Whence, if a number is formed by the continual multiplication of many factors, it cannot be equal to the product of other numbers containing prime factors different from those which enter into the numbers already multiplied together. 116. Every number which can be divided by two or more numbers prime to each other is also divisible by the product of these prime numbers. Let N be divisible by d, d', d'. . . . which are prime to each other; N is divisible by the product dd'd''. Since d divides N, N=dq (q being a whole number); but ď divides N ; therefore divides dq, and d', d are prime to each other; therefore ď divides q (Art. 111). Again, since d' divides q, let q=d'q (q' being a whole number); therefore N, which is equal to dq, is also equal to dď'q=dd' xq; whence N is divisible by dď. Similarly, since d' divides N, it therefore divides dd'q'; but d" is prime with d, d'; and therefore with dxd; whence d'' divides q'. Let q=dq' (q" being a whole number); therefore N=dd'd''q"= dd'd'xq'; whence N is divisible by dďď. The proof is the same when the divisors are more in number than 3. 117. Let d, d', d'"'. . . . be numbers which are prime to each other. Let dxdxd dxdxdxď · or d" or d'n' or d'" be factors which are multiplied together; and let their product be N. The number N is divisible by d" × d'"' ×d''''' ; it is also divisible by all the products which can be formed by multiplying together two by two, three by three, &c., the different powers of d, d', d". between the first power and that marked by n for d, by n' for d', by n'" for d", &c. ̧ d" Since d, ď, d3, d'3 d's are prime to each other. d", ď~, ď''nı are prime to each other (Art. 113); consequently N, which is the product of d", d'"', d''""'. must be divisible by these numbers, or by the products of any of them taken two by two, three by three, &c., in manner aforesaid; for these products are evidently submultiples of the entire product dxď'"' ×ď'''' .., which is equal to N. 118. The possibility of exactly dividing one number by another can always be determined by trial. It may, however, in the case of some numbers, be ascertained more easily by certain tests than by actual division. To investigate these tests this principle (established Art. 105) is required, namely, Every number which is decomposed into two parts, each divisible by a second number, is itself divisible by the second number. To which must be subjoined the following: That if N=P+P', of which d divides P, but not P', then d does not divide N; and the remainder from the division of N by d is equal to that from the division of P' by d. NP P Since N=P+P; d=d+d (see Art. 111). Next, to prove that the remainder from the division P being divisible by d, P=dq (q being a whole number). P' being not divisible by d, P=dq' +r (q′ being a whole number). Whence N=d(q+q)+r. But P+P N. That is, N divided by d gives for quotient q+q', and for remainder r; which r is the remainder from the division of P' by d. Tests of the divisibility of numbers by certain divisors. 119. Every number whose last figure is 0, 2, 4, 6, or 8, is divisible by 2. Separating the number into two parts, namely, the figure expressing simple units, and the number to the left of that figure, the second part (regard being had to its relative value) is some multiple of 10, and therefore divisible by 2; and since each of the figures 0, 2, 4, 6, 8, is also divisible by 2, it follows (Art. 105) that the whole number can be divided by 2. A number ending with one of the figures 1, 3, 5, 7, 9, is not divisible by 2. For the number to the left of the last figure, considered with respect to its relative value, is divisible by 2; but the other part, 1, 3, 5, &c. is not (Art. 118). Numbers divisible by 2 are named even; and numbers which cannot be divided by 2, odd. 120. A number can be divided by 3 when the sum of the digits of the number is divisible by 3. 1 in the place of tens = 9+1 1 in the place of hundreds = 99+1 And, generally, 1 in any place (except the place of simple units) is equal to 1+ a number expressed by 9 repeated as many times as the 1 has zeros after it. Whence any other figure, m, =mx the proper number of nines +m×1, or the figure m itself regarded as simple units. Now, any number 9, 99, 999 . . . is a multiple of 3. Therefore m×999 . . . . is a multiple of 3 (Art. 104). This reduction into a number expressed by a multiple of 9, 99, 999, +the significant figure of the order, considered as simple units, may be made for every figure which is contained in a number. Whence any number can be decomposed into two parts; the one consisting of multiples of 9, 99, 999, ...; the other of the sum of the significant figures of the number, regarded as simple units. The multiples of 9, 99, 999 ... are divisible by 3. Wherefore if the sum of the digits can be divided by 3, the number is itself divisible by 3; but if not, not (Art. 118). 