Ex. 4. Transform 456·16 from the duodenary to the ternary scale. 375. The product of any two consecutive numbers is divisible by 1×2. Of the two numbers one must evidently be even, that is, divisible by 2, therefore their product is divisible by 2. 376. The product of any three consecutive numbers is divisible by 1×2×3, or 6. Every number must be either of the form 3m, or 3m±1, or 3m±2, since it must be either divisible by 3 without remainder, or have a remainder 1 or 2; therefore the product of any three consecutive numbers may be represented by one of the forms 3m(3m+1)(3m+2), (3m-1)3m(3m+1), (3m-2)(3m-1)3m. Now, since by last Art. both (3m+1)(3m+2) and (3m-2)(3m-1) are divisible by 1×2, and 3m is a multiple of 3, therefore the first and third forms are divisible by 1×2×3. Also, if m be an even number, that is, divisible by 2, it is clear that 3m, and therefore the second form is divisible by 1×2×3; and, if m be an odd number, 3m is odd, and 3m+1 even; consequently the second form is divisible by 1×2×3. Hence, in all cases, the product of three consecutive numbers is divisible by 1×2×3. 377. The continued product of any r consecutive numbers is divisible by 1.2.3...r. [This has been already shewn indirectly in Art. 316; but the following is a more independent proof.] Let n be the least of the numbers, and let be represented by P, for all values of n and r. n n(n+1)(n+2).......(n+r−1) 1.2 3 Now assume that the product of any r-1 consecutive integers is divisible by 1.2.3...(r-1), that is, suppose „P is an integer, then „P,P,+ an Integer, for all values of n and r, write n-1 for î‚ ‚‚P‚=„-P,+ an Integer, ...... n-2 T . adding and cancelling, =1+ an Integer = Integer, P-the sum of Integers an Integer; which proves that, if P., be an Integer, then also is P. But we know that P, is an integer, (by the last Art.), therefore also is P.; and if „P, therefore also P,; and so on generally for P.; that is n(n+1)(n+2)...... (n+r-1) is divisible by 1.2.3...r. R Ex. 1. If n be any whole number, then will n(n2—1)(n2—4) be divisible by 120. n(n2—1)(n2—4)=n(n−1)(n+1)(n−2)(n+2) =(n−2)(n−1)n(n+1)(n+2), which is the product of 5 consecutive numbers, and is therefore divisible by 1.2.3.4.5, or 120. Ex. 2. If n be any even number, n3+20n is divisible by 48. then n3+20n=8m3+40m, =8m(m2+5), =8m(m2-1)+48m, 8(m-1)m(m+1)+48m. Now (m-1)m(m+1), being the product of three consecutive numbers, is divisible by 1.2.3 or 6; therefore n3+20n is divisible by 48. 378. Every number which is a perfect square is of one of the forms 5m or 5m±1. For every number is of one of the forms 5m, 5m+1, 5m+2, 5m+3, 5m+4; all of which are included in the forms 5m, 5m±1, 5m±2, since 5m+3=5(m+1)-2=5m'-2, and 5m+4=5(m+1)-1=5m'-1. But (5m)2=5(5m2)=5m', which is of the form 5m; (5m+1)=25m2+10m+1=5(5m2±2m)+1, which is of the form 5m+1; (5m+2)2=25m2±20m+4=5(5m2±4m+1)-1, which is of the form 5m-1; .. every square is of one of the forms 5m, 5m+1, 5m-1. 379. Every "prime number" greater than 2 is of one of the forms 4m+1. For every number is of one of the forms 4m, 4m+1, 4m+2, 4m+3; but neither 4m, nor 4m+2, can represent prime numbers, since each is divisible by 2; therefore all prime numbers greater than 2 are represented by 4m+1 and 4m+3. But 4m+3=4(m+1)−1=4m'-1; therefore the two forms for prime numbers are 4m±1. COR. Since m may be odd or even, that is, of the form 2n+1, or 2n, all prime numbers are represented by 8n±1, or 8n±3. 380. Every prime number greater than 3 is of one of the forms 6m+1. For every number is of one of the forms 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+5, of which the 1st, 3d, 4th, and 5th obviously cannot represent prime numbers; and therefore, all prime numbers greater than 3 are represented by 6m+1 and 6m+5. But 6m+5=6(m+1)-1=6m'-1. Therefore, 6m 1 will include all prime numbers greater than 3. COR. Since m may be odd or even, that is, of the form 2n+1, or 2n, all prime numbers greater than 3, will be included in 12n±1, or 12n+5. [Exercises Zp.] 381. No Algebraical formula can represent prime numbers only. Let p+qx+r+&c. be a general algebraical formula; and let it be a prime number when x=m; therefore (P) the prime number in that case is p+qm+rm2+&c. Now let x=m+nP; then p+q(m+nP)+r(m+nP)2+&c. is the number; and is equal to p+qm+rm2+ &c.