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A book costs $1 plus half its price. How much does it cost?
- Thread starter donphan fan
- Start date
same, i tried to read it but i only reached page 5I gave up on this book.
The part where u make it so they pay 9 dollars each is wrong
This is a very simple question that I don't think is on par with the one in the OP. The one in the OP relies on the obfuscated wording, questioning whether cost is the same as price, and more. The one here barely makes sense given that the equation you provide doesn't even include the $5 refund you mention, so yeah, it's the "wrong" equation in the sense that you gave information that isn't even provided in your question. For example, where did the $9 come from? If anything you should be asking where the $3 in refunds went instead of where the "extra dollar" went.
Here's a better question:
Given a set of numbers adds up to 25, what is the highest number you can make when all of the numbers are multiplied?
Example: Split 25 into 5*5. When all of the terms are multiplied, you'll get 5^5, or 3125 (you can obviously go higher)
I'll do a writeup for this problem later.
Given a set of numbers adds up to 25, what is the highest number you can make when all of the numbers are multiplied?
Example: Split 25 into 5*5. When all of the terms are multiplied, you'll get 5^5, or 3125 (you can obviously go higher)
I'll do a writeup for this problem later.
jerome powell has several thousand guatemalans in his basement doing this on loop every time the fed needs to print money
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if you don't specify "positive" in the requirements for a number to be in the set, the answer is infinite.
my example set is (25, 2, -2, 2, -2, 2, -2 ...)
then even assuming they have to be positive, you should probably also specify integers or not...
if it has to be integers, then the answer is pretty straightforward to use as many 3s as possible, followed by filling out the remainder with 2s, so in this case 7 3s and 2 2s add up to 25 and multiple as 3^7 * 2^2 = 8748
this is because any 4s in the set could break down to 2 2s with an equivalent result and any 5s or higher can be broken down into 2 numbers with a product higher than itself i.e. 5 becomes another 3 and 2, which multiply to 6.
if it can be any number, I don't know off hand how to approach the problem but I would assume you can get considerably higher than 8748. 2.5^10 for example is ~9536.7 and there's surely much better than that. my best guess to the approach is that the maximum would likely come from one of the options where you divide 25 evenly into parts, such as 25/10 -> 2.5, and then raise it back to the 10th, or you could divide it in 9 parts or 11 parts etc. But short of brute force testing them all (or at least testing sequential numbers of partitions to see if it's increasing or decreasing? that might be sufficient), I don't immediately see how to know before hand which would be the biggest.
This method ends up with (25/x)^x, which is easy enough to plug into a plotting software. The result I got from wolfram|alpha is that the maximum occurs at x=25/e (approximately 9.2), with a value of e^(25/e)~=9867.
you can't have 25/e elements of a set though, so then the answer should either be 10 elements of 25/10 or 9 elements of 25/9 since those would come closest to the optimal solution with 25/e and this function shouldn't have other higher maxima farther away from 25/e.
(25/10)^10=~9536.7
(25/9)^9=~9846.4, so this should win
But I'm still not entirely sure how to rigorously prove though that this is the best possible solution and not just the best possible solution of the form you described. It just comes from an intuition that multiplication tends to work this way, i.e. x^2 is bigger than (x+1)(x-1) and extending that idea to higher powers that the most efficient solution is most likely one that uses a similar rule.
What
talk to me when this thread reaches page 5 + half its length.........
see you on page 10
One would think that the answer to this question is $1.50 USD, but the answer to the provided prompt is deceptively tricky to understand underneath the surface.
If a book costs $1 plus half its price, than the book cannot cost $1.50 because half of this price is $0.75, and this plus $1 would equal a price of $1.75, not $1.50. In reality, there is a different finite “correct” answer to this question because the order of operations states that, when X equals the book’s full price, (X divided by 2) would be calculated before adding the additional $1. Therefore, when calculating the equation, ($1 + X / 2 = X), a finite value of X would have to exist that completes the equation on both sides of the equal sign. Upon further inspection, one can discover that 2 is a viable option for the value of X, because 2 divided by 2 equals 1, and 1 plus 1 equals 2. We will ignore the part where I incorrectly voted for “other options” in the poll.
If a book costs $1 plus half its price, than the book cannot cost $1.50 because half of this price is $0.75, and this plus $1 would equal a price of $1.75, not $1.50. In reality, there is a different finite “correct” answer to this question because the order of operations states that, when X equals the book’s full price, (X divided by 2) would be calculated before adding the additional $1. Therefore, when calculating the equation, ($1 + X / 2 = X), a finite value of X would have to exist that completes the equation on both sides of the equal sign. Upon further inspection, one can discover that 2 is a viable option for the value of X, because 2 divided by 2 equals 1, and 1 plus 1 equals 2. We will ignore the part where I incorrectly voted for “other options” in the poll.
Assuming “half its price” means half the books current price, it just keeps adding onto itself forever