Quick and Easy Math

This is the fourth book in the “realm” series of mathematics texts done for Houghton-Mifflin, and the first lacking “realm” in the title. Like the final book in the series, An Easy Introduction to the Slide Rule, Quick and Easy Math has been obsoleted somewhat by the advent of electronic calculators in the early 1970’s, although not nearly as badly. It is not, after all, entirely uncommon to come across a situation where you need to do some quick math and a calculator isn’t handy, whereas it is virtually unheard of for anybody to actually use a slide rule nowadays.

(And I must confess to manufacturing opportunities to do mental math for myself. When I'm stuck in something boring where reading would be impossible or rude, I often set up math problems for myself and solve them as a way to pass the time.)

On the whole, the book is good. The tricks are worthwhile, the rationality behind them is sound, and Asimov explains everything lucidly. I do, however, have some quibbles with the book in terms of what it leaves out.

One is a very useful trick for multiplication, using the fact that (a + b)(a - b) = a² - b². A table of squares can be memorized without much trouble, and squares can be calculated more easily than other multiplications. It’s much faster to see that 14(16) = 224 this way then by repeated doublings.

The more baffling omission is regarding divisibility tricks. Asimov gives tricks for 1, 2, 3, 4, 5, 6, 8, 9, and 10, and then states by fiat that there are no other tricks one can use beyond combinations of these, so you just gotta do the division. That astonished me. There is a fairly trivial divisibility test for 11, and a relatively simple one for 7 which can be adapted for 13, 17, and so on. There is also a very important point to make when testing for divisibility—since you only care about the remainder, even if you’re forced to do the division, you can throw away the quotient as you go along.

(Sorry about this lengthy criticism. It so happens that the mental problems I set up for myself are usually doing prime factorizations of large numbers, so divisibility is a key point to me.)

Anyway, I do enjoy this book a lot, and it’s among my favorites. It has, as I say, a strange weakness but is a generally good and worthwhile volume.

Finally, an aisde: I have been taken to task for mentioning in an earlier version of this review that there are easy tricks for determining if a number of a multiple of seven, or eleven, and so on, without bothering to give the tricks.

Now, first be aware that the trick for 3 (and 9) depends on the fact that 10 is 1 more than a multiple of 3 (and 9). So if `a` is a multiple of 3, then so is `a` x 10, and `a` x 100, and so on. If you have, then, a three digit number, which is `a` x 100 + `b` x 10 + `c`, it's a multiple of 3 (or 9) if and only if `a` + `b` + `c` is one, too. Thus the trick: add all the digits of the number you want to test. If necessary, add all the digits of the result. Keep doing this until you're down to one digit. Then that's a multiple of 3 (or 9) if and only if the original number was.

So for 2909909, we get 2 + 9 + 0 + 9 + 9 + 0 + 9 = 38, 3 + 8 = 11, 1 + 1 = 2, so 2909909 is not a multiple of 3 or 9. 2909907, on the other hand, gives us 2 + 9 + 0 + 9 + 9 + 0 + 7 = 36, 3 + 6 = 9, so 2909907 is a multiple of both 3 and 9.

This same kind of reasoning underlies the tricks for other numbers. I won't go into the math of it, but the rules are:

For 11, alternately add and subtract digits. (This is because 10 is one less than 11.) You can begin from either end, it doesn't matter for the test. 2909909 gives 2 - 9 + 0 - 9 + 9 - 9 + 9 = 2, which isn't a multiple of 11, so 2909909 isn't, either. 2909907, however, gives us 2 - 9 + 0 - 9 + 9 - 9 + 7 = 0, so 2909907 is a multiple of 11.

For 7 the trick is to subtract twice the last digit from the remainder of the number. (2 x 10 is one less than a multiple of 7.) For 2909909 we have 290990 - 18 = 290972, 29097 - 4 = 29093, 2909 - 6 = 2903, 290 - 6 = 284, 28 - 8 = 20, 2 - 0 = 2. So 2909909 isn't a multiple of 7. That 2909907 is is a multiple of 7 we leave as an exercise for the reader.

For 13, we add four times the last digit to the remainder of the number (4 x 10 is one more than a multiple of 13). 2909909 gives us successively 290990 + 36 = 291026, 29102 + 24 = 29126, 2912 + 24 = 2936, 293 + 24 = 317, 31 + 28 = 59, 5 + 36 = 41, 4 + 4 = 8. 2909907, on the other hand, gives us 290990 + 28 = 291018, 29101 + 32 = 29133, 2913 + 12 = 2925, 292 + 20 = 312, 21 + 8 = 39, 3 + 36 = 39 (and we're stuck, but that's OK, as 39 = 13 x 3).

For 17, we subtract five times the last digit from the remainder of the number (5 x 10 is one less than a multiple of 17). 2909909 gives us 290990 - 45 = 290945, 29094 - 25 = 29069, 2906 - 45 = 2861, 286 - 5 = 281, 28 - 5 = 23, which is (by inspection) not a multiple of 17. On the other hand, if we start with 2909907, we get 290990 - 35 = 290955, 29095 - 25 = 29070, 2907 - 0 = 2907, 290 - 35 = 255, 25 - 25 = 0, so 2909907 is a multiple of 17.

For 19, we add twice the last digit to the remainder of the number (2 x 10 is one more than a multiple of 19). 2909909 gives us 290990 + 18 = 291008, 29100 + 16 = 29116, 2911 + 12 = 2923, 292 + 6 = 298, 29 + 16 = 45, 4 + 10 = 14, which is by inspection not a multiple of 19. 2909907 yields 290990 + 14 = 291004, 29100 + 8 = 29108, 2910 + 16 = 2926, 292 + 12 = 304, 30 + 8 = 38, and 3 + 16 = 19.

And so on.

A similar trick for divisibility, and one usually easier to put into practice, takes advantage of the facts that `c` is a multiple of `a` if and only if `c+ka` is for any integer `k`, and that if `a` and `b` are coprime (that is, they have no factors in common), then `bc` is a multiple of `a` if and only if `c` is.

Say we want to know if 2909909 is a multiple of 29. It is if and only if 2909909 - 29 = 2909880 is. Since 29 and 10 are coprime and 2909880 is a multiple of 10, 2909880 is a multiple of 29 if and only if 290988 is, getting rid of a digit. We can then proceed.

(And in this case, of course, we could also subtract 29 x 100000 = 2900000 to get 9909 with which to start our tests.)

The point is that if all you want to do is to check divisibility, that's usually relatively easy to do, even for primes above ten.

(And by the way, while 2909907 = 9 x 7 x 11 x 13 x 17 x 19, 2909909 is prime. I just thought I'd mention that.)

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