = 512. So let's start with that.
2^8 = 256
So we're up to 512 + 256 = 768
2^7 = 128
So we're up to 768 + 128 = 896
We need 4 left... luckily 2^2 = 4.
So we've used, 2^9, 2^8, 2^7, 2^2 to get $900 evenly.
Then we have given out  prizes.
==============
If you don't like this guess-and-check method, there is a fool proof method.
Convert 900 to binary.
900 /2 = 450 r 0 ---> 0
450 /2 = 225 r 0 ---->0
225 / 2 = 112 r 1 ---> 1
112 /2 = 61 r 0 ----> 0
56 / 2 = 28 r 0 --- > 0
28 /2 = 14 r 0 ----> 0
14 /2 = 7 r 0 ----> 0
7 / 2 = 3 r 1 ---> 1
3 /2 = 1 r 1 ----> 1
1 /2 = 0 r 1 -----> 1
Then the binary representation is 111000100.
Count the number of 1s.
There are 4. That is our answer.
Then we have given out prizes.
Answer by Alan3354(69443)  (Show Source):
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! The binary representation of 900 is 111000100. Thus, a minimum of four prizes can be given out, with $512, $256, $128, and $4 prizes.
|
|
|
| |