121. Every number is divisible by 4 when the number composed of its two last figures can be divided by 4. For the number may be separated into two parts; one, in which the places of tens and simple units are occupied by zeros; the other, the number composed of the tens and simple units. The first part is a multiple of 100, and it is divisible by 4, since 100 is so (Art. 104). Whence, if the second part, composed of the two last figures taken with their relative values can also be divided by 4, the whole number must be divisible by 4 (Art. 105). 122. Every number of which the last figure is or 5, is divisible by 5. Let the number be decomposed into parts, the first a multiple of 10, the second a figure expressing simple units. The part which is a multiple of 10 is divisible by 5; and the other part being 0 or 5, the division of this other part by 5 leaves no remainder. Whence (Art. 105) every number ending in 0 or 5 is divisible by 5. If the last figure of a number is not 0 or 5 the number cannot be measured by 5 (Art. 118). 123. Every even number which can be divided by 3 is divisible by 6. For every even number is divisible by 2 (Art. 119). The even number which can be divided by 3 is therefore divisible also by 2; and since the divisors 2, 3, are prime numbers, the number which can be divided by 2 and by 3 is divisible by 2×3, or 6 (Art. 117). 124. To find the conditions of divisibility of any number by 7. Any number, for example 13875964, may be decomposed into the several orders of units of which it consists. And the several orders of units may be again decomposed into the product, for each, of the significant figure of the order by unity followed by the number of zeros required to give that significant figure its relative value. Thus, 4=4×1 60=6 × 10 900 9 x 100 5000 5 x 1000 70000 7 × 10000 800000=8 × 100000 3000000=3 × 1000000 10000000=1x10000000 The 1, 10, 100, 1000, &c. represent 1 of each order of units contained in the proposed number, and the whole number is composed of multiples of these different orders of units. Wherefore, if the conditions of divisibility of these units can be determined, the same conditions will be applicable to the whole number (Art. 118). Now, a unit of each order may be represented by some multiple of 7, augmented or diminished by the quantity required to reproduce this unit by its combination with the multiple of 7. Whence again, if the conditions of divisibility of the quantities by which the multiples of 7 are to be augmented or diminished can be determined, those of the whole number will be found. Now A B 1000000 142857 x7+1.3000000=3× 142857×7+3×1 10000000=1428571×7+3.10000000=1 x 1428571×7+1x3 In the first part of the preceding table 1 of each order of units contained in the given number is divided by 7, and in the second part the results are multiplied by the significant figures of the respective orders of units. 0, 1, 14, the quotients of 1, 10, 100 by 7 are taken, so that the products 0x7, 1x7, 14x7 are less than 1, 10, 100; whence, in reconstructing the original number by the combination of all its parts, the remainders 1, 3, 2, or rather their products by the significant figures 4, 6, 9, must be added to the sum of the other parts. But the quotients 143, 1429, 14286 resulting from the division of 1000, 10000, 100000 by 7 are taken in such manner that the products 143×7, 1429×7, 14286×7 are greater than 1000, 10000, 100000, respectively. Whence, in recomposing the original number, the remainders −1, −3, -2, or their products by the corresponding significant figures, must be subtracted from the sum of the other parts. Proceeding in this manner, it is found that a ternary arrangement of the remainders can be made, and that before multiplication by the significant figures of the respective orders of units the remainders are 1, 3, 2 to be added; 1, 3, 2 to be subtracted; 1, 3, 2 to be added; 1, 3, 2, to be subtracted, and so on alternately. Now, any number decomposed as the preceding, may be considered as composed of two parts represented in the last table by columns A, B. Column A, being composed of multiples of 7, can be divided by 7. In column B let the numbers which are to be added to the other parts be formed into one sum, and those which are to be subtracted from the other parts into another sum. Then, if the difference of these two sums is either equal to zero, or is a multiple of 7, the whole number is divisible by 7 (Art. 105). Whence, in conclusion, if the numbers 1, 3, 2 are multiplied respectively by the 1st, 2d, 3d; 7th, 8th, 9th; 13th, 14th, 15th. ... digits of a given number, and the products are added together; also, if the same numbers 1, 3, 2 are multiplied by the 4th, 5th, 6th; 10th, 11th, 12th; 16th, 17th, 18th.... digits of the same number, and these products are added together; then, if the difference of these results is either 0, or a multiple of 7, the number proposed is divisible by 7. It seems scarcely necessary to add, that this investigation is more curious than useful, since in practice it is much easier to divide any number by 7 