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Efficiently sorting a lot of coinsOf course there is no one single way to sort a lot of coins, and believe me, I have tried almost all of them. What I am here to share with you is the best, most efficient way I have ever found to sort a large number of coins.Most people would automatically assume that sorting them first by decade, then by year, then by mint would be the easiest way to do it - basically a left-to-right sort, in the same order you read the date. Well, I found just the opposite to be true. What you need
How to sortMark each of your containers with each digit from zero through nine. Set the containers out so that you start with zero, and line them up five across to the right until you set out the cup marked 4. Start a new row closer to you with container 5 on the left (just near you from the zero container), and set them up across to the right until you reach the last cup, marked 9.Lay some coins out in front of you and start placing the coins in the containers using ONLY the last digit of the date. For instance, all coins from 1960, 1970, 1980, 1990, and 2000 go into the zero container. Mark your bags from zero through nine, and place the coins into the bags as the containers become full. When you are finished sorting, dump all the containers into the bags. Once you are finished with this process, each of your bags should be full of coins with last digits from zero through nine, and your containers should be empty. Use the containers for decades (you will probably only need the 5, 6, 7, 8, 9, and 0 containers for this step) and resort your first bag (zero) back into the containers by decade. Once you're finished with this step, each container should contain only one date. Sort each container by mint (and date size if applicable), and you're finished with that digit. Only nine to go! How is this method more efficient than others?It's simple, actually. If you sort the decades first, you would be sorting into at least five different containers, then you would need ten more to sort each date from the decades. In effect you would be sorting your largest group (the ones digit, containing ten possibilities) at least four more times than necessary. Sorting that largest group once cuts down the number of possibilities to sort the next time, effectively cutting down the repetitive sorting by at least half.It's more difficult to explain why this is more efficient than it is to simply demonstrate it, and since I cannot easily do that, you would need to take my word that it is easier, faster, and far more efficient. |
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