July 1, 2022
From Andrew Love comes a puzzle about planning ahead and resolving to get things done:
It’s the morning of Jan. 1, 2023, and you have a task you want to complete before the end of the year, on Dec. 31. The task is easily divisible into portions, so you always know exactly what fraction of the task is left.
Your initial plan is to do exactly 1/365 of the task every day. But then you think to yourself, “If I do 2/365 today, I can do a little less every day from now on.” And so, on Jan. 1, you would complete 2/365 of the task. On Jan. 2, 363/365 of the task remains with 364 days left to complete it, so you would do another 2/364 × 363/365 of the task.
If you continue in this fashion every day, dividing the remaining work by the remaining number of days in the year and doing twice that amount, when would you finish?
It will still take 364 days.
Explanation:
The key is to think critically before diving into the algebra, also in fact the algebra needed to solve this puzzle is actually very simple. Remember, the question asks when you would finish, meaning on which day will the entire work remaining be completed. Let's solve for the general case here. Let the work be initially divided into \(D\) days to complete, and \(D_{\text{remaining}}\) be the days remaining to completion. Let the original plan be to work \(1/D\) of the entire task per day. For you to "improve" on the efficiency of your work by working some \(n/D_{\text{remaining}}\) of the remaining amount instead, it is important to consider if on a particular day that all remaining work would be completed. Let \(r\) be the remaining work, and let \(k\) be the day you want the entire work to be done. On that day, you would work \(n/(D - k + 1) \times r\). However, on the \(k\)th day, you need to work exactly \(r\) amount. In other words, you need \(n/D_{\text{remaining}} = 1\) on the \(k\)th day. Therefore, on that day, you need to have \(D_{\text{remaining}} = D - k + 1 = n\), and hence \(k = D - n + 1\).
To re-iterate, let \(D\) be the original number of days to complete the task. Let \(n\) be some positive integer denoting the amount of work per number of days remaining on the remainder of the work to be done, and let day \(k\) be the target day where all the work is complete. Then, \(k = D - n + 1\).
As a sanity check, let \(n = 1\), i.e. the original plan. Then the day of completion is \(k = D - 1 + 1 = D\), as desired.
In the original puzzle, \(D = 365\), \(n = 2\), and so \(k = D - n + 1 = 365 - 2 + 1 = 364\) days.
Remark 1: One way to think about why you only get \(n-1\) days off as compared to the original plan of \(1/D\) per day is that on all days prior to the last day, you will be doing \(ar\) work, where \(r\) is the remaining amount and \(a\) is a fraction of the remaining work that will always be less than 1. Therefore, on all days prior to the last day, you will never tackle the entire work.
Remark 2: Athough you only get \(n-1\) days off as compared to the original plan of \(1/D\) per day, the amount of work you need to do on each day drops drastically as \(n\) increases, such that on the \(k\)th day, the work remaining approaches to zero, unlike the original plan where each day the work is equal.