than to apply the test. 125. If the three last figures of a number, considered with respect to their relative value, can be divided by 8, the number is itself divisible by 8. For the number may be broken into two parts; the one, that in which the three last figures are replaced by zeros; the other, the number expressed by the three last figures. The first part is a multiple of 1000; and since 1000 (=125×8) is divisible by 8, the multiple of 1000 is divisible by 8 (Art. 104). Whence, if the second part, the number expressed by the three last figures, can be divided by 8, the whole number must be divisible by 8 (Art. 105). Since 1000 is a multiple of 25, if the number expressed by the three last figures can be divided by 25, the whole number is divisible by 25. 126. Every number, the sum of whose digits can be divided by 9, is itself divisible by 9. For every number can be decomposed into two parts, the one composed of multiples of 9, 99, 999, . and the other, of the sum of the digits of the number. The multiples of 9, 99, 999, . . are divisible by 9. Whence, if the sum also of the digits can be divided by 9, the whole number is divisible by 9 (Art. 105). 127. Since any number can be divided into two parts of which the one is divisible by 9, if the other part, which is the sum of the digits of the number, cannot be divided by 9, the whole number is not divisible by 9, and the remainder from the division of the sum of the digits by 9 is equal to the remainder from the division of the whole number by 9 (Art. 118). 128. Every number which ends in 0 is divisible by 10. evident. The tests of the divisibility of numbers by the prime numbers 11, 13, 17, &c., might be in like manner investigated; but the investigations depend on principles which are not elementary, and (as in the case of the prime This is self number 7) the application of the test is more tedious and difficult than the actual trial by division. 129. The following are more simple : a. A number is divisible by 18, when that number is even, and the sum of its digits can be divided by 9; for under these conditions the number can be divided by 2 and by 9, and consequently by 2×9, or 18 (Art. 117). b. A number is divisible by 12 or by 36 when the two last figures, taken with respect to their relative value, can be divided by 4, and the sum of the digits of the number by 3 or by 9. For in the one case the number is divisible by 4 and by 3, and therefore by 4× 3, or 12; in the other it is divisible by 4 and by 9, and therefore by 4× 9, or 36. c. A number is divisible by 15 or by 45 when the last figure is 0 or 5, and the sum of the digits of the number is divisible by 3 or by 9. For, in the first instance, the number can be divided by 5 and by 3, and consequently by their product, 15; in the second it is divisible by 5 and by 9, and therefore by their product, 45. 130. The prime numbers between 1 and 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. 131. From the properties of the number 9, established in Articles 126 and 127, are derived easy verifications of multiplication and division. Let N, N' be two numbers whose product is to be found. Let N be divided by 9, and let the quotient be q, and the remainder r. N' 9 Whence N=9x+multiplying these quantities (Art. 76). N N'=9x9xqxq+9xqxr+9xqxr′+rxr'. The product NN' is composed of four partial products. Three of these products are multiples of 9, therefore their sum is a multiple of 9 (Art. 105). Whence NN' is composed of two parts, of which the one is divisible by 9. If the other part, rxr, can be divided by 9, then NN' is also divisible by 9 (Art. 126); but if rx is not divisible by 9, neither is NN'; and the remainder from the division of rxr by 9 is equal to the remainder from the division of NN' by 9. Also, the remainders from the division of NN' and rby 9 are equal respectively to the remainders left from the division of the sum of the digits of NN', rr' by 9 (Art. 127). Whence N, N', being numbers which are to be multiplied together, if the sum of the digits of N, N' are respectively divided by 9, and if the remainders r, are multiplied together, and if the sum of the digits of the product rxr is divided by 9, the remainder from this division of the sum of the digits of rxr' by 9 is equal to the remainder obtained from the division of the sum of the digits of the product NN' by 9. When r, or r', or rxr', is exactly divisible by 9, then the product NN' is also exactly divisible by 9. These principles afford a simple means of verifying the results of multiplication and division. The manner of conducting the verification will be most easily understood from an instance of its application: Let the factors and product of an example in multiplication be 378482,. 3548, and 1342854136.* The sum of the digits is formed and the division by 9 effected simultaneously thus, beginning at the left of the multiplicand, 378482; Division of the sum of the digits of the multiplicand by 9, 3+7=10, 10+9=1+1 rem. F |