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Practical ApplicationI've said a lot of things in Powers of 2 and 3, Logarithmic Forms, and The Multiplication Table, but most of it isn't relevant here, and the relevant parts aren't very well organized. So, let's not worry about all that! Instead, here's my current thinking in two sentences. Define the symbol [a b] to be the number 3^{a}2^{b}. I'm familiar with a lot of those numbers, and they obey the following rule.
[a b] × [c d] = 3^{a}2^{b} × 3^{c}2^{d} = 3^{a+c}2^{b+d} = [a+c b+d] Now let's talk about number puzzles.
KenKen are fun, you should try them, but for the purposes of this essay, the important thing about them is that some of the clues say that the product of a certain group of digits should be equal to a certain number. Now, most digits are powers of two and three—that is, have the form 3^{a}2^{b}—and as a result many of the numbers are powers of two and three as well. When I was solving the 4×4 and 6×6 puzzles, the numbers were small and familiar, but when I moved on to the 8×8 and 9×9 puzzles, the numbers were larger, large enough that I recognized them as powers of two and three but didn't immediately know which powers they were—that is, what the exponents a and b were. This was extremely frustrating, not only because I didn't know but also because I then had to factor the numbers, and my method of factoring is slowest for exactly such numbers. It wasn't long before I was frustrated enough to do what I should have done a long time ago: make and use some flash cards. That was several months ago. Since then, I haven't practiced as much as I meant to, so my knowledge is still imperfect, but it's a lot better than it was. To see how knowing the exponents is useful, let's look at an example. Suppose I'm working an 8×8 puzzle and I see a linear group of four digits with product 288. I immediately know that 288 = [2 5], so I know the digits have to contain two powers of three. The digit 9 isn't used in 8×8 puzzles, so the powers of three have to come from 3s and 6s; and the group is linear, so the digits providing the powers have to be different. Therefore, there must be one 3 and one 6. Those provide [1 0] × [1 1] = [2 1] of the 288, leaving [2 5] / [2 1] = [0 4] = 16, and clearly the only way to write 16 as the product of two different digits is as 8×2. So, in that way I know what the four digits have to be. The digits aren't always completely determined by the product, but the same kind of thinking is always useful. On the other hand, if I've figured out the digits in some other way, and want to check that the product is what it ought to be, I can just add up the exponents.
When I started thinking about logarithmic forms, I had no idea that using them would be faster and easier than simple multiplication! That's most of what I wanted to say; everything else is just various notes. Here are two quick ones.
Next I'd like to compare logarithmic forms (the symbols [a b]) and regular logarithms. I really like the concreteness of logarithmic forms. With logarithms, it seems to confuse people that you can “take the logarithm” and transform a number into some totally different number that obeys different rules, but with logarithmic forms there's nothing to be confused about. The symbol [a b] means the number 3^{a}2^{b}, and that's that. Could we do the same thing with regular logarithms? Sure! Define [x] = 10^{x} and observe that [x] × [y] = [x+y], and we're all set. In that framework, taking and untaking logarithms amounts to nothing more than the strangely plausible statement that [x] = [y] if and only if x = y. (Plausible and also true, as long as complex numbers aren't involved.) The one time we still need logarithms is when we have a number n and want to write it as [x] but don't know the correct value of x. That is, we want n = [x] = 10^{x}; but then we can just take the logarithm of both sides and get the answer, x = log n. In other words, we have this equation.
n = [log n] The corresponding equation for logarithmic forms is
n = [f_{3}(n) f_{2}(n)], where f_{p}(n) is the number of times that p goes into n. There's no single function that corresponds to the regular logarithm unless you want to define a function that returns an ordered pair. Finally, in case anyone else is interested, here are some notes on flash cards and mnemonics. (I talked about flash cards in relation to my method of factoring in Three Digits and its threesubessay footnote.) If you haven't already looked at the table in Logarithmic Forms, please do that now. First, mainly for my own reference, here's a table of the current set of flash cards, i.e., the powers I'm trying to learn.
The set also includes two cards that don't fit into the table, [4 8] and [6 6]. If you check those numbers against the southeast quadrant of the table in Logarithmic Forms, you'll see that the only powers I don't aspire to learn are [3 8], [4 6], and [6 3]. There are 68 cards in the southeast quadrant. I'm not really interested in the northeast quadrant at the moment, because numbers of the form 3^{a}5^{b} are much less common in KenKen and much easier to factor on the fly. (Easier because pulling out a factor of five is just a left shift operation and because in practice there are at most two powers of five in any KenKen clue.) I did make some cards though before I figured that out. So, there are 15 cards in the northeast quadrant, not counting the overlap along [a 0]. The most useful mnemonic is based on a lovely equation that I discussed in Powers of 2 and 3.
1024 = 1000 + 24 We can rewrite that as [0 10] = 1000 + [1 3], but actually here we don't care about the last term; the important thing is that [0 10] is approximately 1000, or, more generally, that [a 10+b] is approximately [a b] × 1000. Then, for example, if I've forgotten what the value of [1 13] is, all I have to do is remember that [1 3] = 24 and then use my amazing existing knowledge of powers to think of the one that's approximately 24000, which is 24576. (Actually, what I think of is more like “the one that starts with 24000”, which probably says something about my internal representation of numbers.) You can use a related method to remember [a 9]. For example, [2 10] is about 9000, so [2 9] is half of that, about 4500; in fact it's 4608. You can do the same thing with smaller values of b, but I don't find that worthwhile. You could also cut out the intermediate step and jump directly from [2 9] to [2 1] = 9/2 = 4.5. To put the whole thing another way, the powers of two and three that are single digits make a little structure that looks like this.
There's a copy of that structure (and more) at [0 10] because 1024 is close to 1000, but there's also another copy at [3 2] because 108 is close to 100. Unfortunately, that second copy causes more trouble than it's worth. To use it as a mnemonic you need to subtract [3 2], which isn't as easy as subtracting [0 10]. And, it really does cause trouble, because it's the source of a bunch of confusingly similar numbers. From any cell in the structure, you can take one step east and two steps north and find the original number with its digits rotated left. There are even two places where the structure and the translated structure overlap, creating chains of length three. Taking one step east and two steps north is the same as multiplying by [1 2] = 3/4, so what we have here is actually just a different (and better) view of the “multiplying by 4/3” pattern that I discussed in Logarithmic Forms. We have now reached the final note! Here's a corollary to the rule about multiplication of symbols.
[a b]^{n} = [na nb] Since I know the small powers of numbers like 12 better than I know the large powers of two and three, I can use things like this as mnemonics.
[3 6] = [1 2]^{3} = 12^{3} = 1728

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