sity divide b, and the proof is the same as in the prop osition. (11.) Also, if P is the product of any number of whole numbers which is not divisible by p, and c another whole number which is not divisible by p, then, by the proposition, Pe is not divisible by p. Hence, if the integers a and b are not divisible by p, since their product ab is not divisible by p, we may put ab for P, and we shall have abc not divisible by p. In the same way, if the integer d is not divisible by p, the product abcd is not divisible by p, and so on for any number of integers. (12.) Reversely, if p divides the product abcd, etc., of any number of integers, it must divide one or more of the integers; for if it does not, it can not divide their product, which is against the hypothesis. (13.) If each of the numbers a, b, c, d, etc., is a prime number, and A another prime number different from unity, which does not enter into the product abcd, etc., then A can not divide the product; for it can not divide either of the prime numbers a, b, c, d, etc., and of course can not divide their product. 17th (14.) Still supposing a, b, c, d, etc., to be prime numbers, and n, n', n'', n'"', etc., any positive integers, one or more of which may equal unity, should the case require it; then a", br', cn'', d'''', ', etc., which is the product of the nth power of a by the n'th power of b, by the n' power of C, and so on, can not be divided by the prime number A. For since a" the product aaaa, etc., to n factors, and that A does not divide either of these factors, it does not divide their product a"; and in the same way A does not divide b", cn'', d'''', etc., consequently it does not divide their product. = (15.) Again, the product a"b"'", etc., is of course divisible by a", but it is not divisible by a" if m is greater than n; for let m = n + 1, then am = an+1 = ana; and to divide by ara is the same as to divide by a", and then to divide the quotient by a; dividing the given product by a" reduces it to be", etc., which can not be divided by a, since a does not enter it as a factor, and because b, c, etc., being prime numbers, be", etc., can not be divided by the prime number a, which does not enter their product; consequently a"b"'"", etc., can not be divided by a" if m is greater than n, but if m equals n, or is less than n, then the product a"b"b"", etc., is clearly divisible by am. And similar remarks are applicable to the division of anbon", etc., by any integral power of any one of the prime numbers b, c, d, etc. = = (16.) If we put anb"cn", etc., P1, and suppose that the product of the integers A, B, C, D, etc., is equal to P1, we shall get ABCD, etc., P1. Then it is evident, from what has been shown, that A, B, C, D, etc., can not contain any prime numbers that are different from the prime numbers a, b, c, d, etc.; also the sum of the exponents of a in A, B, C, D, etc., must equal n, the sum of the exponents of b must equal n', and so on for the exponents of c, d, etc. (17.) It follows, from what has been done, that any integral number, P1, can be put under the form a"b", etc. For if we divide P1 by a, and the quotient by a, and so on, as often as it can be done, then the number of divisions will give the value of n, the exponent of a; and in the same way by dividing by b we get the exponent of b, and so on (for the exponents of c, d, etc.; the simplest way will be to divide the last quotient of the division by a by b, and the quotient by b, and so on), as often as it can be done; and then to divide the last quotient of these divisions by c, in the same way as before, and so on for all the prime divisors. Thus, to put 324 under the above form, we divide by 2 and get 162 for the quotient, which, divided by 2 again, gives 81 for the quotient, then we have 324 = 22 x 81. = And dividing 81 by the prime 3, we get 27 for the quotient, which divided by 3 gives 9 for the quotient, and 9 divided by 3 gives 3 for the quotient, and 3 divided by 3 gives 1 for the quotient; and since there are four divisions, we have 81 3', consequently we have 324 = 2a × 3*; and a is represented by 2, and its exponent n is also represented by 2; also b is expressed by 3, and its exponent is expressed by 4, and the other primes, c, d, etc., are each equal to unity, so that we have 324 = 22 × 31 expressed in the form a"b", etc. In the same way, we can express 1890 by the form 1890 = 2 × 33 × 5 × 7; and 2592 is expressed by 2592 = 25 × 31 ; we also have 234 = 2 × 32 × 13. (18.) Now, since 2 and 32 = 9 are factors of all the numbers, it follows that 2 and 32 9 are common measures or divisors of the numbers; and 2 × 32 18 is the greatest common measure or divisor of the same numbers. (19.) Also, 25 × 3 × 5 × 7 × 13 = 1179360 is the least common multiple of the same numbers; which is found by taking 2 with the greatest exponent that it has in any one of the numbers, and 3 with the greatest exponent that it has in any one of the numbers, and so on; then the product of all the