Copasetic Flow

D.r Harold Daw In the late 1950’s, my uncle, then a teenager, found himself at odds with the Law in Las Cruces, NM.  Simply put, the police had decided he was, in fact, his brother, wanted for fleeing a drag race in the same car my uncle had just driven home.  Despite his repeated pleas that they were arresting the wrong Carter boy, the police persevered, first handcuffing my uncle, and then shuffling him into the police car.  Unbeknownst to them, the woman who lived across the street from my dad’s family had observed everything.  Turning to her husband, she said Harold, they go the wrong boy!  You go down to the station and help them straighten this out!”  That’s how my uncle found himself riding back to his house in the car of Dr. Harold Daw, head of the New Mexico State University physics department.  The Hot Rod in Question The reason I bring this story up now, isn’t because of its somewhat topical nature in relation to the state of ...

Today’s math fun involves portions of yesterday's, but with a few more steps.  It may also—dare I say it?—involve partitioning!  I might be using the word partitioning in an incorrect way, and if so, then pardons please, (also, please let me know).  The question is, what are the fewest number of coins you need to make change for up to a dollar. Here's how I worked, using an iterative algorithm, (fancy words for: "I'm going to use the same trick over and over").  It was all about granularity of coins, and getting quickly from one amount to the next.  The quickest way, (where quick is defined by using the smallest number of coins), to get to a large amount of change is with large coins.  So, as we move around within 99 cents, the biggest step we can make is with a half dollar. Using yesterday’s method, we can fit one half dollar into 99 cents.  That leaves us with 49 cents left move around in.  The quickest way to make progress within that interva...

Working through the problems in Niven's book on combinatorics, I came across the following one that cleverly introduces least-common-multiples without saying any of those words.  The book asks the following question: How many numbers that are evenly divisible by 11 exist between 1 and 2000?  How many that aren't also evenly divisible by 3?  How many numbers that are evenly divisible by 6, but not by 4 exist between 1 and 2000? The hastily scribbled answer can be seen below, with each of the answers boxed in succession down the screen. By simply dividing 2000 by 11, we find out how many integers between 1 and 2000 are evenly divisible by 11.  In other words, we ask how many multiples of 11 can fit between 1 and 2000.  When we want to eliminate the multiples of 3, that's when the least common multiple comes in.  We already have the answer for all numbers divisible by 11, but how to eliminate those also divisible by 3?  By first asking what number...