Thursday, January 17, 2008

REALLY basic example

I got my hands on a copy of Short-Cut Math recently and I'm loving it. The author presents methods for simplifying almost any tough problem involving addition, subtraction, multiplication, or division. One example he gives is how multiplying by .25, 2.5, 25, or 250 can be done (sometimes more easily) by dividing by 4 and then multiplying by 1, 10, 100, or 1000, respectively. Obviously this will be faster or slower depending on how easy the number is to divide by 4 and how well you do with division, but it's an option available to speed things up in some cases.

He also discusses how multiplying by a number comprised of the same digit (for example, 888) is easier because you only have to take the partial answer from the first digit and adjust by a factor of 10 rather than getting a different partial value for each digit. Likewise, numbers comprised of multiples (like 842 or 168) are easier than others because you get the answer for one digit and multiply it by the appropriate factor for each other digit or set of digits.

With that background information out of the way, on to the problem I'm solving here. The other day I was trying to subtract some numbers from each other (48-16, 56-28, 132-24, for example) and felt like there was a pattern that might lead to a non-standard method for solving them. In case you didn't catch it right away, the minuend and subtrahend in each case share a common factor (8, 7, 12 being the most immediately apparent to me at the time). The aforementioned multiplication trick led me to think that there might be a trick I could use involving this property. If we examine the answer for each (32, 28, 108), it may be obvious that the answer also shares the factor. What's interesting is that the answer is the result of the following equation: (minuend/factor - subtrahend/factor)*factor. I happen to like this method because division and multiplication come much more naturally to me than subtraction.

Of course, in subtraction this might not always be terribly practical to do. It's very nice when one immediately notices that the subtrahend in 56-28 is itself a factor of the minuend, but it's not always that simple. The time it takes to identify the common factor, divide by it, subtract, and multiply the result by the factor may be longer than what it would take to do the subtraction with the original numbers (especially if you have other short-cuts at hand), but can it be generalized? For addition, it seems equally applicable and equally problematic. We still have to divide at least twice and multiply at the end. In multiplication, a similar method can be used with similarly questionable improvement. It would be of the form ((multiplicand/factor)*(multiplier/factor))*factor^2, which certainly does not appear simpler than some of the other short-cuts available. It can, of course, be even more problematic if no factor is shared and we wasted time trying to find one.

Now, can we find a group of cases for which this method is especially well-suited? What if we are dealing with a long list of numbers that is known beforehand to share a common factor? One such example might be if you were trying to calculate the amount of memory available in a cluster of computers with different amounts of RAM. We might know ahead of time that their RAM comes in increments of 256MB. Even if you're not geeky enough to be accustomed to thinking quickly in increments of 256, this problem is very well suited to our new method. You can add up the total in smaller numbers and multiply at the end by 256 or just divide by 4 and get the total in GB.

So, we see that this method might be overly complex for small cases or ones in which the factors are previously unknown or foreign to us, but is fairly well-suited to problems with a long list of numbers which all share a factor that is known in advance and perhaps familiar to work with.