powers of 2, 3, etc., thus found, is the least common multiple of all the numbers, as is evident from the nature of the least common multiple, as has heretofore been defined. (20.) In like manner, if we resolve any whole numbers into products of powers of prime numbers, and select the greatest power of each prime that is common to all the products, then the product of the powers of the primes thus obtained will be the greatest common measure or divisor of the numbers. (21.) If 1 is the only prime that is common to the products, the numbers will of course have no other common divisor than 1; and if no two products have any other divisor than 1, the numbers are said to be prime to each other. (22.) Again, supposing the numbers resolved into products of powers of primes as before, if we select the greatest power of each prime in the products, then the product of all the powers thus obtained will be the least common multiple of the numbers; observing that the same power of any prime is to be taken but once in the product. It hence follows, if the numbers are prime to each other, that their product will be their least common multiple. (23.) It is clear that we may proceed in a similar way to find the greatest algebraic divisor and least common multiple of algebraic expressions of integral forms, by resolving them into products of powers of algebraic primes, and then proceeding as before. EXAMPLES. Ex. 1.-To find the greatest common divisor and least common multiple of 27 and 72. = Here we have 27 33, and 72 = 3a × 23, and 32 is the greatest power and the only power of any prime (except 1) which is common to the resolved numbers; consequently 329 is the greatest common divisor of the numbers, as required. Also 33 is the greatest power of the prime 3 that enters into the resolved numbers; and 23 is the only power of the prime 2 that enters into the resolved numbers; consequently 23 × 33 216 the least common multiple of the numbers, as required. Ex. 2.-To find the greatest common divisor and least common multiple of 91, 819, 2145, and 2431. Here we have 91 13 × 7, 819 = 32 × 7 × 13, 2145 = 3 × 5 × 11 × 13, 2431 = 11 × 13 × 17. The only power of any prime that is common to all the resolved numbers is 131 13; consequently 13 is the greatest common divisor of the numbers, as required. = Also, since 32, 51, 7',11', 131, 17' are the greatest powers of the primes that enter into the resolved numbers, we have 32 x 5 x 7 x 11 x 13 x 17 = 765765, which equals the least common multiple of the numbers, as required. Ex. 3. To find the greatest common divisor and least common multiple of 7, 19, 36, 37. Since the numbers have no common divisor except 1, it is of course their greatest common divisor. And to find the least common multiple of the numbers, we have 7 × 19 × 36 × 37 = 177156 for the least common multiple, as required. Ex. 4.-To find the greatest common divisor and least common multiple of abc, 3abd, 4a2b3d2e. Since a2 and b are the only factors common to all the quantities, we have ab for their greatest common measure. And selecting the highest powers of the prime factors of the quantities, we have 1 × 3 × 4 × a3 × b2 × c × ď2 × e = 12abcde, which is the least common multiple of the quantities. Ex. 5.-To find the greatest common divisor and least common multiple of 42a2b3c1, 54a3b3c3, and 6a3b3Ã3. Since 426 x 7, 54 = 6 x 9, the greatest common divisor of the coefficients is 6, and abs being the only powers of prime literal factors that are common to all the quantities, we of course have 6a2b for the greatest common measure. And since 21 × 33 × 71 = 378 = the least common multiple of the coefficients of the quantities, and that abcd the least common multiple of the literal parts of the quantities, we evidently have 378abc1d for the required least common multiple of the quantities. Ex. 6. To find the greatest common measure and least common multiple of a2 — b3, a3 + b3, (a + b)3. From what was formerly shown, we know that a + b is the greatest common divisor of all the quantities, for a2-b2 = (a + b) (a - b), and a3 + b3 = (a + b) (a2 — ab + b2), (a + b)3, = (a + b) (a + b). Also, (a + b)2 × (a — b) (a2 — ab + b2) = (a2 — b3) (a3 + b3) is evidently the least common multiple, as required. EXAMPLES FOR EXERCISE. 1. Find the greatest common divisor of 119, 323, and 1088. Ans. 17. 2. Find the least common multiple of 13, 17, and 19. Ans. 4199. 3. Find the least common multiple of 13x'y', 39x'y', and 26x*a*b. Ans. 78aabx1y1z2. 4. Find the greatest common divisor of 114abc2, 437a°b°c3, 133a10b14c15d. Ans. 19abc2. 5. Find the greatest common divisor of 3(a - b2), 6(a3 — b3), 9ab3 — 9b3. Ans. 3(a - b). (24.) Since it is necessary to have prime numbers, in order to find the greatest common measure and least common multiple of integers by the methods that have been given, we shall here insert the common way of determining them. Write down the odd numbers as follows, viz., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99. 101; then |