Number Crunching

Had a million dollars worth of nickels and dimes,
She sat around and counted them all a million times,
Cab Calloway (and many others) – Minnie the Moocher

It occurred to me this morning, just how long would that take?

The first thing to establish is just how many coins would a million dollars in nickles and dimes add up to? A dime is ten cents, so a million dollars worth of dimes would add up to 10,000,000 coins (which incidentally would weigh 22.68 tonnes). A nickle is five cents, so a million dollars worth of them would be 20,000,000 coins (weighing 100 tonnes). We don’t have any information on exactly how Minnie’s cash is divided, but we can average it out and assume that she has 15,000,000 coins to count (weighing about 61.34 tonnes – better hope that gold and steel house she got from the King of Sweden has reinforced floors).

We can make a not unreasonable assumption that it takes about a second to count a single coin. Furthermore over the course of counting 15,000,000 coins Minnie would probably get pretty good at it, so let’s assume that it takes her only half a second per coin. This means that to count the whole lot would take Minnie 7,500,000 seconds. This is 2,083.3333 hours, or 86.8 days – assuming that Minnie takes a whole bunch of stimulants and never sleeps. It’s actually more reasonable to assume that she spends only 8 hours a day counting coins, so she has time to enjoy racing those horses and riding around in her diamond car.

Taking that into account, counting all the coins once would take 260.41 days, or 0.71 years.

That doesn’t seem too bad – but remember, she counts them a million times. To achieve this Minnie would need to spent eight hours a day counting coins for 713,001.49 years.

Poor Min indeed.

I still got it!

Anyone recognise the formula?

I just spent 20 minutes converting the formula…

d = M-((M/2)+7)


M = 2(d+7)

I still got it baby!

(Even if it would have taken me about a 10th of the time back in high school :))

The steps are…

d = M-((M/2)+7)
M = d+((M/2)+7)
M = d+7+(M*0.5)
M-(M*0.5) = d+7
(This one took me ages)
0.5*M = d+7
M = 2(d+7)

Now, does anyone recognise the base formula?

The Laughter of Mr Rose

A tale from my disreputable past

I was thinking the other day of an incident that happened to me in high school. Not a hugely important or earth-shattering incident, just one that sort of illustrates a point about how you can sometimes be too intelligent for your own good.

The incident occurred in year eight maths. My teacher was Mr Rose (not his real name by the way), a youngish and slightly arrogant fellow with the looks of someone who’d much rather be strutting up and down the beach in a speedo than stuck inside forcing mathematics down the throats of a bunch of unwilling thirteen year olds. This particular day, towards the end of the school year he posed us a problem to do with a clock face.

He gave us an angle, and claimed that at only one time of the day did the hour and minute hands of the clock form said angle. Our job was to determine what time of day that was.

A simple question you might think. But for the life of me I couldn’t figure it out! I did all the maths I could, and even resorted to rigging up a crude clock face with a protractor and a couple of pencils, but I couldn’t for the life of me find the answer. What was particularly disturbing was that all around me my classmates – even the particularly thick ones – were apparently figuring it out and going up to Mr Rose to be marked. The best I could do was a rough estimate (around 5:42 I seem to recall) which Mr Rose totally rejected. How were they doing it!?

The class finished without my finding an answer, and I got 0 marks on that particular exercise.

It was some years later – after I’d left high school – that I figured out what I’d been doing wrong.

You see, there was an unspoken assumption about the exercise. The hands were assumed to instantaneously jump between set points on the clock face without crossing the space in between. Much like electrons jumping between valence shells within an atom, they just plain didn’t exist between these points. This meant that the minute hand could only occupy 60 positions on the clock face, and the hour hand only 720 – a fairly manageable number of angles to account for with a well structured mathematical relationship between them.

I on the other hand was assuming an analogue clock face where every division of every angle counted and hence – although I didn’t realise it at the time – the positions and angles of the hands were infinite. The problem as I understood it was unsolvable without inventing differential calculus, which was a bit beyond me at the time as I was only thirteen years old and wasn’t Sir Isaac Newton.

So yeah, that’s the story. If I’d been a bit dumber I would have assumed that the hands could only point to round minutes from the start and would have solved the problem in no time. As it was I outwitted myself by thinking the problem was about a real clock face, and not the numbers represented by one.

Mr Rose is probably still laughing at me.

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