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What Is Memorable?
Now, with those few preliminaries out of the way, I can tell you the real reason that the vast majority of products are memorable: because I found some clever way to multiply the factors, and the clever way is memorable. So, all I need to do is tell you the clever ways to multiply, and then you'll really know everything. In that sense, this essay is the inverse of Other Methods, which is about clever ways to divide. Anyhow, here are the ways, starting from the most general.
- Standard multiplication. I can run the standard multiplication algorithm in my head, but it's quite slow, and doesn't have a nice visual representation; hence it's not very memorable. I don't remember any products by way of the standard algorithm.
- FOIL. In junior high I learned the excellent mnemonic “FOIL”, short for “First Outer Inner Last”. It was designed for multiplication of binomials, but it applies equally well to multiplication of two-digit numbers (which are actually binomials over the fixed base b). For example,
31×41 = 1200 + 30 + 40 + 1 = 1271.
The number 1200 comes from the product of the first digits, and so on. I don't remember any three-digit products by way of this algorithm, but it's the next most general after standard multiplication. By the way, if the first digits or last digits are the same, you can optimize by combining the outer and inner multiplications.
- Split factor I. Any product can be computed in four different ways, by first choosing a factor to split and then choosing whether or not to apply a second complement to it. Here are the ways for the product 611 = 13×47.
|(10+3)||× 47||= 470||+||141|
|(20-7)||× 47||= 940||-||329|
|13||× (40+7)||= 520||+||91|
|13||× (50-3)||= 650||-||39|
The first and third ways are just standard multiplications, but the second and fourth are new. As you can see, the four ways are not equally easy to use; by picking the easiest (which depends on the factors), one can get something of an improvement over the standard.
In particular, since here the last digit of the split factor has to be 1, 3, 7, or 9, by choosing wisely as regards the complement, one can make it become either 1 or 3. If that reminds you of backward modulation, good! This method is really just backward modulation in reverse. (For example, to see if 611 is divisible by 13, the first step would be to add 39.) As a result, I was tempted to call the method “reverse backward modulation”, or maybe “double secret reverse backward pseudo-modulation”.
- Split factor II. This method is just a common special case of the above: 75% of the time, one can arrange for the split last digit to be 1, and then that term doesn't require any multiplication at all. The product 551 = 19×29 is a neat example, because it can be expressed as either 570 - 19 or 580 - 29.
Another neat thing is that when a prime pair surrounds a multiple of 10, the products split like little atomic levels.
|17×29 = 17 × (30-1)||= 510||-||17 = 493|
|17×31 = 17 × (30+1)||= 510||+||17 = 527|
- Small factor I. If one of the factors is (very) small, it's sometimes possible to compute a product by concatenation. Here the symbol “&” represents the concatenation operator.
7×41 = 7 × (4 & 1) = 28 & 7 = 287
It's slightly rarer, but the digits in the product can also split the other way.
7×113 = 7 × (1 & 13) = 7 & 91 = 791
- Small factor II. When there's a carry from the right half, you can't use true concatenation, but you can still think of the left half as a base that the right half adds on to.
7×137 = 7 × (130 + 7) = 910 + 49 = 959
That's just standard multiplication all over again, of course; what makes this method different is the idea that several products can use the same base. Here the number 910 is by far the best example … there are three products 7×13_ and two products 13×7_ that use it as a base.
The prime pair examples I mentioned in split factor II can also be thought of in terms of bases, albeit with negative offsets for the smaller primes.
- Small factor III. In small factor I, we saw one- and two-digit right halves; in small factor II, we saw a one-digit right half with carry; what about a two-digit right half with carry? Well, most of the time, that's too complex … the result isn't memorable. There is one special case where it works, however: if the two digits form a prime, the right-half multiplication is easy, because it's another product! For example,
7×173 = 7 × (100 + 73) = 700 + 511 = 1211.
This method isn't significant for three-digit products, but it does come in handy later on.
- Difference of squares. This is basically just the Babylonian multiplication algorithm that I mentioned in In Mathematics; the only difference is that since we're multiplying (odd) primes, we don't have to worry about the center being a half-integer. Here's the fundamental identity, rephrased in terms of the center c and difference d (delta).
(c-d)(c+d) = c2 - d2
This method is memorable if the factors are close together, …
19×23 = (21-2) × (21+2) = 212 - 22 = 441 - 4 = 437
… or if they add up to a nice round number (a multiple of 10).
17×43 = (30-13) × (30+13) = 302 - 132 = 900 - 169 = 731
Of course, the method only works if you know the squares!
- Multiplication by 11. To multiply by 11, you just shift to the left and then add the original number back in. That's trivial when there's no carry, as in 451 = 11×41, and still pretty easy when there is one, as in 649 = 11×59.
- Multiplication by 49. To multiply by 49, you just shift right in binary, shift left twice in decimal, and then subtract the original number. But why would anyone want to do that? Well, there are four three-digit products of the form 72×p, and they're hard to remember until you see the pattern, which is a consequence of the multiplication algorithm.
For thoughts about how to shift left and right, see Operations.
Multiplication in Base 10
Multiplication Table, The
@ November (2004)