This is the first in a series of recreational math problems that I will be asking and then answering (to the best of my ability) in this space.\u00a0 Hello?\u00a0 Is anyone still reading this?<\/p>\n
I have been doing recreational math for thirty-some years, but not so much lately.\u00a0 I have had several articles published in Journal of Recreational Mathematics<\/em> (which seems to have suspended publication several years ago).\u00a0 But I still enjoy posing and then solving math problems of the type I present below.<\/p>\n \u2022 \u2022 \u2022<\/p>\n I hate to carry pocket change.\u00a0 [This is not a complaint — this is the setup to the problem.]\u00a0 I hate it so much that I want to minimize the number of coins I carry when I have to make a cash transaction.\u00a0 If I want to carry the fewest possible number of coins, defined as the average number of coins before and after the transaction, what coins should I start with?<\/p>\n Here are some assumptions that will help you solve the problem.\u00a0 [Recreational math always involves assumptions.]\u00a0 Assume that I have an unlimited supply of dollar bills and the only thing I care about here are the coins.\u00a0 Assume that the only coins in circulation are pennies, nickels, dimes and quarters.\u00a0 Assume that I can’t ask the cashier to convert one denomination to another as a side transaction (e.g., give her five pennies in exchange for a nickel).\u00a0 Assume that the cashier makes change in a way that returns the fewest possible number of coins.\u00a0 Finally, assume that it is equally likely for the transaction price to end in $.01, $.02, $.03, etc.\u00a0 In other words, we want to minimize the number of coins we carry, on average, given some arbitrary number of independent, randomly-priced transactions.<\/p>\n Take as long as you would like to work on this.\u00a0 The answer is given below.\u00a0 Good luck.<\/p>\n \u2022 \u2022 \u2022<\/p>\n Wow, that was fast.\u00a0 Congratulations on your mental agility.\u00a0 Now let’s see if you are right.<\/p>\n I initially guessed that I should carry at least one coin of each denomination.\u00a0 The first case I studied was 2Q – 1D – 1N – 3P.\u00a0 If you want to get rid of change, this is not a bad choice.\u00a0 On average, you will unload 2.30 coins per transaction.\u00a0 But your goal here is to minimize the average number of coins carried in your pocket or purse, and in this case the result is 5.85 coins (you carry 7 coins before the transaction and 4.70 coins afterward).<\/p>\n I suspected that this result could be improved upon.\u00a0 Out of curiosity, I looked at the case where one carries no change at all and pays for every transaction with dollar bills.\u00a0 This is in fact the best answer.\u00a0 The average number of coins in your pocket in this case is 2.35. While you are gaining 4.70 coins per transaction, you are starting out with zero coins.<\/p>\n If you wonder why you accumulate change, this is the reason.\u00a0 If you are like me, you offer only currency for 90% of your cash purchases.\u00a0 For every ten currency-only transactions, you will come home with 47 coins (15 quarters, 8 dimes, 4 nickels and 20 pennies).\u00a0 I had thought about verifying this result by counting the coins in my own change dish, but since I fish out the quarters and dimes to use in parking meters, my ratios would be off.<\/p>\n To close, let’s consider the effect of the “take-a-penny” dish next to the cashier.\u00a0 If I carry no coins with me and use one penny from the dish whenever the price ends in 1 or 6, then the number of coins I accumulate per transaction drops from 4.70 to 3.90.\u00a0 If I were also to “balance the books” by putting a penny in the dish whenever the price ends in 4 or 9, then I would accumulate only 3.70 coins per transaction.<\/p>\n So, here’s the bottom line — if you hate change, only carry currency, use the penny dish, and vote for conservatives.\u00a0 Asked and answered.<\/p>\n","protected":false},"excerpt":{"rendered":" This is the first in a series of recreational math problems that I will be asking and then answering (to the best of my ability) in this space.\u00a0 Hello?\u00a0 Is anyone still reading this? I have been doing recreational math … Continue reading