Calculating amount left at the end of a repetetive cutting down cycle.

In summary: No, it's not. The process actually works as follows:You start with an original amount of X's and each time you apply the find-and-replace process, the amount of X's that are replaced is a smaller amount. So, in this example, the first step would replace the first X with a 1, the second step would replace the 2 with a 2, and so on.
  • #1
Zekes
4
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Let's say I have a group of X's characters that I want to cut down. I use a tool similar to 'find-and-replace-all', which takes an amount of X's each time, deletes them, and replaces them with another, smaller amount. For example, I can state that I want an amount of 8 X's to be replaced with 5 X's from an original amount of 14 X's. This cycle happens forever, or until there isn't a sufficient remainder left to replace anymore. To visualise the example I just provided:


  • (XXXXXXXX)XXXXXX --> [XXXXX]XXXXXX
    (XXXXXXXX)XXX --> [XXXXX]XXX
    (XXXXXXXX) --> [XXXXX]

So, you start with 14 X's and end with 5 X's, and the process carried out here can be summarised as '14, 11, 8, 5'. The process that was underwent in terms of finding and replacing can be described as (8,5) as 8 X's are replaced with 5 each time. Now, the question here is:


  1. When you initially have a string of 1,000,000 X's, and apply the find-and-replace process of (99,90), what is the summary of the process here?
    By summary of the process, I mean the '14, 11, 8, 5' which shows how many X's are left after each step.

I tried solving this manually, starting the process summary as '1,000,000, 909,091, 826,453, 751,321, 683,020, 620,929, 564,481, 513,172, 466,525' however I realized this was a very inefficient way of solving this, and prone to mistakes. Can anyone help me do it in a better, easier way?

ALSO FOLLOW UP QUESTION

  • For the (7,3) process, find all start lengths which eventually result in an end length of 6.

I can see 10 being applicable for this, however manual testing every number to find a suitable match is also quite inefficient. If you can help me for this question too, please do. Also, please explain in more simple math terms, rather than using symbols (as I am still a beginner) Thank you!
 
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  • #2
Zekes said:
Let's say I have a group of X's characters that I want to cut down. I use a tool similar to 'find-and-replace-all', which takes an amount of X's each time, deletes them, and replaces them with another, smaller amount. For example, I can state that I want an amount of 8 X's to be replaced with 5 X's from an original amount of 14 X's. This cycle happens forever, or until there isn't a sufficient remainder left to replace anymore. To visualise the example I just provided:


  • (XXXXXXXX)XXXXXX --> [XXXXX]XXXXXX
    (XXXXXXXX)XXX --> [XXXXX]XXX
    (XXXXXXXX) --> [XXXXX]

So, you start with 14 X's and end with 5 X's, and the process carried out here can be summarised as '14, 11, 8, 5'. The process that was underwent in terms of finding and replacing can be described as (8,5) as 8 X's are replaced with 5 each time.
Isn't this simply repeatingly subtracting 3 from 14 ?
14 - 3 = 11 - 3 = 8 - 3 = 5 ...whoa!
 

FAQ: Calculating amount left at the end of a repetetive cutting down cycle.

How do you calculate the amount left at the end of a repetitive cutting down cycle?

To calculate the amount left at the end of a repetitive cutting down cycle, you will need to know the initial amount, the cutting rate, and the number of cycles. The formula is: amount left = initial amount - (cutting rate * number of cycles).

What is the significance of calculating the amount left at the end of a repetitive cutting down cycle?

Calculating the amount left at the end of a repetitive cutting down cycle is important for understanding the impact of continuous cutting on a resource. It can help determine if the resource is being depleted at a sustainable rate or if it is being overexploited.

How does the cutting rate affect the amount left at the end of a repetitive cutting down cycle?

The cutting rate directly affects the amount left at the end of a repetitive cutting down cycle. The higher the cutting rate, the more of the resource will be depleted after each cycle, resulting in a smaller amount left at the end.

Can the formula for calculating the amount left at the end of a repetitive cutting down cycle be applied to any resource?

Yes, the formula can be applied to any resource that is being continuously cut down. However, the accuracy of the calculation may vary depending on the specific characteristics of the resource and its rate of regeneration.

Are there any limitations to using the formula for calculating the amount left at the end of a repetitive cutting down cycle?

One limitation of the formula is that it assumes a constant cutting rate and does not account for any fluctuations or changes in the rate over time. It also does not consider other factors such as natural disasters or human interventions that may impact the resource. Therefore, it should be used as a general guideline rather than an exact prediction.

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