Elements of Algebra

18 ELEMENTS SECT. I. 64. Farther, as the general expression $abcd$ is not only divisible by $a$, and $b$, and $c$, and $d$, but also by $ab, ac, ad, bc, bd, cd$, and by $abc, abd, acd, bcd$, and lastly by $abcd$, that is to say, its own value; it follows that $60$, or $2 \times 2 \times 3 \times 5$, may be divided not only by these simple numbers, but also by those which are composed of any two of them; that is to say, by $4, 6, 10, 15$: and also by those which are composed of any three of its simple factors; that is to say, by $12, 20, 30$, and lastly also, by $60$ itself. 65. When, therefore, we have represented any number assumed at pleasure, by its simple factors, it will be very easy to exhibit all the numbers by which it is divisible. For we have only, first, to take the simple factors one by one, and then to multiply them together two by two, 4. A number is divisible by $11$, when the sum of the first, third, fifth, &c. digits is equal to the sum of the second, fourth, sixth, &c. digits. It would be easy to explain the reason of these rules, and to extend them to the products of the divisors which we have just now considered. Rules might be devised likewise for some other numbers, but the application of them would in general be longer than an actual trial of the division. For example, I say that the number $53704689213$ is divisible by $7$, because I find that the sum of the digits of the number $64004245433$ is divisible by $7$; and this second number is formed, according to a very simple rule, from the remainders found after dividing the component parts of the former number by $7$. Thus, $53704689213 = 50000000000 + 3000000000 + 700000000 + 0 + 4000000 + 600000 + 80000 + 9000 + 200 + 10 + 3$; which being, each of them, divided by $7$, will leave the remainders $6, 4, 0, 0, 4, 2, 4, 5, 4, 3, 3$, the number here given. Bernoulli. If $a, b, c, d, e$, &c. be the digits composing any number, the number itself may be expressed universally thus; $a + 10b + 10^2c + 10^3d + 10^4e$, &c. to $10^n z$; where it is easy to perceive that, if each of the terms $a, 10b, 10^2c$, &c. be divisible by $n$, the number itself $a + 10b + 10^2c$, &c. will also be divisible by $n$. And, if $\frac{a}{n}, \frac{10b}{n}, \frac{10^2c}{n}$, &c. leave the remainders $p, q, r$, &c. it is obvious, that $a + 10b + 10^2c$, &c. will be divisible by $n$, when $p+q+r$, is divisible by $n$; which renders the principle of the rule sufficiently clear. The reader is indebted to that excellent mathematician, the late Professor Bonnycastle, for this satisfactory illustration of M. Bernoulli's note.