+MP, (M signifying "some multiple of,”)=P+PM, which is divisible by P, and therefore not a prime; consequently the formula does not represent prime numbers only. 382. The number of primes is indefinitely great. For, if not, let there be a fixed number of them, and let P greatest: then 1.2.3.5.7.11......p is divisible by each of them, .... ... ..not one of them. be the and 1.2.3.5.7.11......p +1 If this latter number, then, be divisible by a prime number, it must be one greater than p; if not, it is itself a prime, (since every number is either a prime, or capable of being resolved into factors which are prime) and is greater than p. Therefore, in either case, there is a prime greater than p; that is, we may not assume any prime to be the greatest; or, the number of primes is indefinitely great. 383. To determine whether a proposed number be a prime or not. It is obvious that this may be done by dividing the proposed number by every number less than itself, beginning with 2, until we have either proved it to be divisible by some one of them without remainder, or that it is a prime, from not being divisible by any one of them. But there is no necessity to proceed so far, as may thus be shewn. If the proposed number (p) be not a prime, then p=ab, the product of two other numbers. If then a>√p, b<√p; and if a<√p, b>√p. Hence in both cases p is divisible by a number less than the square root of itself. Or it may be that the proposed number is an exact square, in which case it is divisible by its square root. If, therefore, a proposed number be not divisible by some number not greater than the square root of itself, it must be a prime. 384. To find the number of divisors of a given number. Let a, b, c, &c. represent the prime factors of which the given number is composed; and let a be repeated p times; b, q times; c, r times; &c.; so that the number =abc", &c.; then it is evident that it is divisible by each of the quantities a", p+1 in number. ... ... ...... ... ... &c. &c. and also by the product of any two or more of them, that is, by every term of the continued product of (1+a+a2+...+a3)(1+b+b2+...+b2)(1+c+c2+...+c3).&c.; for (1+a+a+...+a3)(1+b+b2+...+b2)=1+a+a2+......+a? which are all the different divisors (including 1) of ab2. pro Similarly, if this result be multiplied by (1+c+c+...+c), the duct will consist of all the different divisors of abc'; and so on, if there be more factors of the given number. Now the number of these divisors is obviously p+1 in a; in aol the number is p+1 taken as many times as there are terms in the second series, that is, (p+1)(q+1); in a2bc" it is (p+1)(q+1) taken r+1 times, or (p+1)(q+1)(r+1); and generally the number of divisors in a'b'c'. &c.=(p+1)(q+1)(r+1).&c. including 1 and the number itself. Ex. Find the number of divisors of 2160. Here 2160=2×1080=22 × 540=23× 270=2a × 135=2*× 3 × 45=2'x3'×15 =21×33× 5. Therefore the number of divisors =(4+1)(3+1)(1+1) or 40. COR. 1. The number of divisors will always be even unless each of the quantities p, q, r, &c. be even, that is, unless the proposed number be a perfect square. COR. 2. It is also obvious from what has been said above that the sum of all the different divisors of a b'c. &c. is the sum of all the terms in the continued product of (1+a+a2+...+a2)(1+b+b2+...+b3)(1+c+c3 +...+c) &c. 385. If m be any prime number, and N a number not divisible by m, then N-1-1 is divisible by m. (FERMAT'S THEOREM). For, it is easily seen by the Binomial Theorem, that in the expansion of (a+b+c+&c.) m is a factor of every term except aTM, bTM, cTM, &c.; and that the coefficients are always whole numbers. But m, being a prime number, will not be divisible by any factor in the denominator of a coefficient, and will therefore remain as a factor of each term, when the coefficients are reduced. Therefore we may assume, (m being a prime number), have (a+b+c+&c.)"=a+b+cTM+&c.+mP. Let, then, a=b=c=&c.=1, and the number of them be N; and we N"=N+mP, .. N"-N=mP, or N(N-1-1)=mP. But, by supposition, N is not divisible by m, .. NTM-1 is a multiple of m, or is divisible by m. |